Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap6 Trusses and frames 8 Tension, shear and bending moment

Chapter 7 Transmissions and mechanisms

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7.0Introductionstudy PDF
7.1Springsstudy PDF + solutions
7.2Force amplificationstudy PDF + solutions
7.3Mechanismsstudy PDF + solutions

All chapters and individual parts

Sometimes solid parts are assembled to cause force or torque in one place when a different force or torque is applied at another place. Such assemblies include levers, gear boxes, presses, pliers, clippers, chain drives, and crank-drives. Besides solid parts connected by pins, a few special-purpose parts are commonly used, including springs, strings and gears. Tricks for amplifying force are usually based on principles idealized by pulleys, levers, wedges and toggles. Force-analysis of transmissions and mechanisms is done by drawing free-body diagrams of the parts, writing equilibrium equations for these, and solving the equations for desired unknowns.

Here we consider collections of parts assembled to transmit motion or force. We are not going to address the conversion of thermal, chemical (or biological) or electrical source to a useful force. Rather we discuss the transmission of that force. We are concerned with the passive parts of machines, or with passive machines that have no energy source within them. Most often there is an input force or torque and a desired output which does the machine’s job.

The categorization of an assembly of parts as a structure or as a machine is mostly a matter of intent. Is the main job to hold or support something still (a structure) or to move something. Besides its intended use, there is no way to look at an assembly of parts and assuredly characterize it as a structure or a machine^†^†margin: A candidate delineation between machine and structure might be whether the assembly allows any motion (mechanism) or not (structure). Even that concept is ambiguous. Common machines, like presses, wrenches and clamps, are as motion-restricted in their end-use as many structures. And, conversely, many structures are adjustable and thus are designed to move. . So the statics analysis of mechanisms and transmissions is the same as for frames. Our main concern, as in the rest of statics, is:

Given some information about the forces on or in a mechanism, find out more about the forces.

The practice of mechanism-design is often dominated by kinematic analysis, the study of the geometry of the interacting motions of the parts as the mechanism configuration changes. Such is not our concern here. Rather we focus on the relations between the various forces in a given configuration of the mechanism^†^†margin: The dynamics portion of this book is largely an introduction to the kinematics of mechanisms. .

Building blocks

In the same way that machines and buildings are built from bricks, gears, beams, bolts and other standard pieces, elementary mechanics models of the world are made from a few elementary building blocks. Conspicuous so far, roughly categorized, are: • Special objects: – Point masses. – Rigid bodies: * Two-force bodies, * Three-force bodies, * Springs, * Pulleys, and * Wheels. • Special connections: – Hinges, – Welds, – Sliding contacts, and – Rolling contacts.

Products and models Some of these things have dual lives, as products and as models. On the one hand, a mechanical hinge corresponds to a product you can buy in a hardware store called, naturally, a hinge. On the other hand a hinge in mechanics represents a constraint that restricts certain motions and freely allows others. A hinge in a mechanics model may or may not correspond to hardware called a hinge. For example, when considering a box balanced on an edge, we may model the contact as a hinge meaning we would use the same equations for the forces of contact as we would use for a hinge. Although you can buy a pulley, you might model a rope sliding around a post as a rope on a pulley even though there was no literal pulley in sight.

The relation between product and model can even sound contradictory. Although ‘like a rock’ means ‘solid’ in English, one may model a rock as a spring (which is done for foundation engineering, understanding waves in rocks, and calculating the energy of earthquakes). A coil spring may be modeled as a rigid rod for a simple structure-like study of a machine. And a hinge might be modeled as a spring if its deformation is important. That is, the appropriate mechanics model for a thing and its common name don’t always agree.

What’s new in this chapter The new content in this chapter is

•

Detailed discussion of a few components used in mechanisms and transmissions that are not used commonly in simple ‘structures’. These include springs, pulleys, wheels, and gears.
•

Introduction to a variety of design tricks to, say, cause a big force when only a small force is available.

We start the chapter by discussing a few special parts and assemblies of those parts. Then we consider more general assemblies.

7.1 Springs

Box 7.1 ‘Zero-length’ springs

Zero rest-length springs

A special case of linear springs, that has remarkable mechanical consequences, is a zero-rest-length spring (also called a ‘zero-length’ spring for short) with relaxed length $\ell_{0}=0$ . Zero-rest-length-spring ideas are useful for design, but are not essential for basic understanding of statics.

The defining equations for a zero-rest-length spring, in scalar and vector form, are

T=k\ell\hbox to36.135pt{\hfil and\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{B}}=k\cdot{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{AB}}.

The tension versus length curve for a zero-length spring is shown in fig. 7.4b.

At first blush, such a spring seems non-physical; it seems to represent a thing that you can’t build. If you take a coil spring and try to collapse it to zero length, when the two ends would touch, all the metal gets in the way and prevents that. In fact, however, there are many ways to make zero-rest-length springs. For example, the tension versus length curve of a rubber band (or piece of surgical tubing) looks something like that shown in fig. 7.4c. Over some portion of the curve, the zero-length spring approximation is reasonable (a sign of this is that the vibration frequency is almost independent of stretch for some range of stretch). For other physical implementations of zero-length springs, read on.

The mathematics in many mechanics problems is simpler for $\ell_{0}=0$ springs than for $\ell_{0}\neq 0$ springs.

Rubber bands. As shown in fig. 7.4c, straps of rubber behave like zero-length springs over some of their length. If this is the working length of your mechanism, then the zero-length spring approximation may be good.

A stretchy conventional spring. Some springs are stretched much beyond their rest lengths. Thus the approximation that $k(\ell-\ell_{0})=k\ell\left(1-{\ell_{0}\over\ell}\right)\approx k\ell$ may be reasonable.

A pre-stressed coil spring. Some door springs and many springs used in desk lamps are made ‘close-wound’ so that each coil of wire is pressed against the next one. It takes some tension just to start to stretch such a spring. The tension versus length curve for such springs can look very much like a zero-length spring once stretch has started. In fact, the original elegant 1930’s patent, which common present-day parallelogram-mechanism lamps imitate, specifies that the spring should behave as a zero-length spring. Such a pre-stressed, zero-length coil spring was a central part of the design of the long period seismometer featured on a 1959 Scientific American cover.

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A spring, string, and pulley. If a spring is connected to a string that is wrapped around a pulley then the end of the string can feel like a zero force spring if the attachment point is at the pulley when the spring is relaxed.

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A string pulled from the side. If a taut string is pulled from the side, it acts like a zero-length spring in the plane orthogonal to the string.

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A ‘U’ clip. If a springy piece of metal is bent so that its unloaded shape is a pinched ‘U’ then it acts very much like a zero length spring. This is perhaps the best example in that it needs no anchor (unlike the pulley) and can be relaxed to almost zero length (unlike a pre-stressed coil).

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A spring is a deformable solid that regains its original shape after being compressed, extended or otherwise deformed.

The word spring has a dual personality.

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Figure 7.1: Some commercially available springs (Viking Products).

1)

Spring as product. Springs, in various forms, most characteristically as helices made of steel wire, can be purchased from hardware stores and mechanical parts suppliers (fig. 7.1). Springs are used to hold things in place (in a clothes pin), to store energy (in a clock or wind-up toy), to reduce contact forces (bumpers), to isolate something from vibrations (a car’s suspension), and to modulate the feel for human interaction (under keyboard keys). You will find springs in almost any complicated machine. Take apart a disposable camera, a laser printer, a gas lawn mower, a bicycle, a washing machine or a cruise missile, and you will find springs.
2)

Spring as model. On the other hand, springs are used in mechanical ‘models’ of many things that are not, by name, springs (see page 0.3 for discussion of ‘models’). For much of this book we approximate solids as rigid. But sometimes the flexibility or elasticity of an object is an important part of its mechanics. The simplest accounting for this is to think of the object as a spring. So a tire may be modeled as a spring as might be the near-contact-point material of a bouncing ball, a strut in a truss, the snapping-back part of the earth’s crust in an earthquake, your Achilles tendon, or the give of soil under a concrete slab. Biomechanics Professor Tom McMahon idealized the give of the ground under a person as that of a spring. He used the idea to design the record-breaking running track used in the Harvard stadium.

For simplicity we only concern ourselves with tension and compression springs here. These are springs that only have axial loads applied and only at the ends.

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Figure 7.2: Spring connection. The tension in a spring is usually assumed to be proportional to its change in length, with proportionality constant

k

T=k\Delta\ell

If the tension in a spring is a function of its length alone, independent of its rate of lengthening, the spring is said to be ‘elastic.’ Many materials are well-modeled as elastic for small-enough deformation. If the tension in the spring is proportional to its stretch, the spring is said to be ‘linear.’ Most elastic materials are close to linear in their behavior. Thus the word spring is often short for linear elastic spring.

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Figure 7.3: An ideal spring with rest length

\ell_{0}

and stretched length

\ell_{o}+\Delta\ell

. The tension in the spring is

T

and the vector forces at the ends are

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{A}}

and

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}

The stretch of a spring is the amount by which the spring is longer than when it is relaxed. This relaxed length is also called the un-stretched length, the rest length, or the reference length. If the relaxed length (the length at zero tension) is $\ell_{0}$ and the present length is $\ell$ , then the stretch of the spring is

\Delta\ell=\ell-\ell_{0}=\mbox{Increase in length from rest length}

An ideal spring is a massless two-force body characterized by its rest length $\ell_{0}$ (also called the relaxed length, or reference length), its spring constant $k$ , and the defining equation (or constitutive law), Hooke’s Law:

T=k\cdot(\ell-\ell_{0})\hbox to36.135pt{\hfil or\hfil}T=k\cdot\Delta\ell

where $\ell$ is the present length and $\Delta\ell$ is the increase in length or stretch (see fig. 7.3).

The spring constant $k$ is also sometimes called the spring rate, the spring stiffness or the spring proportionality constant.

The ideal spring is called linear because of the formula $k\Delta\ell$ and not, say, $k(\Delta\ell)^{3}$ . The defining spring formula is sometimes, although we don’t recommend this, memorized as ‘ $F=kx$ ’

Note: the formula ‘ $F=kx$ ’ can lead to errors: the direction of the force is not evident, and some people are unclear about the meaning of $x$ in this formula.

The safest way to avoid sign errors when dealing with springs is to

•

Draw a free-body diagram of the spring;
•

Write the increase in length $\Delta\ell$ in terms of geometry variables in your problem (even if you know that this increase is going to be a negative number);
•

Use $T=k\Delta\ell$ to find the tension in the spring (even if you know the tension will turn out negative); and then
•

Use the principle of action and reaction to find the forces on the objects to which the spring is connected.

The main idea is to pick a sign convention (tension and lengthening are positive) and stick with it, accepting the arithmetic of negative numbers if it arises.

A plot of tension versus length for an ideal spring is shown in fig. 7.4a.

A comment on the notation $\Delta\ell$ Often in engineering we write $\Delta\mbox{(}something)$ to mean the change of ‘ $s o m e t h i n g$ .’ Most often one also has in mind a small change. In the context of springs, however, $\Delta\ell$ is allowed to be a rather large change. A useful way to think about springs is that increments of force are proportional to increments of length change, whether the force or length is already large or small:

\Delta T=k\Delta\ell\hbox to36.135pt{\hfil or\hfil}{dT\over d\ell}=k

Compliance. A spring with a large stiffness is called stiff or hard. The reciprocal of stiffness $1\over k$ is called the compliance. A spring with a small stiffness and large compliance is called compliant or soft and has a lot of ‘give’.

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Figure 7.4: a) Tension versus length for an ideal spring, b) for a zero-length spring, and c) for a strip of rubber.

The force vector on one end of a spring. Because the spring force is along the spring, which is in a known direction, we can write a vector formula for the force on the B (say) end of the spring as (see fig. 7.3)

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}=k\cdot% \underbrace{\left(|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB% }}|-\ell_{0}\right)}_{\Delta\ell}\par\underbrace{\left({{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}\over|{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}|}\right)}_{\hat{\boldsymbol{% \lambda}}_{\mathrm{AB}}},

(7.1)

where $\hat{\boldsymbol{\lambda}}_{\mathrm{AB}}$ is a unit vector along the spring. This explicit formula is useful for, say, numerical calculations. This formula becomes especially simple if the rest-length of the spring is zero ( $\ell_{0}=0$ ) so

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}=k{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}.

Absurd as this seems (how could a spring have zero rest length?) the idea is useful both as a model and for engineering design (see box 7.1 on page 7.1).

Assemblies of springs

Here we see how springs are put together with other springs in parallel and in series. For starters we’ll put together just two springs with rest lengths $\ell_{01}$ and $\ell_{02}$ . The extensions and tensions of the two springs are $\Delta\ell_{1},\Delta\ell_{2},T_{1},$ and $T_{2}$ .

The assembly of springs also acts like a single spring. The central issue is finding the properties (the total stiffness and rest length) of the combined spring.

Much of what you need to know about the words ‘in parallel’ and ‘in series’ is this:

In parallel,	forces	and	stiffnesses	add.
In series,	displacements	and	compliances	add.

which we discuss in detail below.

Springs in parallel

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Figure 7.5: a) Schematic of parallel springs, b) genuinely parallel springs, c) a simple implementation that gives a reasonable approximation of parallel springs. The triangle of cables limits the rotation of the end bars if the triangles are long.

Two springs that share the burden of a load and stretch the same amount are said to be in parallel.

Figure 7.5a shows the standard schematic for springs in parallel. This schematic is a non-physical cartoon because the applied tension would likely cause the end-bars to rotate. What is meant by the schematic in fig. 7.5a is the somewhat clumsy constrained mechanism of fig. 7.5b. In engineering practice one rarely builds such a structure.

For the purposes of discussion here, we assume that any of fig. 7.5abc represent a situation where the springs both stretch the same amount.

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Figure 7.6: Free-body diagrams of the components of a parallel spring arrangement.

For each spring we have the defining constitutive relation:

T_{1}=k_{1}\Delta\ell_{1}\hbox to36.135pt{\hfil and\hfil}T_{2}=k_{2}\Delta\ell% _{2}.

(7.2)

Using the free-body diagrams in fig. 7.6, force balance for one of the end supports shows that

T=T_{1}+T_{2}.

(7.3)

This is what is meant by the two springs sharing the load. Springs in parallel stretch the same amount thus we have the kinematic relation:

\Delta\ell_{1}=\Delta\ell_{2}=\Delta\ell.

(7.4)

For simplicity we have assumed that the two springs have the same rest length.

Put the two results above together and we have

\begin{array}[]{cccccc}T&=&\quad T_{1}&+&T_{2}\\ &=&k_{1}\Delta\ell_{1}&+&k_{2}\Delta\ell_{2}\\ &=&k_{1}\Delta\ell&+&k_{2}\Delta\ell\\ &=&\lx@intercol\underbrace{(k_{1}+k_{2})}_{k}\Delta\ell\hfil\lx@intercol.\\ \end{array}

Thus the effective spring constant of the pair of springs in parallel is, naturally enough:

k=k_{1}+k_{2}.

(7.5)

The loads carried by the springs are

T_{1}={k_{1}\over k_{1}+k_{2}}T\hbox to36.135pt{\hfil and\hfil}T_{2}={k_{2}% \over k_{1}+k_{2}}T

which add up to $T$ as they must.

Example: Two springs in parallel.

Take $k_{1}=99\,{\mbox{N}}/\,\mathrm{cm}$ and $k_{2}=1\,{\mbox{N}}/\,\mathrm{cm}$ . The effective spring constant of the parallel combination is:

k=k_{1}+k_{2}=99\,{\mbox{N}}/\,\mathrm{cm}+1\,{\mbox{N}}/\,\mathrm{cm}=100\,{% \mbox{N}}/\,\mathrm{cm}.

Note that $T_{1}/T=.99$ so even though the two springs share the load, the stiffer one carries 99% of it. For practical purposes, or for the design of this system, it would be reasonable to remove the much less stiff spring.

The reasoning above with two springs in parallel is easy enough to reproduce with 3 or more springs. The result is:

k_{\rm tot}=k_{1}+k_{2}+k_{3}+\dots\hbox to36.135pt{\hfil and\hfil}T_{1}=Tk_{1% }/k_{\rm tot},\quad T_{2}=Tk_{2}/k_{\rm tot}\dots

That is,

•

The net spring constant is the sum of the constants of the separate springs; and
•

The load carried by springs is in proportion to their spring constants.

Some comments on parallel springs

Once you understand the basic ideas and calculations for two side-by-side springs connected to common ends, there are a few things to think about for context.

The simplest redundant truss For the purposes of drawing pictures (e.g., fig. 7.5a) parallel springs are drawn side by side. But, in the mechanics analysis we treated them as if they were on top of each other. A pair of parallel springs is like a two-bar truss where the bars are on top of each other and connected at their ends. With 2 bars ( $b=2$ ) and 2 joints ( $j=2$ ), we have $2j<b+3$ , and a redundant truss. This is the simplest redundant truss with one spring (read bar) doing exactly the same job as the other (carries the same loads, resists the same motions). With statics alone, we can not find the tensions in the springs since the statics equation $T_{1}+T_{2}=T$ has non-unique solutions.

Statically indeterminate problems. Calculating the forces in a set of parallel springs is solving (using more than just statics, namely the spring constitutive law) the simplest statically-indeterminate problem.

Parallel springs and the three pillars of mechanics The laws of statics allow multiple solutions to redundant problems. But a bar in a real physical structure has, at one instant of time, some unique bar tension determined by the deformations and material properties. This is the first, and perhaps most conspicuous, occasion in this book that you see a problem where the three pillars of mechanics (see page 0.1) are assembled in such clear harmony, namely, material properties (eq. 7.2), the laws of mechanics (eq. 7.3), and the geometry of motion and deformation (eq. 7.4). In strength of materials calculations, where the distribution of stress is not determinable by statics alone, this threesome (geometry of deformation, material properties and statics) clearly comes together in almost every calculation.

Parallel springs are not necessarily geometrically parallel. In the discussion above ‘in parallel’ corresponded to the springs being geometrically parallel. In common mechanics usage the words ‘in parallel’ are more general and mean that the net load is the sum of the loads carried by the two springs, and the stretches of the two springs are the same (or in a ratio restricted by kinematics). You will see cases where ‘in parallel’ springs are not the least bit parallel (e.g., see fig. 7.7).

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Figure 7.7: Parallel springs are not always geometrically parallel. The deformation of the structure above into a diamond is resisted by the two springs. They share the load and they have stretches that are linked by the kinematics. Thus these two perpendicular springs are ‘in parallel.’ (Detailed analysis of this structure is a little beyond the coverage here.)

Springs in series

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Figure 7.8: Schematic of springs in series.

Two springs that share a displacement and carry the same load are in series.

A schematic of two springs in series is shown in fig. 7.8a where the springs are aligned serially, one after the other. To determine the net stiffness of this simple spring network we again assemble the three pillars of mechanics, using the free-body diagram of fig. 7.8b.

\begin{array}[]{lcccc}\mbox{Constitutive law:}&T_{1}=k_{1}(\ell_{1}-\ell_{1_{0% }}),&T_{2}=k_{2}(\ell_{2}-\ell_{2_{0}}),\\ \mbox{Kinematics:}&\ell_{0}=\ell_{1_{0}}+\ell_{2_{0}},&\ell=\ell_{1}+\ell_{2},% \\ \mbox{Force Balance:}&\hfil\ \ T_{1}=T,\mbox{\quad and}&T_{2}=T.\end{array}

(7.6)

(where,e.g., $\ell_{1_{0}}$ is the rest length of spring 1). We can manipulate these equations much as we did for the similar equations for springs in parallel. The manipulation differs in structure the same way the equations do. For springs in parallel the tensions add and the displacements are equal. For springs in series the displacements add and the tensions are equal:

\begin{array}[]{cccccc}\Delta\ell&=&\ell&-&\ell_{0}\\ &=&(\ell_{1}+\ell_{2})&-&(\ell_{1_{0}}+\ell_{2_{0}})\\ &=&(\ell_{1}-\ell_{1_{0}})&+&(\ell_{2}-\ell_{2_{0}})\\ &=&\Delta\ell_{1}&+&\Delta\ell_{2}\\ &=&T_{1}\over k_{1}&+&T_{2}\over k_{2}\\ &=&T\over k_{1}&+&T\over k_{2}\\ &=&\lx@intercol\underbrace{\left({1\over k_{1}}+{1\over k_{2}}\right)}_{1\over k% }T\hfil\lx@intercol.\\ \end{array}

Thus we get that the net compliance is the sum of the compliances:

{1\over k}={1\over k_{1}}+{1\over k_{2}}\hbox to36.135pt{\hfil or\hfil}k={1% \over{1/k_{1}}+{1/k_{2}}}={k_{1}k_{2}\over k_{1}+k_{2}},

which you should compare with the case of springs in parallel (Eqn. 7.5).

The sharing of the net stretch is in proportion to the compliances:

\Delta\ell_{1}={1/k_{1}\over 1/k_{1}+1/k_{2}}\Delta\ell\hbox to36.135pt{\hfil and% \hfil}\Delta\ell_{2}={1/k_{2}\over 1/k_{1}+1/k_{2}}\Delta\ell

which add up to $\Delta\ell$ as they must.

Example: Two springs in series.

Take $1/k_{1}=99\,\mathrm{cm}/\,{\mbox{N}}$ and $1/k_{2}=1\,\mathrm{cm}/\,{\mbox{N}}$ . The effective compliance of the parallel combination is:

{1\over k}={1\over k_{1}}+{1\over k_{2}}=99\,\mathrm{cm}/\,{\mbox{N}}+1\,% \mathrm{cm}/\,{\mbox{N}}=100\,\mathrm{cm}/\,{\mbox{N}}.

Note that $\Delta\ell_{1}/\Delta\ell=0.99$ so even though the two springs share the displacement, the more compliant one has 99% of it. For design purposes, or for modeling this system, it would be fair to replace the much stiffer spring with a rigid link.

Box 7.2 How stiff a spring is a solid rod

Here we derive the formula for stiffness $k$ of a rod:

k=EA/\ell

in terms of the material stiffness $E$ , length $\ell$ and cross-sectional area $A$ .

This foreshadowing of Strength of Materials concepts is not central to the study of statics.

Let’s take a reference bar with cross sectional area $A_{0}$ and rest length $\ell_{0}$ and pull it with tension $T$ and measure the elongation $\Delta\ell_{0}$ (figure below). The stiffness of this reference rod is $k_{0}=T_{0}/\Delta\ell_{0}$ . Now put two such rods side-by-side and you have parallel springs. You might imagine this sequence: two bars are near each other, then side by side, then touching each other, then glued together, then melted together into one rod with twice the cross section. The same tension in each causes the same elongation, or it takes twice the tension to cause the same elongation when you have twice the cross sectional area. Likewise with three side by side bars and so on, so for bars of equal length

k={A\over A_{0}}k_{0}.

On the other hand, we could put the reference rods end-to-end in series. Then the same tension causes twice the elongation. We could put three or more rods together in series, thus for bars with equal cross sections:

k={\ell_{0}\over\ell}k_{0}.

Putting these together we get:

k=\left({A\over A_{0}}\right)\left({\ell_{0}\over\ell}\right)k_{0}=\left({k_{0% }\ell_{0}\over A_{0}}\right){A\over\ell}.

Now presumably if we took a rod with a given material, length, and cross section the stiffness would be $k$ , no matter what the dimensions of the reference rod. So $\left({k_{0}\ell_{0}\over A_{0}}\right)$ has to be a material constant. It is called $E$ , the modulus of elasticity or Young’s modulus. For all steels $E\approx 30\cdot 10^{6}\,{\rm lbf}/\in^{2}\approx 210\cdot 10^{9}\,{\mbox{N}}/% \,\mathrm{m}^{2}$ (consistent with fig. 7.10c). Aluminum has about a third this stiffness. So, a solid bar is a linear spring, obeying the spring equations:

k={EA\over\ell}\hbox to36.135pt{\hfil or\hfil}\Delta\ell={T\ell\over EA}\hbox to% 36.135pt{\hfil or\hfil}T={\Delta\ell\ EA\over\ell}\ .

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Box 7.3 Stiffer but weaker

This is an aside for those who wonder how one thing can be both stiffer and weaker than another.

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The structure on the left is made with 4 identical springs. The structure on the right is made with 5 of the same springs. All 9 springs have stiffness

k_{0}

and break when the tension in them reaches

T_{0}

. We now want to compare the stiffness and strength of the two structures. Because of the mixture of parallel and series springs, the net stiffness of the structure in (a) is

k_{\rm net}=k_{0}\qquad\mbox{and}\qquad\mbox{strength}=2T_{0}

because none of the springs reaches its breaking tension until $F=T_{0}$ .

By doubling up one of the springs in (a) to get (b) we get

k_{\rm net}=7k_{0}/6\qquad\mbox{and}\qquad\mbox{strength}=21T_{0}/12.

The structure is made 16% stiffer but spring AB now reaches its breaking point $T_{0}$ when the applied load is 12.5% smaller.

What’s going on? The second structure is made stiffer by reducing the deflection of point A. But this causes spring AB to stretch more and thus carry more of the total load. In some sense, the load is concentrated in spring AB. This load concentration, where the total load is unevenly carried, is one reason that stiffness and strength need to be considered separately. Load concentration (or stress concentration) is a major cause of structural failure.

Consider the extreme case: put hundreds of springs in parallel where the pair of springs is now to the left of A in (b). Effectively this welds point A to the left wall. In (b) the load is concentrated in spring AB so (b) is about 50% stiffer than (a) and only has about 75% of (a)’s strength.

In common experience, stiffness and strength do correlate. Something that feels rickety (is very compliant) also tends to fail with a small load. But this common experience can be misleading: 1) A stiff structure can be weak because of stress concentrations, and 2) a given material, say baked clay, may be both stiffer and weaker than the other, say tar.

Box 7.4 A puzzle with two springs and three ropes.

This tricky puzzle is an aside.

A weight hangs from 3 strings (BD,BC, and AC) and 2 springs (AB and CD). Point B is above point C and all ropes are taut.

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When rope BC is cut does the weight go (a) down?, (b) up?, or (c) stay put? (Three dots represents a time for you to stop and think.)

Do the experiment. In 15 minutes you can do this experiment with 3 pieces of string, 2 rubber bands and a soda bottle. Hang the partially filled soda bottle from a door knob (or the top corner of a door, or a ruler cantilevered over the top of a refrigerator). Adjust the string lengths and amount of weight so that no strings or rubber bands are slack and make sure point B is above point C. The two points A can coincide as can the two points D. You can separate the strings a little with, say, a small wad of paper so you can see which string is which.

Look at your experimental setup, but don’t pull and poke at it. Try to predict whether your bottle will go up down or not move when you cut the middle string.

Spoiler, Answer:

If you did the experiment you know the answer, the weight goes up a little when you cut. That’s what really happens.

This puzzle was published as one for which people have bad intuitions. And that’s true, as you probably just found out. Why does the experiment come out the way it does? If you got it wrong (like most people do), can you find the error in your ways?

Clue: All simple explanations are based on the assumption that the lengths of the two strings AC and BD are constant at $\ell_{1}$ .

Explanation 1: To simplify the reasoning assume that springs AB and CD are identical and carry the same tension $T_{s}$ and that the ropes AC and BD carry the same tension $T_{r}$ . As usual, start with free-body diagrams (below). With the symmetry we have assumed diagrams (a) and (c) provide identical information. The three free-body diagrams can be considered before and after the middle string removal by having $T_{d}>0$ or $T_{d}=0$ , respectively. Vertical force balance gives (approximating $T_{d}$ as vertical):

T_{s}+T_{r}=W\hbox to36.135pt{\hfil and\hfil}2T_{r}+T_{d}=W\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}T_{s}=(W+T_{d})/2.

Because we approximate AC as rigid with length $\ell_{1}$ , the downwards position of the weight is the string length $\ell_{1}$ plus the rest length of the spring $\ell_{0}$ plus the stretch of the spring $T_{s}/k$ :

\ell=\ell_{1}+\ell_{0}+T_{s}/k=\ell_{1}+\ell_{0}+(W+T_{d})/(2k).

In the course of this experiment $\ell_{1},\ell_{0},W$ and $k$ are constants. So as the tension $T_{d}$ goes from positive to zero (when the rope BC is cut) and $\ell$ decreases. So the weight goes up.

Explanation 2: More intuitively, start with the configuration with the rope already cut and apply a small upwards force at C. It has no effect on the tension in spring CD thus the weight does not move. Now apply a small downwards force at B. This does stretch spring AB and thus lower point B, thus lowering the weight since $\ell_{1}$ is constant. Applying both simultaneously is like attaching the middle rope. Thus attaching the middle rope lowers the weight so cutting the middle rope raises the weight.

Explanation 3: Here is another intuitive approach. Point C can’t move. Point B moves up and down just as much as the weight does. Point B is a distance $d$ above point C. Since the rope BC is taut, releasing it will allow B and C to separate, thus increasing $d$ and raising the weight.

A wrong explanation: What about springs in parallel and series?

Here is a quick but wrong explanation for the experimental result, though it happens to predict the right direction of motion.

“Before rope BC is cut the two springs are more or less in series because the load is carried from spring through BC to spring. Afterwards they are more or less in parallel because they have the same stretch and share the load. Two springs in parallel have 4 times the stiffness of the same two springs in series. So in the parallel arrangement the deflection is less. So the weight goes up when the springs switch from series to parallel.”

What is the error in this thinking? The position of the weight comes from spring deflection added to the position when there is no weight. For the argument just presented to make sense, the rest-position of the mass (with gravity switched off) would have to be the same for the supposed ‘series’ and ‘parallel’ cases, which it is not ( $\ell_{1}+\ell_{0}\neq\ell_{0}+d+\ell_{0}$ ).

Filename:tfigure-weightpuzzlefbds

Another way to see the fallacy of the ‘parallel versus series’ argument is that the incremental stiffness of the system is, assuming inextensible ropes, infinite. That is, if you add or subtract a small load, the bottle only moves because of a small stretch of the ropes (which is neglected in the correct simple explanations above). If the springs were in series or parallel we would expect an incremental stiffness that was related to spring stretch not rope stretch.

Consequences of series and parallel springs for modeling

As the previous two examples illustrate, springs can sometimes be replaced with ‘air’ (nothing) or with rigid links without changing the system or model behavior much. One way to think about this is that in the limit as $k\rightarrow\infty$ a spring becomes a rigid bar and in the limit $k\rightarrow 0$ a spring becomes air.

These ideas are used by engineers, often intuitively or even subconsciously and with no substantiating calculations, when making a model of a mechanical system.

•

If one of several pieces in series is much stiffer than the others it is often replaced with a rigid link.
•

If one of several pieces in parallel is much more compliant than the others it is often replaced with air (nothing, sailboat fuel).

For example:

•

When a coil spring is connected to a linkage, the other pieces in the linkage, though undoubtedly somewhat compliant, are typically modeled as rigid. They are stiffer than the spring and in series with it.
•

A single hinge resists rotation about axes perpendicular to the hinge axis. But a door connected at two points along its edge is stiffly prevented against such rotations. Thus the hinge stiffness is in parallel with the greater rotational stiffness of the two connection points and is thus often neglected (see the discussion and figures in section 2.3 starting on page 2.3).
•

Welded joints in a determinate truss are modeled as frictionless pins. The rotational stiffness of the welds is ‘in parallel’ with the axial stiffness of the bars. To see this, look at two bars welded together at an angle. Imagine trying to break this weld by pulling the two far bar ends apart. Now imagine trying to break the weld if the two far ends are connected to each other with a third bar. The third bar is ‘in parallel’ with the weld material. See the first few sentences of section 6.1 for a do-it-yourself demonstration of the idea.
•

Human bones are often modeled as rigid because, in part, when they interact with the world they are in series with more compliant flesh.

Filename:sfig8-2-3b

Figure 7.9: a) The two springs shown are in series because they carry the same load and their displacements add. b) These two springs are in parallel because they have a common displacement and their forces add.

Note, again, that the mechanics usage of the words ‘in parallel’ and ‘in series’ don’t always correspond to the geometric arrangement. For example the two springs in fig. 7.9a are in series and the two springs in fig. 7.9b are in parallel.

Filename:sfig8-3-1

Figure 7.10: a) a steel rod in tension, b) tension versus length curve, c) zoom in on the tension versus length curve.

Strength and stiffness

Most often when you build a structure you want to make it stiff and strong. The ideas of stiffness and strength are so intimately related that it is sometimes hard to untangle them. For example, you might examine a product in discount store by putting your hand on it, applying small forces and observing the motion. Then you might say: “pretty shaky, I don’t think it will hold up” meaning that the stiffness is low so you think the thing may break if the loads get high.

Although stiffness and strength are often correlated, they are distinct concepts. Something is stiff if the force to cause a given motion is high. Something is strong if the force to cause any part of it to break is high. In fact, it is possible for a structure to be made weaker by making it stiffer (see box 7.1 on page 7.1)

Why aren’t springs in all mechanical models?

All things deform a little under load. Why don’t we take this deformation into account in all mechanics calculations by, for example, modeling solids as elastic springs?

Because many problems have solutions that would be little affected by such deformation. In particular, if a problem is statically determinate then very small deformations only have a very small effect on the equilibrium equations and calculated forces.

Linear springs are just one way to model ‘give’

If it is important to consider the deformability of an object, the linear spring model is just one simple model. It happens to be a good model for the small deformation of many solids. But the linear spring model is defined by the two words ‘linear’ and ‘elastic’. For some purposes one might want to model the force due to deformation as being non-linear, like $T=k_{1}(\Delta\ell)+k_{2}(\Delta\ell)^{3}$ . And one may want to take account of the dissipative or inelastic nature of something. The most common example is a linear dashpot $T=c\dot{\ell}$ . Various mixtures of non-linearity and inelasticity may be needed to model the large deformations of a yielding metal, for example.

Box 7.5 2D geometry of spring stretch

The material here is used in advanced sample 7.21 on page 7.21 and some of the later homework problems.

The key result concerns a spring with one end fixed at A and the other at moving point B. When point B moves from ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}$ to ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}+\Delta{\accentset% {\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}$ then the spring length changes from $\ell$ to $\ell+\Delta\ell$ with

\Delta\ell\approx\hat{\boldsymbol{\lambda}}_{AB}\cdot\Delta{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}

(7.7)

where $\hat{\boldsymbol{\lambda}}_{AB}={\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {\!\mathrm{AB}}/|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}|$ is a unit vector in the direction AB.

Filename:tfigure-springgeometry

People use either $\Delta\ell$ or $\delta\ell$ to represent the increase of a spring’s length from its rest length. Before we derive the result above a few ways, let’s discuss its relevance.

Fixed-configuration statics: the usual approach

The forces and moments on a system in static equilibrium satisfy force and moment balance. In these equations the force magnitudes and directions, the moments and the locations of points of application of these are those in the equilibrium configuration. The equilibrium of the deformed state is expressed in terms of the geometry of that deformed state. Where the structure was before loading doesn’t appear in the equilibrium equations.

However, often we know the geometry of a structure before the loads are applied, not after. To avoid calculation and confusion, we assume that the deformations cause negligible changes in positions. This is one reason people mistakenly think of statics as being limited to rigid bodies. Rather, for bodies that don’t deform much, we can use the before-load geometry of a structure for reasonably accurate estimation of the deformed geometry.

Statics, taking account of deflection

In principle, the statics of deformable solids is the same as for rigid solids. You just need to use the deformed geometry in the statics calculations. Unfortunately, to find that geometry one needs the forces and their points of application. And one can’t find all the locations without finding the deformation which depends on the forces, etc. This dizzying circle is escapable using the ‘three pillars’ (page 0.1).

Example: A structure made of springs.

Filename:tfigure-springstructure

Assume all the lengths and geometry of the two-bar truss are known when there is no load at C. We can find all the tensions and deflections as follows (See page Look for equations containing unknowns. Don’t look for formulas that evaluate unknowns. for the general strategy):

1.

Assume that the equilibrium loaded location of C is displaced from the rest location by $\delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}=\delta x_{C% }\hat{\boldsymbol{\imath}}+\delta y_{C}\hat{\boldsymbol{\jmath}}$ , where $\delta x_{C}$ and $\delta y_{C}$ are unknowns;
2.

Calculate the lengths of the springs in terms of $\delta x_{C}$ and $\delta y_{C}$ (this will be a complex expression with squares and square roots);
3.

Find the tensions in the springs in terms of their new lengths and thus in terms of $\delta x_{C}$ and $\delta y_{C}$ ;
4.

Draw a free-body diagram of C, using the spring orientations and tensions you have found (still in terms of unknowns $\delta x_{C}$ and $\delta y_{C}$ );
5.

Write the force-balance equations. These are two equations for two unknowns $\delta x_{C}$ and $\delta y_{C}$ .
6.

Solve for $\delta x_{C}$ and $\delta y_{C}$ .
7.

Use $\delta x_{C}$ and $\delta y_{C}$ to find the lengths and thus the tensions in the springs.

What’s wrong with taking account of the deflection?

The trap — having to know the deflection to find the tensions but having to know the tensions to find the deflection — is avoided by setting up and solving simultaneous non-linear equations.

Although this non-linear-equation approach is correct, given our spring model, it is generally not used in structural mechanics because:

•

Confusing. The equations are a mess.
•

Hard. It is hard to solve non-linear equations, sometimes even hard on a computer.
•

Non-uniqueness. There may be more than one solution. For example in the math problem above, if $F$ is not too large, there will be two solutions. One solution with C deflected up and to the right, and another with C way to the left of the wall. To get rid of such off-the-wall solutions you need to either use judgment after you find them, or further specify your math problem to eliminate them.
•

Linear equations are good enough. There are simpler methods that give approximate solutions that are accurate for small-enough loads. That is, if the deflection is small compared to the size of the structure, then linear equations exist which reasonably approximate the non-linear equations above.

Small-deflection theory: the structural-mechanics approach.

So long as $F$ is not too large, the motion of point C will be small compared to the lengths of the springs. Especially since, in practice, those springs are often solid metal rods. The usual small deformation assumption is that

•

The deflection is small enough so that the spring angle changes have negligible effect on the equilibrium equations, and
•

The deflection is small enough for the approximate formula for spring length change, eqn. (7.7), to be adequate.

The recipe for finding the deflection of C in the example above is greatly simplified with these approximations:

1.

Assume that the equilibrium loaded location of C is displaced from the rest location by $\delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}=\delta x_{C% }\hat{\boldsymbol{\imath}}+\delta y_{C}\hat{\boldsymbol{\jmath}}$ , where $\delta x_{C}$ and $\delta y_{C}$ are unknowns (unchanged);
2.

Calculate the lengths of the springs in terms of $\delta x_{C}$ and $\delta y_{C}$ using eqn. (7.7) (simplified);
3.

Find the tensions in the springs in terms of their new lengths (unchanged) and thus in terms of $\delta x_{C}$ and $\delta y_{C}$ (much simpler expressions);
4.

Draw a free-body diagram of C, using the original undeflected geometry (much simplified);
5.

Write the force-balance equations. These are two equations for two unknowns $\delta x_{C}$ and $\delta y_{C}$ . (These will now be linear equations instead of a non-linear mess.)
6.

Solve for $\delta x_{C}$ and $\delta y_{C}$ . (This is now the solution of linear instead of non-linear equations.)
7.

(simplified) Use $\delta x_{C}$ and $\delta y_{C}$ to find the lengths and thus the tensions in the springs. (This now uses eqn. (7.7) instead of complicated relations with square roots, etc.)

This simplified recipe depends on the simplified formula for the spring length change eqn. (7.7), derived below four different ways.

Derivations of equation 7.7

Derivation 1 of eqn. (7.7). The law of cosines (page 1.5) says

\left(\ell+\delta\ell\right)^{2}=\ell^{2}+|\delta{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{AB}}|^{2}+2\ell|\delta{\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{\!\mathrm{AB}}|\cos\theta

( $\theta$ here is negative of that used in the statement of the law of cosines). Expanding the left side and dropping terms in $\delta\ell^{2}$ and $|\delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}|^{2}$ on both sides (assuming $\delta\ell/\ell\ll 1$ and $|\delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}|/\ell\ll 1$ ), and dividing both sides by $\ell$ we get

\delta\ell\approx|\delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{AB}}|\cos\theta=\hat{\boldsymbol{\lambda}}_{AB}\cdot\delta{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}

where the last equality comes from the definition of the dot product (Section 1.2).

Derivation 2 of eqn. (7.7). Use the Pythagorean theorem to determine the lengths of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}$ and of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}+\delta{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}$ :

	$\displaystyle\ell$	$\displaystyle=$	$\displaystyle\sqrt{\left(x_{AB}^{2}+y_{AB}^{2}\right)}$
	$\displaystyle\ell+\delta\ell$	$\displaystyle=$	$\displaystyle\sqrt{\left(\left(x_{AB}+\delta x_{AB}\right)^{2}+\left(y_{AB}+% \delta y_{AB}\right)^{2}\right)}$

Subtracting the first from the second, dividing both sides by $\ell$ , and expanding the contents of the square root we get

\delta\ell/\ell=\sqrt{1+2\left(x_{AB}\delta x_{AB}+y_{AB}\delta y_{AB}\right)/% \ell^{2}+\delta x_{AB}^{2}\delta x_{AB}^{2}/\ell^{2}}-1.

Neglecting $\delta x_{AB}^{2}$ and $\delta y_{AB}^{2}$ (assuming $\delta\ell\ll\ell$ ) and expanding the square root ( $\sqrt{1+\epsilon}\approx 1+\epsilon/2$ ), and multiplying through by $\ell$ , we get

\delta\ell\approx(x_{AB}/\ell)\delta x_{AB}+(y_{AB}/\ell)\delta y_{AB})

which is eqn. (7.7) because $\hat{\boldsymbol{\lambda}}_{AB}=(x_{AB}/\ell)\hat{\boldsymbol{\imath}}+(y_{AB}% /\ell)\hat{\boldsymbol{\jmath}}$ .

Derivation 3 of eqn. (7.7). Using vector notation throughout:

	$\displaystyle\ell^{2}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}$
	$\displaystyle(\ell+\delta\ell)^{2}$	$\displaystyle=$	$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}+% \delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}})\cdot({% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}+\delta{\accentset% {\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}})$

Expanding the second equation, neglecting second order terms and subtracting the first we get

\ell\delta\ell\approx{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{% AB}}\cdot\delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}

dividing by $\ell$ and noting that $\hat{\boldsymbol{\lambda}}_{AB}={\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {\!\mathrm{AB}}/\ell$ we again get eqn. (7.7).

Derivation 4 of eqn. (7.7). Finally, and most intuitively, look at this sketch.

Filename:tfigure-springgeomintuitive

The line AB and its deflected self are nearly parallel. Thus the triangle at the end is nearly a right triangle. So, approximately, $\delta\ell=|\delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}% }|\cos\theta$ which is $\delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}\cdot\hat{% \boldsymbol{\lambda}}_{AB}$ , again giving eqn. (7.7).

What the pros do

To find the loads in metal structures, the pros treat solid bars as springs, as per box 7.1 on page 7.1. Then they use eqn. (7.7) in the small-deflection theory described above. Generally this is all automated in computer code called a finite-element program. Such programs are standard commercial products used daily by hundreds of thousands of engineers around the world.

Solid bars are linear springs

When a structure or machine is built with literal springs (e.g., a wire helix), it is common to treat the other parts as rigid. But when a structure has no literal springs, the small amount of deformation in rigid-looking objects can be important, especially for determining how loads are shared in redundant structures.

Let’s consider a $1\,\mathrm{m}$ (about a yard) steel rod with a $5\,\mathrm{cm}$ square (about $(2\in)^{2}$ ) cross section (fig. 7.10a). If we plot the tension versus length we get a curve like fig. 7.10b. The length just doesn’t visibly change (unless the tension got so large as to damage the rod, not shown). But, when you pull on anything, it does deform at least a little. If we zoom in on the tension versus length plot we get fig. 7.10c. To change the length by one part in a thousand (a millimeter, a twenty-fifth of an inch) we have to apply a tension of about $500,000\,{\mbox{N}}$ (about 60 tons). Nonetheless the plot reveals that the solid steel rod behaves like a (very stiff) linear spring.

Surprisingly perhaps, this little bit of compliance is important to structural engineers.

Modeling solid metal rods as linear springs is essential for finding internal forces in statically indeterminate structures. Because it is hard to picture steel deforming, your intuition may be helped by exaggerating the deformation. Think of all solids as being rubber. Or, if you want to look inside the solid in your mind, think of every solid as if it was a piece of deforming Jello. ^†^†margin: Jello is colored sugar water held together, jelled, by long springy gelatin molecules extracted from animal hooves. For their deformation fantasies, vegetarians can use sea-weed based agar jell (‘Kosher gelatine’).

How does a solid bar’s stiffness depend on its shape and composition?

In box 7.1 on page 7.1 we show that the stiffness of a solid elastic bar is

k=\frac{EA}{\ell}

where $E$ is a material property called the Young’s modulus. It’s that $E$ is big that keeps most solids from deforming visibly

^†^†margin: Why aren’t springs in all mechanical models? All things deform a little under load. Why don’t we take this deformation into account in all mechanics calculations by, for example, modeling solids as elastic springs? Because many problems have solutions which would be little-affected by such deformation. In particular, if a problem is statically determinate, then very small elastic deformations only have a very small effect on the equilibrium equations and calculated forces. So the deformations are neglected, even though they are there.

SAMPLE 7.1

Filename:sfig8-3-1a

Figure 7.11:

Springs in series versus springs in parallel: Two springs with spring constants $k_{1}=100\,{\rm N}/{\rm m}$ and $k_{2}=150\,{\rm N}/{\rm m}$ are attached together in their relaxed state (with no applied force) as shown in Fig. 7.11. In case (a), a vertical force $F=10\,{\mbox{N}}$ is applied at point A, and in case (b), the same force is applied at the end point B. Find the force in each spring for static equilibrium. Also, find the equivalent stiffness for (a) and (b).

Solution In static equilibrium, let $\Delta y$ be the displacement of the point of application of the force in each case. We can figure out the forces in the springs by writing force-balance equations in each case.

•

Case (a):

Filename:sfig8-5-wiper

Figure 7.12: Free-body diagram of joint A.

The free-body diagram of point A is shown in Fig. 7.12. As point A is displaced downwards by $\Delta y$ , spring 1 gets stretched by $\Delta y$ whereas spring 2 gets compressed by $\Delta y$ . Therefore, the forces applied by the two springs, $k_{1}\Delta y$ and $\Delta k_{2}y$ , are in the same direction. Then, the force balance in the vertical direction, $\hat{\boldsymbol{\jmath}}\cdot(\sum{\accentset{\rightharpoonup}{\boldsymbol{F}% }}={\accentset{\rightharpoonup}{\boldsymbol{0}}})$ , gives:

$\displaystyle F$	$\displaystyle=$	$\displaystyle F_{1}+F_{2}=(k_{1}+k_{2})\Delta y$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\Delta y$	$\displaystyle=$	$\displaystyle\frac{F}{k_{1}+k_{2}}=\frac{10\,{\mbox{N}}}{(100+150)\,{\rm N}/{% \rm m}}=0.04\,\mathrm{m}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F_{1}$	$\displaystyle=$	$\displaystyle k_{1}\Delta y=100\,{\rm N}/{\rm m}\cdot 0.04\,\mathrm{m}=4\,{% \mbox{N}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F_{2}$	$\displaystyle=$	$\displaystyle k_{2}\Delta y=150\,{\rm N}/{\rm m}\cdot 0.04\,\mathrm{m}=6\,{% \mbox{N}}.$

The equivalent stiffness of the system is the stiffness of a single spring that will undergo the same displacement $\Delta y$ under $F$ . From the equilibrium equation above, it is easy to see that,

k_{e}\equiv\frac{F}{\Delta y}=k_{1}+k_{2}=250\,{\mbox{N}}/\,\mathrm{m}.

Answer: $F_{1}=4\,{\mbox{N}},\quad F_{2}=6\,{\mbox{N}},\quad k_{e}=250\,{\mbox{N}}/\,% \mathrm{m}$

•

Case (b):

Filename:sfig8-5-wiper-a

Figure 7.13: Free-body diagrams

The free-body diagrams of the two springs are shown in Fig. 7.13 along with that of point B. In this case both springs stretch as point B is displaced downwards. Let the net stretch in spring 1 be $y_{1}$ and in spring 2 be $y_{2}$ . $y_{1}$ and $y_{2}$ are unknown, of course, but we know that

$y_{1}+y_{2}=\Delta y.$

Now, using the free-body diagram of point B and writing the force balance equation in the vertical direction, we get $F=k_{2}y_{2}$ , and from the free-body diagram of spring 2, we get $k_{2}y_{2}=k_{1}y_{1}$ . Thus the force in each spring is the same and equals the applied force, i.e.,

$F_{1}=k_{1}y_{1}=F=10\,{\mbox{N}}\quad\qquad\mbox{and}\qquad\quad F_{2}=k_{2}y% _{2}=F=10\,{\mbox{N}}.$

The springs in this case are in series. Therefore, their equivalent stiffness, $k_{e}$ , is

$k_{e}=\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}\right)^{-1}=\left(\frac{1}{100\,{% \rm N}/{\rm m}}+\frac{1}{150\,{\rm N}/{\rm m}}\right)^{-1}=60\,{\rm N}/{\rm m}.$

Note that the displacements $y_{1}$ and $y_{2}$ are different in this case. They can be easily found from $y_{1}=F/k_{1}$ and $y_{2}=F/k_{2}$ .

Answer: $F_{1}=F_{2}=10\,{\mbox{N}},\quad k_{e}=60\,{\rm N}/{\rm m}$

Comments: Although the springs attached to point A do not visually seem to be in parallel, from a mechanics point of view they are parallel. Springs in parallel have the same displacement but different forces. Springs in series have different displacements but the same force.

SAMPLE 7.2

Filename:sfig8-5-2disks

Figure 7.14:

Stiffness of three springs: For the spring networks shown in Fig. 7.14(a) and (b), find the equivalent stiffness of the springs in each case, given that each spring has a stiffness of $k=20\,{\rm N}/{\rm m}$ .

Solution

1.

In Fig. 7.14(a), all springs are in parallel since all of them undergo the same displacement $\Delta x$ in order to balance the applied force $F$ . Each of the two springs on the left stretches by $\Delta x$ and the spring on the right compresses by $\Delta x$ . Therefore, the equivalent stiffness of the three springs is

$k_{p}=k+k+2k=4k=80\,{\rm kN}/\,\mathrm{m}.$

Pictorially,

Filename:sfig8-4-1

Figure 7.15:

Answer: $k_{\rm equiv}=80\,{\rm kN}/\,\mathrm{m}$
2.

In Fig. 7.14(b), the first two springs (on the left) are in parallel but the third spring is in series with the first two. To see this, imagine that, in equilibrium, point A moves to the right by $\Delta x_{A}$ and point B moves to the right by $\Delta x_{B}$ . Then each of the first two springs has the same stretch $\Delta x_{A}$ while the third spring has a net stretch $=\Delta x_{B}-\Delta x_{A}$ . Therefore, to find the equivalent stiffness, we can first replace the two parallel springs by a single spring of equivalent stiffness $k_{p}=k+k=2k$ . Then the springs with stiffnesses $k_{p}$ and $2k$ are in series and therefore their equivalent stiffness $k_{s}$ is found as follows:

$\displaystyle\frac{1}{k_{s}}$ $\displaystyle=$ $\displaystyle\frac{1}{k_{p}}+\frac{1}{2k}=\frac{1}{2k}+\frac{1}{2k}=\frac{1}{k}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}k_{s}$ $\displaystyle=$ $\displaystyle k=20\,{\rm kN}/\,\mathrm{m}.$

Filename:sfig8-4-1a

Figure 7.16:

Answer: $k_{\rm equiv}=20\,{\rm kN}/\,\mathrm{m}$

SAMPLE 7.3 Stiffness vs strength: Which of the two structures (network of springs) shown in the figure is stiffer and which one has more strength if each spring has stiffness $k=10\,{\rm kN}/{\rm m}$ and strength $T_{0}=10\,{\rm kN}$ .

Filename:sfig8-6-2

Figure 7.17:

Solution In structure (a), all the three springs are in parallel. Therefore, the equivalent stiffness of the three springs is

k_{a}=k+k+k=3k=30\,{\rm kN}/\,\mathrm{m}.

Filename:sfig8-6-2a

Figure 7.18:

For figuring out the strength of the structure, we need to find the force in each spring. From the free-body diagram in Fig. 7.18 we see that,

$\displaystyle k\Delta x+k\Delta x+k\Delta x$	$\displaystyle=$	$\displaystyle F$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\Delta x$	$\displaystyle=$	$\displaystyle\frac{F}{3k}.$

$\displaystyle F_{\rm s}$	$\displaystyle=$	$\displaystyle k\Delta x=\frac{F}{3}.$

But the maximum force that a spring can take is $(F_{\rm s})_{\rm max}=T_{0}=10\,{\rm kN}$ . Therefore, the maximum force that the structure can take ( i.e., the strength of the structure), is

F_{\rm max}=3T_{0}=30\,{\rm kN}.

Answer: Stiffness $=30\,{\rm kN}/{\rm m},$ Strength = $30\,{\rm kN}$

Now we carry out a similar analysis for structure (b). There are four parallel chains in this structure, each chain containing two springs in series. The stiffness of each chain, $k_{c}$ , is found from

	$\displaystyle\frac{1}{k_{c}}$	$\displaystyle=$	$\displaystyle\frac{1}{k}+\frac{1}{k}=\frac{2}{k}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}k_{c}$	$\displaystyle=$	$\displaystyle\frac{k}{2}=5\,{\rm kN}/\,\mathrm{m}.$

So, the stiffness of the entire structure is

k_{b}=k_{c}+k_{c}+k_{c}+k_{c}=4k_{c}=20\,{\rm kN}/\,\mathrm{m}.

Filename:efig1-2-28

Figure 7.19:

From the free-body diagram shown in fig. 7.19, we find the force in each spring to be F/4 . Therefore, the maximum force that the structure can take is

F_{\rm max}=4T_{0}=40\,{\rm kN}.

Answer: Stiffness $=20\,{\rm kN}/{\rm m},$ Strength = $40\,{\rm kN}$

Thus, structure (a) is stiffer but structure (b) is stronger (higher strength).

SAMPLE 7.4

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Figure 7.20:

Zero length springs are special. A rigid and massless rod OAB of length 2 $\,\mathrm{m}$ supports a weight $W=100\,\mathrm{kg}$ hung from point B. The rod is pinned at O and supported by a zero length (in relaxed state) spring attached at mid-point A and point C on the vertical wall. Find the equilibrium angle $\theta$ and the force in the spring.

Solution

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Figure 7.21:

The free-body diagram of the rod is shown in Fig. 7.21 in an assumed equilibrium state. Let $\hat{\boldsymbol{\lambda}}=-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat% {\boldsymbol{\jmath}}$ be a unit vector along OB. The spring force can be written as ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{s}}=k{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}$ (since AC is a zero-length spring, the stretch in the spring is $|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}|$ ). We need to determine $\theta$ and $F_{s}$ .

Let us write moment equilibrium equation about point O, i.e., $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{O}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ ,

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/O}}\times{% \accentset{\rightharpoonup}{\boldsymbol{W}}}+{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{A/O}}\times{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{s}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

Noting that

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/O}}$	$\displaystyle=$	$\displaystyle\ell\hat{\boldsymbol{\lambda}},\quad{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{A/O}}=\frac{\ell}{2}\hat{\boldsymbol{\lambda}},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{s}}$	$\displaystyle=$	$\displaystyle k{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}=% k({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}})$
	$\displaystyle=$	$\displaystyle k\left(h\hat{\boldsymbol{\jmath}}-\frac{\ell}{2}\hat{\boldsymbol% {\lambda}}\right),$

we get,

	$\displaystyle\ell\hat{\boldsymbol{\lambda}}\times(-W\hat{\boldsymbol{\jmath}})% +\frac{\ell}{2}\hat{\boldsymbol{\lambda}}\times k\left(h\hat{\boldsymbol{% \jmath}}-\frac{\ell}{2}\hat{\boldsymbol{\lambda}}\right)$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle-W\ell(\hat{\boldsymbol{\lambda}}\times\hat{\boldsymbol{\jmath}})% +kh\frac{\ell}{2}(\hat{\boldsymbol{\lambda}}\times\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Dotting this equation with $(\hat{\boldsymbol{\lambda}}\times\hat{\boldsymbol{\jmath}})$ , we get,

	$\displaystyle-W\ell+kh\frac{\ell}{2}$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}kh$	$\displaystyle=$	$\displaystyle 2W.$

Thus the result is independent of $\theta$ ! As long as the spring stiffness $k$ and the height $h$ of point C are such that their product equals $2W$ , the system will be in equilibrium at any angle. This, however, is in general not possible if AC is not a zero-length spring.

Answer: Equilibrium is satisfied at any angle if $kh=2W$

SAMPLE 7.5

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Figure 7.22:

Deflection of an elastic structure: For the two-spring structure shown in the figure, find the deflection of point C when

1.

${\accentset{\rightharpoonup}{\boldsymbol{F}}}=1\,{\mbox{N}}\hat{\boldsymbol{% \imath}}$ ,
2.

${\accentset{\rightharpoonup}{\boldsymbol{F}}}=1\,{\mbox{N}}\hat{\boldsymbol{% \jmath}}$ ,
3.

${\accentset{\rightharpoonup}{\boldsymbol{F}}}=30\,{\mbox{N}}\hat{\boldsymbol{% \imath}}+20\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$ ,

The spring stiffnesses are $k_{1}=10\,{\rm kN}/m$ and $k_{2}=20\,{\rm kN}/m$ .

Solution

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Figure 7.23:

Let $\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}=\Delta x\hat{\boldsymbol{% \imath}}+\Delta y\hat{\boldsymbol{\jmath}}$ be the displacement of point C of the structure due to the applied load. We can figure out the deflections in each spring as follows. Let $\hat{\boldsymbol{\lambda}}_{\mathrm{AC}}$ and $\hat{\boldsymbol{\lambda}}_{\mathrm{BC}}$ be the unit vectors along AC and BC, respectively (see fig. 7.24). Then, the change in the length of spring AC due to the (assumed small) displacement of point C is (see page 7.1 for a discussion)

$\displaystyle\Delta_{\rm AC}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\lambda}}_{\mathrm{AC}}\cdot\Delta{\accentset{% \rightharpoonup}{\boldsymbol{r}}}\quad\mbox{(this is the key equation)}$
	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\imath}}\cdot(\Delta x\hat{\boldsymbol{\imath}}+% \Delta y\hat{\boldsymbol{\jmath}})=\Delta x.$

$\displaystyle\Delta_{\rm BC}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\lambda}}_{\mathrm{BC}}\cdot\Delta{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$
	$\displaystyle=$	$\displaystyle(\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{\boldsymbol{% \jmath}})\cdot(\Delta x\hat{\boldsymbol{\imath}}+\Delta y\hat{\boldsymbol{% \jmath}})=\Delta x\cos\theta-\Delta y\sin\theta.$

Now we can find the force in each spring since we know the deflection in each spring.

	$\displaystyle\mbox{Force in spring AC}=F_{1}$	$\displaystyle=$	$\displaystyle k_{1}\Delta x$		(7.8)
	$\displaystyle\mbox{Force in spring BC}=F_{2}$	$\displaystyle=$	$\displaystyle k_{2}(\Delta x\cos\theta-\Delta y\sin\theta).$		(7.9)

The forces in the springs, however, depend on the applied force, since they must satisfy static equilibrium. Thus, we can determine the deflection by first finding $F_{1}$ and $F_{2}$ in terms of the applied load and substituting in the equations above to solve for the deflection components.

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Figure 7.24:

1.

Deflections with unit force in the $x$ -direction:

Let ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=f_{x}\hat{\boldsymbol{\imath}}=1% \,{\mbox{N}}\hat{\boldsymbol{\imath}}$ , (we have adopted a special symbol $f_{x}$ for the unit load). Then, from the free-body diagram of the springs and the end pin shown in fig. 7.23 and the force equilibrium ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ ), we have,

$\displaystyle f_{x}\hat{\boldsymbol{\imath}}-F_{1}\hat{\boldsymbol{\imath}}+F_% {2}(-\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})=% {\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Dotting this eqn. with $\hat{\boldsymbol{\jmath}}$ and $\hat{\boldsymbol{\imath}}$ , respectively, we get,

$\displaystyle F_{2}$ $\displaystyle=$ $\displaystyle 0$

$\displaystyle F_{1}$ $\displaystyle=$ $\displaystyle f_{x}=1\,{\mbox{N}}.$

Substituting these values of $F_{1}$ and $F_{2}$ in eqns. (7.8) and (7.9), and solving for $\Delta x$ and $\Delta y$ we get,

$\left(\begin{array}[]{c}\Delta x\\ \Delta y\end{array}\right)_{{\accentset{\rightharpoonup}{\boldsymbol{F}}}=f_{x% }\hat{\boldsymbol{\imath}}}=\left(\begin{array}[]{c}\frac{1}{k_{1}}\\ \frac{1}{k_{1}}\cot\theta\end{array}\right)f_{x}.$ (7.10)

Substituting the given values of $\theta,k_{1}$ , and $f_{x}=1\,{\mbox{N}}$ , we get

$\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}=\Delta x\hat{\boldsymbol{% \imath}}+\Delta y\hat{\boldsymbol{\jmath}}=(100\hat{\boldsymbol{\imath}}+173% \hat{\boldsymbol{\jmath}})\times 10^{-6}\,\mathrm{m}.$

Answer: $\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}=(100\hat{\boldsymbol{% \imath}}+173\hat{\boldsymbol{\jmath}})\times 10^{-6}\,\mathrm{m}$

Deflections with unit force in the $y$ -direction: We carry out a similar analysis for this case. We again assume the displacement of point C to be $\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}=\Delta x\hat{\boldsymbol{% \imath}}+\Delta y\hat{\boldsymbol{\jmath}}$ . Since the geometry of deformation and the associated results are the same, eqns. (7.8) and (7.9) remain valid. We only need to find the spring forces from the static equilibrium under the new load. From the free-body diagram in Fig. 7.25 we have,

$\displaystyle(-F_{1}-F_{2}\cos\theta)\hat{\boldsymbol{\imath}}+(F_{2}\sin% \theta+F)\hat{\boldsymbol{\jmath}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}$			(7.11)
$\displaystyle\{\mbox{eqn.~{}(\ref{eq:compliance.fy})}\}\cdot\hat{\boldsymbol{% \jmath}}$	$\Rightarrow$	$\displaystyle F_{2}=-\frac{F}{\sin\theta}$
$\displaystyle\{\mbox{eqn.~{}(\ref{eq:compliance.fy})}\}\cdot\hat{\boldsymbol{% \imath}}$	$\Rightarrow$	$\displaystyle F_{1}=-F_{2}\cos\theta=F\cot\theta.$

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Figure 7.25:

Substituting these values of $F_{1}$ and $F_{2}$ in terms of $F=f_{y}$ in eqns. (7.8) and (7.9), we get

$\displaystyle f_{y}\cot\theta$	$\displaystyle=$	$\displaystyle k_{1}\Delta x\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\Delta x=% \frac{f_{y}}{k_{1}}\cot\theta$
$\displaystyle-\frac{f_{y}}{\sin\theta}$	$\displaystyle=$	$\displaystyle k_{2}(\Delta x\cos\theta-\Delta y\sin\theta)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\Delta y$	$\displaystyle=$	$\displaystyle\frac{1}{\sin\theta}\left(\Delta x\cos\theta+\frac{f_{y}}{k_{2}% \sin\theta}\right)$
	$\displaystyle=$	$\displaystyle f_{y}\left(\frac{1}{k_{1}}\cot^{2}\theta+\frac{1}{k_{2}}\csc^{2}% \theta\right).$

Thus,

\left(\begin{array}[]{c}\Delta x\\ \Delta y\end{array}\right)_{{\accentset{\rightharpoonup}{\boldsymbol{F}}}=f_{y% }\hat{\boldsymbol{\jmath}}}=\left(\begin{array}[]{c}\frac{1}{k_{1}}\cot\theta% \\ \frac{1}{k_{1}}\cot^{2}\theta+\frac{1}{k_{2}}\csc^{2}\theta\end{array}\right)f% _{y}.

(7.12)

Substituting the values of $\theta,k_{1},k_{2}$ , and $f_{y}=1\,{\mbox{N}}$ , we get

\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}=\Delta x\hat{\boldsymbol{% \imath}}+\Delta y\hat{\boldsymbol{\jmath}}=(173\hat{\boldsymbol{\imath}}+500% \hat{\boldsymbol{\jmath}})\times 10^{-6}\,\mathrm{m}.

Answer: $\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}=(173\hat{\boldsymbol{% \imath}}+500\hat{\boldsymbol{\jmath}})\times 10^{-6}\,\mathrm{m}$

Deflection under a general load: Since we have already got expressions for deflections in the $x$ and $y$ -directions under unit loads in the $x$ and $y$ -directions, we can now combine the results (using superposition, see page 5.2) to find the deflection under any general load ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=F_{x}\hat{\boldsymbol{\imath}}+F% _{y}\hat{\boldsymbol{\jmath}}$ as follows.

	$\displaystyle\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle\left(\begin{array}[]{c}\Delta x\\ \Delta y\end{array}\right)=F_{x}\cdot\left(\begin{array}[]{c}\Delta x\\ \Delta y\end{array}\right)_{{\accentset{\rightharpoonup}{\boldsymbol{F}}}=1% \hat{\boldsymbol{\imath}}}+F_{y}\cdot\left(\begin{array}[]{c}\Delta x\\ \Delta y\end{array}\right)_{{\accentset{\rightharpoonup}{\boldsymbol{F}}}=1% \hat{\boldsymbol{\jmath}}}$
		$\displaystyle=$	$\displaystyle\left[\begin{array}[]{cc}k_{1}^{-1}&k_{1}^{-1}\cot\theta\\ k_{1}^{-1}\cot\theta&k_{1}^{-1}\cot^{2}\theta+k_{2}^{-1}\csc^{2}\theta\end{% array}\right]\left(\begin{array}[]{c}F_{x}\\ F_{y}\end{array}\right).$

Once again, substituting all given values and $F_{x}=30\,{\mbox{N}}\qquad\mbox{and}\qquad F_{y}=20\,{\mbox{N}}$ , we get

\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}=(6.4\hat{\boldsymbol{% \imath}}+15.2\hat{\boldsymbol{\jmath}})\times 10^{-3}\,\mathrm{m}.

Answer: $\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}=(6.4\hat{\boldsymbol{% \imath}}+15.3\hat{\boldsymbol{\jmath}})\times 10^{-3}\,\mathrm{m}$

Note: The matrix obtained above for finding the deflection under a general load is called the compliance matrix of the structure. Its inverse is the stiffness matrix $K$ of the structure. $K$ is used to find forces given deflections.

Problems for 7.1 Springs

Preparatory Problems

7.1.1 Find the force $F$ required to push the massless block by $1\,\mathrm{cm}$ to the right if $k=500\,{\rm N}/{\rm m}$ . Answer: $10\,{\mbox{N}}$

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Figure 7.26:

7.1.2 A force $F=20\,{\mbox{N}}$ is applied on the massless block shown in the figure. Find the displacement of the block for equilibrium if $k=100\,{\rm N}/{\rm cm}$ . Answer: $0.08\,\mathrm{cm}$

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Figure 7.27:

7.1.3 A network of relaxed springs holds a massless block as shown in the figure where $k_{1}=100\,{\rm N}/{\rm cm}$ and $k_{2}=400\,{\rm N}/{\rm cm}$ . If the block is pushed to the right by $2\,\mathrm{cm}$ , find the force $F$ to hold the block in equilibrium. Answer: $1160\,{\mbox{N}}$

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Figure 7.28:

7.1.4 A block of mass $m=300\,\mathrm{kg}$ hangs from the ceiling with the help of a network of springs in series and parallel as shown. Taking $k=20\,{\rm kN}/{\rm m}$ and $g=10\,\mathrm{m}/\mathrm{s}^{2}$ , find the stretch in the two side (the left and right) springs. Answer: $5\,\mathrm{cm}$

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Figure 7.29:

7.1.5 For the arrangement of springs shown in the figure, $k_{1}=50\,{\rm N}/{\rm cm}$ and $k_{2}=100\,{\rm N}/{\rm cm}$ . Find

(a)

the equivalent spring stiffness of the arrangement,
(b)

the displacement of the block if a force $F=50\,{\mbox{N}}$ acts on the block.

Answer: $k_{e}=66.7\,{\rm N}/{\rm cm},\,\delta=0.75\,\mathrm{cm}$

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Figure 7.30:

7.1.6 Find $F$ in terms of some or all of $k_{1}$ , $\ell_{1}$ , $k_{2}$ , $\ell_{2}$ , $\ell_{0}$ and $\delta$ . Note that $F$ is generally not zero even if $\delta$ is zero.

(a)

Springs in parallel.
(b)

Springs in series.

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Figure 7.31:

More-Involved Problems

7.1.7 A massless block is held in position by a network of springs shown in the figure. If the block is displaced to the right by $1\,\mathrm{cm}$ from the relaxed position of the springs, a force of $F=50\,{\mbox{N}}$ is required to keep the block in equilibrium. Find the value of $k$ . Answer: $k=20\,{\rm N}/{\rm cm}$

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Figure 7.32:

7.1.8 A box weighing 1000 N is hung from the ceiling using a network of springs, each with stiffness $k=500\,{\rm N}/{\rm cm}$ . Find the stretch in each spring. Answer: Middle spring: $\delta=1\,\mathrm{cm}$ ; side-springs $\delta=0.5\,\mathrm{cm}$

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Figure 7.33:

7.1.9 For the network of springs, find the net stiffness and strength of each network if the stiffness and strength of individual springs are $k=10\,{\rm kN}/{\rm m}$ and $T_{0}=2\,{\rm kN}$ , respectively. The load is applied to the central mass and the springs are all relaxed when no load is applied.

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Figure 7.34:

7.1.10 Find the stretch in each spring to hold the pin in equilibrium for $F=10\,{\rm kN}$ if the relaxed length (in the horizontal position) of each spring is $\ell_{0}=15\,\mathrm{cm}$ and $k=10\,{\rm kN}/{\rm m}$ .

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Figure 7.35:

7.1.11 A mass slides with negligible friction in a rigid horizontal track. It is also pulled by a zero-rest-length spring ( $\ell_{0}=0$ ) of stiffness $50\,{\rm kN}/{\rm m}$ . Find the horizontal position $x$ of the pin if it is in equilibrium with an applied force $F=1000\,{\mbox{N}}$ .

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Figure 7.36:

7.1.12 A zero length spring (relaxed length $\ell_{0}=0$ ) with stiffness $k=5\,{\rm N}/{\rm m}$ supports the pendulum shown. Assume $g=10\,\,{\mbox{N}}/\,\mathrm{m}$ . Find $\theta$ for static equilibrium. Answer: Surprise! This pendulum is in equilibrium for all values of $\theta$ .

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Figure 7.37:

7.1.13 In the figure shown, the two springs with $k_{1}=50\,{\rm N}/{\rm cm}$ (left spring) and $k_{2}=100\,{\rm N}/{\rm cm}$ (right spring) are in relaxed position when $h=30\,\mathrm{cm}$ and $\ell=40\,\mathrm{cm}$ (and, of course, $F=0$ ). For $F=200\,{\mbox{N}}$ , find the position of the pin on the horizontal frictionless track, and change in length of each spring,

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Figure 7.38:

7.1.14 In the mechanism shown, the relaxed length of the spring is $\ell/2$ and the length of the bar AB is $\ell=2\,\mathrm{m}$ . For $F=500\,{\mbox{N}}$ , find the equilibrium angle $\theta$ of the rod and the stretch in the spring.

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Figure 7.39:

7.1.15 The ends of three identical springs are rooted at the corners of a $10\,\mathrm{cm}$ equilateral triangle with base that is in the $\hat{\boldsymbol{\imath}}$ direction. Find the force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ needed to hold the ends of the springs $5\,\mathrm{cm}$ to the right of the triangle center if

(a)

$\ell_{0}=0$ , $k=10\,{\mbox{N}}/\,\mathrm{cm}$ ?
(b)

$\ell_{0}=10/\sqrt{3}\,\mathrm{cm}$ , $k=10\,{\mbox{N}}/\,\mathrm{cm}$ ?

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Figure 7.40:

7.1.16 The hoop is rigid, round and frictionless and the force is tangent to the hoop.

(a)

In terms of some or all of $k$ , $\ell_{0}$ , $r$ , and $\theta$ , find $F$ .
(b)

How does the answer above simplify in the special case that $\ell_{0}=0$ ? [You can do this by simplifying the expression above, or by doing the problem from scratch assuming $\ell_{0}=0$ . In the latter case, an answer can be generated quickly if vector methods are used.]

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Figure 7.41:

7.1.17 The square box mechanism shown consists of three identical bars and two identical diagonal springs in their relaxed configuration. Each bar is $0.4\,\mathrm{m}$ long. A horizontal force $F=100\,{\mbox{N}}$ acts at C. Find the change in length of each spring if $k=10\,{\rm kN}/{\rm m}$ . Answer: $\Delta L_{1}=-\Delta L_{2}=F/(\sqrt{2}\,k)$

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Figure 7.42:

7.1.18 In the mechanism shown, the pin is held in the center of the square frame of side $1\,\mathrm{m}$ with relaxed springs of stiffness $k=5\,{\rm kN}/{\rm m}$ in the absence of any force. Find the change in length of each spring when an applied horizontal force $F=50\,{\mbox{N}}$ keeps the pin in equilibrium at a position slightly to the right of the center.

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Figure 7.43:

7.2 Force amplification devices: Levers, wedges, toggles, gears, and pulleys

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Figure 7.44: One half of a pair of pliers is one of many classic lever examples.

Simple objects can be connected in various arrangements for various purposes. Here we describe 5 machine fragments that can be used to amplify force. Most machines use these ideas in combination. It might help intuitive understanding of machines to recognize one of these methods in use, although precise categorization of every machine part as one or another of these devices is not possible.

A lever

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Figure 7.45: A lever can have the pivot in various places. The free-body diagram looks the same whether the pivot is at the left, middle, or right.

One of the simplest machines, long understood, and even longer used by humans, is a lever (e.g., , fig. 7.44 and the top of fig. 7.45). Although now we think of statics as a special case of dynamics, statics is, and was, a well-founded subject, independent of dynamics. The statics of a lever was well understood 50 generations before Newton discovered modern dynamics.

An ideal lever is a rigid body held in place with a frictionless hinge and with two other applied loads.

The free-body diagram fig. 7.45 is the same whether the hinge is at point A, B or C

^†^†margin: People studied levers before mechanics was a quantitative subject. So they have a specialized antiquated vocabulary, classifying them according to which of A, B, or C is the pivot point and which of the other two forces you think of as input and which as the output: A ‘class one’ lever has the pivot in between; a ‘class two’ lever has the pivot at one end and the input force at the other; and a ‘class three’ lever has the pivot at one end and the input force in the middle.

Lots of things can be viewed as levers including, for example, a wheelbarrow, a hammer extracting a nail, a boat oar, one half of a pair of tweezers, a brake lever, a gear, and, most generally, any three-force body. Using the equilibrium relations on the free-body diagram in fig. 7.45 you get that

{F_{A}\over a}={F_{B}\over b}={F_{C}\over c}

from which you could find the relation between any pair of the forces. In practice it is easier to use moment balance about an appropriate point than to memorize and recall this formula.

Refer to caption — Figure 7.46: A hatchet splitting wood is a classic wedge. In this case it is used dynamically, with the force coming from the deceleration of mass. But other more complex wood splitting devices push the wedge in slowly.

An ideal wedge

Wedges are a kind of machine. You see them used in exactly their wedge configuration, to split and separate things (see fig. 7.46). But screws are also, effectively wedges, with torque replacing the downwards force.

For an ideal wedge, one neglects friction, effectively replacing sliding contact with rolling contact (see fig. 7.47ab). Although this approximation may not be accurate, it is helpful for building intuition. Because the key idea depends on force balance and not moment balance, for the free-body diagrams of fig. 7.47c, we have not specified the exact location of the contact forces . Neglecting gravity,

\begin{array}[]{lccc}\mbox{For block A},\left\{\sum{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{\!\!\mathrm{i}}={\accentset{\rightharpoonup}{\boldsymbol{0% }}}\right\}\cdot\hat{\boldsymbol{\jmath}}&\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}&-F_{A}+F\sin\theta&=0\\ \mbox{For block B},\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{% \!\!\mathrm{i}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\right\}\cdot\hat% {\boldsymbol{\imath}}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&-F_{B}+F\cos% \theta&=0\\ \end{array}

\mbox{eliminating $F$}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F_{B}={1\over% \tan\theta}F_{A}.

Using the small-angle approximation that $\tan\theta\approx\theta$ , we have that

The force amplification of a wedge is roughly the reciprocal of the wedge angle (in radians).

To multiply the force $F_{A}$ by 10 takes a wedge with a taper of $\theta=\tan^{-1}0.1\approx 6^{{}^{\circ}}$ . With this taper, an ideal wedge could also be viewed as a device to attenuate the force $F_{B}$ by a factor of 10, although wedges are never used for force attenuation in practice, because friction often causes them to bind. A wedge with friction is considered in 7.68 on page 7.68.

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Figure 7.47: a) A wedge, b) treated as frictionless in the ideal case, c) free-body diagrams (assuming no friction)

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Figure 7.48: A simple toggle mechanism. Moment balance about point A gives that

N=F/\tan\theta

so if

\theta

is small

N

is large.

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Figure 7.49: Two bars in a toggle mechanism.

A toggle

The classic toggle mechanism for amplifying force tends to have a ‘snap-through’ or bi-stable aspect which is used in the design of some electrical switches. Hence, perhaps, the two dictionary meanings of the word toggle: 1) a force amplifying mechanism, 2) a switch between two states (see fig. 7.50).

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Figure 7.50: Cutaway drawing of a toggle switch and a simplified schematic of the idea. The lever, spring and plunger rotate about C. The spring is pre-stressed so that the plunger presses hard against the rocker. The plunger slides on the rocker (input line) which is pressed against the contact at outputs A or B. The relevant idea here is that the force of the plunger on the rocker can be much bigger than the input force on the lever (cutaway from NKK Switches online catalogue).

The simplest version of the toggle mechanism is shown in fig. 7.48.

The force amplification is $N/F=1/\tan\theta$ .

Usually the toggle concept is not used with a wall but with a pair of bars (fig. 7.49) Simple truss analysis shows the bar compressions are $-T=N/2\sin\theta$ and $N=F/2\tan\theta$ . The toggle-like force amplification occurs for tension as well as compression. But, because of the oft-desirable snap-through and because the amplification increases as the applied force $F$ moves down, the toggle is most often used in compression.

A toggle as lever and wedge. The distinction between toggles and wedges and levers is not precise. On the one hand the toggle is a lever where the lever arm of $F$ is $\ell\cos\theta$ and the lever arm of $N$ is $\ell\sin\theta$ . On the other hand the toggle is sort of a rotary wedge with wedge angle $\theta$ .

Pulleys and Gears

Here we discuss a few more common machine components which are used to transmit and amplify or attenuate a force or moment.

Gears

One type of transmission is based on gears (fig. 7.52a). If we think of the input and output as the moments on the two gears, we find from the free-body diagram in fig. 7.52b that

\begin{array}[]{lccc}\mbox{For gear A},\left\{\sum{\accentset{\rightharpoonup}% {\boldsymbol{M}}}_{\mathrm{i/A}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}% \right\}\cdot\hat{\boldsymbol{k}}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&-R% _{\rm A}F+M_{\rm A}&=0\\ \mbox{For gear B},\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{% \mathrm{i/B}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\right\}\cdot\hat{% \boldsymbol{k}}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&-R_{\rm B}F+M_{\rm B% }&=0\\ \end{array}

\mbox{eliminating $F$}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}M_{B}={R_{B}% \over R_{A}}M_{A}\hbox to36.135pt{\hfil or\hfil}M_{A}={R_{A}\over R_{B}}M_{B}

depending on which you want to think of input and which as output. The force amplification or attenuation ratio is just the radius ratio, just like for a lever.

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Figure 7.51: One gear may be thought of as a lever.

Because the spacing of gear teeth for both of a meshed pair of gears is the same, a gear’s circumference, and hence its radius is proportional to the number of teeth. And formulas involving radius ratios can just as well be expressed in terms of ratios of numbers of teeth. The tooth ratio is not just used as an approximation to the radius ratio. Averaged over the passage of several teeth, it is exactly the reciprocal ratio of the turning rates of the meshed gears.

Two gears pulled out of a bigger transmission are shown in fig. 7.52c. Gear A has an inner part with radius $R_{A_{i}}$ welded to an outer part with radius $R_{A_{o}}$ . Gear B also has an inner part welded to an outer part.

Moment balance about A in the first free-body diagram in fig. 7.52d gives that $R_{A_{i}}F_{A}=R_{A_{o}}F$ . You can think of the one gear as a lever (see fig. 7.51).

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Figure 7.52: a) Two gear pairs pulled out of a transmission with forces on the teeth, b) Free-body diagrams, c) The same gear pair, but loaded with tooth-forces from unseen gears, d) the consequent free-body diagrams.

Moment balance about B in the second free-body diagram gives that $R_{B_{i}}F=R_{B_{o}}F_{B}$ . Combining, we get

F_{B}={R_{A_{i}}\over R_{A_{o}}}{R_{B_{i}}\over R_{B_{o}}}F_{A}\hbox to36.135% pt{\hfil or\hfil}F_{A}={R_{A_{o}}\over R_{A_{i}}}{R_{B_{o}}\over R_{B_{i}}}F_{B}

depending on which force you want to find in terms of the other. The transmission attenuates the force if you think of $F_{A}$ as the input and amplifies the force if you think of $F_{B}$ as the input. If the inner gears have one tenth the radius of the outer gears then the multiplication or attenuation is a factor of 100.

Trains of gears can build up large net gear ratios. The ratio of the fastest to slowest gear in a common clock or mechanical watch is on the order of 10,000.

In some gear trains, like the example above, large torque amplification comes from a large ratio of concentrically welded gears. A large amplification can also come from differences rather than ratios. The designs based primarily on differences rather than ratios are called ‘differentials’, ‘harmonic drives’, or ‘planetary gears’.

Example: Planetary gear with a large ratio

fig. 7.54 shows a gear design where the ratio of the input torque on the drive gear, to the output torque, on the spider can be huge. In particular, for the design shown, the torque ratio is approximately:

\frac{M_{out}}{M_{in}}\approx\frac{2}{R_{D}/R_{R}-1}

where $R_{D}$ is the ratio of the outer drive-gear to inner drive-gear radius and $R_{R}$ is the ratio of the outer ring-gear to inner ring gear radius. Thus if the inner and outer drive gears have 49 and 50 teeth, respectively, and the inside and outside of the ring gear have 50 and 51 teeth, then the torque multiplication is nearly 5000. (See homework 7.86).

Pulleys

We have already studied a pulley as a single object (see page 5.2). Now we show, as you probably have learned a few times before in school, how to use pulleys to amplify or attenuate force. We assume pulleys are round, massless, and have frictionless bearings.

The key fact for statics analysis is

For an ideal round pulley with negligible mass (or negligible angular acceleration) the tension on the cable is the same on both sides of the pulley:

T_{1}=T_{2}

The classic problem is shown in fig. 7.53a where you would like to use a pulley to make the task easier. Figures 7.53b-d show three possible uses of pulleys. If, at a glance, you can’t see that these three designs are quite different in their effects you should puzzle them out slowly now.

Because the two tensions in the rope that wraps around the pulley are the same on both sides, the central rope has twice the tension. Design (b) gives no mechanical advantage but does allow one to pull down in order to lift the weight. Design (c) halves the required pulling force. Design (d), which might look superficially similar to (c) doubles the required pulling force, requiring 4 times the force of (c).

By using pulleys in combination, one can get various force attenuations and gains. The 10-pulley design in fig. 7.55 multiplies the force by about 1000.

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Figure 7.53: Simple pulley arrangements. a) Lifting a weight with no pulley, b) a pulley lets you pull down instead of up, c) a pulley halves the needed pull, d) a pulley doubles the needed pull.

Filename:pfigure-blue-127-2

Figure 7.54: A high-gain planetary gear system, so named because the ‘planet’ gears go around the ‘sun’ gear. Often planetary gears depend on the sun or the spider or the ring not rotating. In this design, all three rotate. A CCW torque is applied to the inner drive gear which is welded to the outer drive gear (they are concentric and have nearly the same radius). The outer drive gear turns the ring gear CW. The inner drive gear turns the connection gear CW which turns the sun gear CCW. The three planet gears spin between the inside of the ring gear and the outside of the sun gear. If the inside of the ring gear moves a little faster down than the outside of the sun gear moves up, then the spider turns CW with the center of the planet gears. As drawn, this motion would be much slower than the motion of the drive gears.

Filename:pfigure-s94h13p4

Figure 7.55: Archimedes pulley. A pulley arrangement sometimes commonly attributed to Archimedes. With 10 pulleys, he could lift a weight of

W

with a pull of about

T=2^{-10}W\approx W/1000

. With 20 pulleys the force amplification is by more than a million. With 67 pulleys he, assuming he was as strong as the average person, could hoist a rock the size of the moon.

General force amplification concepts

If a mechanism generates a large force ratio (output/input), this usually corresponds to a large ratio in some geometric quantities. For a lever we have the ratio of two lever arms. For a wedge, the small wedge angle, and for a toggle also a small angle.

More precisely

For a frictionless transmission, the ratio of the input force to the output force is the reciprocal of the ratio of input motion to output motion.

For a high-gain lever, the handle moves much further than the load. For a narrow wedge, the slip distance is much bigger than the spreading distance. For a toggle the motion of the compressed end is much smaller than that of the applied load. That the force amplification is identical to the motion attenuation follows from energy conservation: the work into the mechanism equals the work out.

So, when Archimedes pulls in a kilometer of rope while lifting a rock, the moon, with 67 pulleys, that huge rock moves $2^{-67}$ km or about one hundred millionth of a nanometer. A big force, and a small motion. This is as it must be, because Archimedes only supplies so much work. If the transmission greatly amplifies force it must greatly attenuate motion.

Similarly with levers. When Archimedes famously said:

Give me a lever long enough and a fulcrum on which to place it, and I shall move the world,

he was careful not to say how far he would move it. People might have been less impressed if they knew that the distance of movement was far below the resolving power of all past, present and future microscopes.

SAMPLE 7.6

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Figure 7.56:

A wheeled suitcase of length 60 cm, height 30 cm and ‘weighing’ 20 $\,\mathrm{kg}$ on the airport check-in counter, has a telescopic handle of length 40 cm. The suitcase is dragged at an angle $\theta=30^{\circ}$ . Assuming good wheels (negligible friction), find the force applied on the handle in order to wheel the suitcase steadily. (Take $g\approx 10\,\mathrm{m}/\mathrm{s}^{2}$ ).

Solution

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Figure 7.57: Free-body diagram of the suitcase. Unit vectors

\hat{\boldsymbol{\lambda}}

and

\hat{\boldsymbol{n}}

are along the length and height of the suitcase, respectively. Thus,

\hat{\boldsymbol{\lambda}}=\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}}

and

\hat{\boldsymbol{n}}=-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{% \boldsymbol{\jmath}}

The free-body diagram of the suitcase is shown in fig. 7.57. The reaction force at the wheel is almost vertical because of negligible friction. So, we can also assume the force $F$ applied at the handle to be almost vertical. We assume that the center of mass G is located at the geometric center of the rectangular suitcase. Now the moment-balance equation about point A, $\sum M_{A}=0$ , gives

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}\times(-mg\hat{% \boldsymbol{\jmath}})+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm% {B/A}}\times F\hat{\boldsymbol{\jmath}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}.

Substituting ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}=\frac{\ell_{1}}% {2}\hat{\boldsymbol{\lambda}}+\frac{h}{2}\hat{\boldsymbol{n}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}=(\ell_{1}+\ell_% {2})\hat{\boldsymbol{\lambda}}$ , and noting that $\hat{\boldsymbol{\lambda}}\times\hat{\boldsymbol{\jmath}}=\cos\theta\hat{% \boldsymbol{k}}$ and $\hat{\boldsymbol{n}}\times\hat{\boldsymbol{\jmath}}=-\sin\theta\hat{% \boldsymbol{k}}$ , we have

	$\displaystyle-\frac{1}{2}mg(\ell_{1}\cos\theta-h\sin\theta)\hat{\boldsymbol{k}% }+F(\ell_{1}+\ell_{2})\cos\theta\hat{\boldsymbol{k}}=0$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F=\frac{1}{2}mg\left[% \frac{\ell_{1}}{\ell_{1}+\ell_{2}}-\frac{h}{\ell_{1}+\ell_{2}}\tan\theta\right].$

Substituting $\ell_{1}=60\,\mathrm{cm},\ell_{2}=40\,\mathrm{cm},h=30\,\mathrm{cm},\theta=30^% {\circ}$ and $mg=200\,{\mbox{N}}$ , we get

F=\frac{1}{2}\cdot 200\,{\mbox{N}}\left[\frac{60\,\mathrm{cm}}{100\,\mathrm{cm% }}-\frac{30\,\mathrm{cm}}{100\,\mathrm{cm}}\tan 30^{\circ}\right]=42.7\,{\mbox% {N}}.

Answer: $F=42.7\,{\mbox{N}}$

SAMPLE 7.7

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Figure 7.58: Toggle mechanism.

The figure shows a basic toggle mechanism. (a) If the applied force is $P=20\,{\mbox{N}}$ and the mechanism is in equilibrium at $\theta=5^{\circ}$ , find the force applied by the spring. (b) If doubling the load P (to $P=40\,{\mbox{N}}$ ) causes a decrease of $\theta$ by $1^{\circ}$ (to $\theta=4^{\circ}$ ), does the spring force at C double too?

Solution

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Figure 7.59: Free-body diagrams.

The free-body diagrams of the pin connecting the two rods and the rod BC are shown in fig. 7.59. From the static equilibrium of the pin B, we have

	$\displaystyle\sum F_{x}=0$	$\Rightarrow$	$\displaystyle T_{2}\cos\theta-T_{1}\cos\theta=0\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}T_{1}=T_{2}$
	$\displaystyle\sum F_{y}=0$	$\Rightarrow$	$\displaystyle-(T_{1}+T_{2})\sin\theta-P=0\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}T_{2}=-\frac{P}{2\sin\theta}.$

which follows from setting $T_{1}+T_{2}=2T_{2}$ since $T_{1}=T_{2}$ . Now, we consider the free-body diagram of rod BC. The force-balance equation in the $x$ -direction ( $\sum F_{x}=0$ ) gives

-T_{2}\cos\theta-F=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F=-T_{2}\cos% \theta=\frac{P\cos\theta}{2\sin\theta}.

Since $\theta$ is small, we have $\sin\theta\approx\theta$ and $\cos\theta\approx 1$ . Thus $F=P/2\theta$ where $\theta$ is in radians. Substituting $P=20\,{\mbox{N}}$ and $\theta=5\pi/180$ , we get

F=\frac{20\,{\mbox{N}}}{2\pi/36}=115\,{\mbox{N}}

which is almost 6 times $P$ .

(b) If $P$ is doubled, we might expect $F$ to double because $F\approx P/2\theta$ . But if $\theta$ also decreases to $\theta=4^{\circ}$ , repeating the calculation above with $P=40\,{\mbox{N}}$ , and $\theta=4^{\circ}$ we get $F=286\,{\mbox{N}}$ which is 2.5 times the previous spring force.

Answer: $\mbox{For }P=20\,{\mbox{N}},\theta=5^{\circ},F=115\,{\mbox{N}};\mbox{and for }% P=40\,{\mbox{N}},\theta=4^{\circ},F=286\,{\mbox{N}}$

SAMPLE 7.8

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Figure 7.60:

A gear train: In the compound gear train shown in the figure, the various gear radii are: $R_{A}=10\,\mathrm{cm},R_{B}=4\,\mathrm{cm},R_{C}=8\,\mathrm{cm}$ and $R_{D}=5\,\mathrm{cm}$ . The input load $F_{i}=50\,{\mbox{N}}$ . Assuming the gears to be in static equilibrium find the machine load $F_{o}$ .

Solution You may be tempted to think that a free-body diagram of the entire gear train will do since we only need to find $F_{o}$ . However, it is not so because there are unknown reactions at the axle of each gear and, therefore, there are too many unknowns. On the other hand, we can find the load $F_{o}$ easily if we go gear by gear from the left to the right.

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Figure 7.61:

The free-body diagram of gear A is shown in Fig. 7.61. Let $F_{1}$ be the force at the contact tooth of gear A that meshes with gear B. From the moment balance about the axle-center O, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{O}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , we have

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{M}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{N}}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle-F_{i}R_{A}\hat{\boldsymbol{k}}+F_{1}R_{A}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F_{1}$	$\displaystyle=$	$\displaystyle F_{i}.$

Similarly, from the free-body diagram of gear B and C (together) we can write the moment balance equation about the axle-center P as

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Figure 7.62:

$\displaystyle F_{1}R_{B}\hat{\boldsymbol{k}}+F_{2}R_{C}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F_{2}$	$\displaystyle=$	$\displaystyle\frac{R_{B}}{R_{C}}F_{1}$
	$\displaystyle=$	$\displaystyle\frac{R_{B}}{R_{C}}F_{i}.$

Finally, from the free-body diagram of the last gear D and the moment equilibrium about its center R, we get

$\displaystyle-F_{2}R_{D}\hat{\boldsymbol{k}}+F_{o}R_{D}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F_{o}$	$\displaystyle=$	$\displaystyle F_{2}$
	$\displaystyle=$	$\displaystyle\frac{R_{B}}{R_{C}}F_{i}$
	$\displaystyle=$	$\displaystyle\frac{4\,\mathrm{cm}}{8\,\mathrm{cm}}\cdot 50\,{\mbox{N}}=25\,{% \mbox{N}}.$

Answer: $F_{o}=25\,{\mbox{N}}$

SAMPLE 7.9

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Figure 7.63:

Find the force $F$ to hold the 100 kg box shown in the figure in equilibrium. Assume $g\approx 10\,\mathrm{m}/\mathrm{s}^{2}$ .

Solution

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Figure 7.64: Free-body diagram of the two pulleys. We use C and D to designate both the pulleys and their respective centers.

The free-body diagrams of the two pulleys are shown in fig. 7.64 where the tension in the rope running over the two pulleys has been assumed as $T$ . For the lower pulley D, the force balance in the $y$ -direction, $\sum F_{y}=0$ , requires

2T-mg=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T=\frac{mg}{2}.

The free-body diagram of the upper pulley C contains an unknown reaction force $R$ at the attachment point C. However, if we write moment balance about point C, $\sum M_{C}=0$ , this unknown force contributes nothing. Let the radius of pulley C be $r$ . Thus, the moment-balance equation about C gives

	$\displaystyle Tr-Fr$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F$	$\displaystyle=$	$\displaystyle T=\frac{mg}{2}\approx 500\,{\mbox{N}}.$

Answer: $F\approx 500\,{\mbox{N}}$

SAMPLE 7.10

Filename:sfig4-6-4a

Figure 7.65:

A container box weighing 1 kN is dragged slowly and steadily along the floor with force $F$ as shown in the figure. The coefficient of friction between the box and the floor is 0.6. Find the force required to pull the box and the force amplification obtained by the pulley arrangement.

Solution

Filename:sfig4-6-3

Figure 7.66:

It is clear from the figure that the same rope passes over the two pulleys used in the arrangement to pull the box. Let the tension in the rope be $T$ . A partial free-body diagram (that includes forces acting only in the $x$ -direction) of the box along with the pulley attached to it is shown in fig. 7.66. The same figure also shows the free-body diagram of pulley A at the force end. From the force-balance equation for the box in the $x$ -direction, we get

f-3T=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T=\frac{f}{3}=\frac{\mu mg}{3}.

Now, from the force balance of pulley A in the $x$ -direction, we get

$\displaystyle 2T-F$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F$	$\displaystyle=$	$\displaystyle 2T=\frac{2\mu mg}{3}$
	$\displaystyle=$	$\displaystyle\frac{2\cdot(0.6)\cdot(1\,{\rm kN})}{3}=400\,{\mbox{N}}.$

Since the force of friction on the box while sliding is $f=\mu mg=0.6(1\,{\rm kN})=600\,{\mbox{N}}$ and the force applied at A to overcome this friction is $400\,{\mbox{N}}$ , the force amplification is 1.5. That is, the pulley arrangement amplifies the input force ( $400\,{\mbox{N}}$ ) 1.5 times at the output end.

Answer: $F=400\,{\mbox{N}},\mbox{ Force amplification}=1.5$

SAMPLE 7.11

Filename:sfig4-6-3a

Figure 7.67:

A differential hoist is used to lift a crate of mass 500 kg. The hoist pulley uses two discs of radius 30 cm and 25 cm. Find the force $F$ required to lift the crate steadily. Take $g\approx 10\,\mathrm{m}/\mathrm{s}^{2}$ .

Solution

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Figure 7.68:

The free-body diagrams of the upper pulley and the lower pulley are shown in fig. 7.68. Since the lower pulley is slightly smaller than the upper pulley, the chain passing over the two pulleys is not exactly vertical but makes a small angle with the vertical. Thus the tension forces shown in the free-body diagrams are slightly off from the vertical direction. However, since the angle is very small, we can treat $T$ to be essentially vertical.

For the lower pulley, the force balance in the $y$ direction gives

	$\displaystyle 2T-mg$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T$	$\displaystyle=$	$\displaystyle\frac{mg}{2}.$

Now the moment balance about point C, $\sum M_{C}=0$ , for the upper pulley gives

$\displaystyle Fr_{o}+Tr_{i}-Tr_{o}$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F$	$\displaystyle=$	$\displaystyle\left(\frac{r_{o}-r_{i}}{r_{o}}\right)T$
	$\displaystyle=$	$\displaystyle\left(1-\frac{r_{i}}{r_{o}}\right)\frac{mg}{2}$
	$\displaystyle=$	$\displaystyle\left(1-\frac{25\,\mathrm{cm}}{30\,\mathrm{cm}}\right)\frac{5000% \,{\mbox{N}}}{2}$
	$\displaystyle=$	$\displaystyle 417\,{\mbox{N}}$

Thus the force amplification in this case is about 12 (5000 N/417 N). From the analysis above, it is also clear that the ratio of the radii of the two disks used in the upper pulley decides this force amplification. One can get a big force amplification, at least theoretically, by making $r_{i}\approx r_{o}$ . In this problem, for example, if $r_{i}=29\,\mathrm{cm}$ rather than the given $25\,\mathrm{cm}$ , we get $r_{i}/r_{o}\approx 0.97$ giving $F\approx 83\,{\mbox{N}}$ which corresponds to a force amplification of 60.

Answer: $F=417\,{\mbox{N}}$

SAMPLE 7.12

A wedge with friction.

Filename:pfigure-blue-38-2

Figure 7.69: A wedge with friction

\mu

between the blocks. Treat the walls as frictionless.

Consider the wedge described in the text (page 7.2), but now with friction $\mu$ between the blocks.

Consideration of friction qualitatively changes the behavior of the machine. For simplicity still take the wall and floor interactions to be frictionless.

1.

What is the relation between $F_{A}$ and $F_{B}$ when block A is sliding down?
2.

What is the relation between $F_{A}$ and $F_{B}$ when block A is sliding up?
3.

Under what conditions is it impossible for $F_{B}$ to slide block A up, even when $F_{A}$ is vanishingly small? Such a case is called ‘non-back-driveable’ or ‘self-locking’.

How do your answers simplify for small wedge angle $\theta$ and small friction angle $\phi$ (with $\tan\phi=\mu$ ).

Solution

Filename:pfigure-blue-49-2

Figure 7.70: Free-body diagrams of a wedge a) assuming A slides down, b) assuming A slides up

Figure 7.70 shows free-body diagrams of wedge blocks. We draw separate free-body diagrams for the case when (a) block A is sliding down and block B to the right, and (b) block A is sliding up and block B to the left. In both cases the friction resists relative slip and obeys the sliding friction relation

F_{f}=\underbrace{\tan\phi}_{\mu}N,

where fig. 7.70 shows the resultant contact force (normal component plus frictional component) and its angle $\phi$ to the surface normal.

1.

Block A sliding down: Assuming block A is sliding down we get from free-body diagram 7.70a that

$\begin{array}[]{lccc}\mbox{For block A},\left\{\sum{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{\!\!\mathrm{i}}={\accentset{\rightharpoonup}{\boldsymbol{0% }}}\right\}\cdot\hat{\boldsymbol{\jmath}}&\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}&-F_{A}+F\sin(\theta+\phi)&=0\\ \mbox{For block B},\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{% \!\!\mathrm{i}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\right\}\cdot\hat% {\boldsymbol{\imath}}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&-F_{B}+F\cos(% \theta+\phi)&=0.\\ \end{array}$

Eliminating $F$ we get,

$F_{B}={1\over\tan(\theta+\phi)}F_{A}.$ (7.15)

Answer: When A slides down $\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\displaystyle F_{B}={1\over\tan(% \theta+\phi)}F_{A}.$

If we take a taper of $6^{{}^{\circ}}$ and a friction coefficient of $\mu=.3$ ( $\Rightarrow\phi\approx 17^{{}^{\circ}}$ ) we get that $F_{B}/F_{A}\approx 2.5$ instead of 10 as we got when neglecting friction. The wedge still serves as a way to multiply force, but substantially less so than the frictionless idealization led us to believe.
2.

Block A sliding up: Now let’s consider the case when force $F_{B}$ pushes block B to the left, pinching block A, and forcing it up. The only change in the calculation is the change in the direction of the friction interaction force. From free-body diagram 7.70b

$\begin{array}[]{lccc}\mbox{For block A},\left\{\sum{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{\!\!\mathrm{i}}={\accentset{\rightharpoonup}{\boldsymbol{0% }}}\right\}\cdot\hat{\boldsymbol{\jmath}}&\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}&-F_{A}+F\sin(\theta-\phi)&=0\\ \mbox{For block B},\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{% \!\!\mathrm{i}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\right\}\cdot\hat% {\boldsymbol{\imath}}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&-F_{B}+F\cos(% \theta-\phi)&=0\\ \end{array}$

Eliminating $F$ we get,

$F_{A}=\tan(\theta-\phi)F_{B}.$ (7.16)

Answer: When A slides up $\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\displaystyle F_{B}=F_{A}/{\tan(% \theta-\phi)}.$

Again using $\theta=6^{{}^{\circ}}$ and $\phi=17^{{}^{\circ}}$ we see that if $F_{B}=100\,{\rm lbf}$ then $F_{A}=\tan(-11^{{}^{\circ}})\cdot 100\,{\rm lbf}\approx-20\,{\rm lbf}$ . That is, the 100 pounds doesn’t push block A up at all, but even with no gravity, you need to pull up with a 20 pound force to get it to move.

If we insist that the downwards force $F_{A}$ is positive or zero, that the pushing force $F_{B}$ is positive, and that block A is sliding up then there is no solution to the equilibrium equations whenever $\phi>\theta$ .

(Actually we didn’t need to do this second calculation at all. Eqn 7.15 shows the same paradox when $\theta+\phi>90^{{}^{\circ}}$ . Trying to squeeze block B to the right for large $\theta$ is exactly like trying to squeeze block A up for small $\theta$ .)
3.

Self-locking: This self-locking situation is intuitive. In fact it’s hard to picture the contrary, that pushing a block like B would lift block A. If you view this wedge mechanism as a transmission, it is said to be non-back-drivable whenever $\phi>\theta$ . Even though pushing down on A can ‘drive’ block B to the right, pushing to the left on block B cannot ‘back-drive’ block A up. Non-backdrivability is a feature or a defect depending on context^†^†margin: Standard car transmissions are backdriven when they are push-started and when a driver downshifts to slow the car instead of using the brakes. On the other hand, most electric hand-mixers cannot be backdriven; you can’t turn the motor by forcing the beater blade (Unplug before trying.) .

The borderline case of backdrivability is when $\theta=\phi$ and $F_{B}=F_{A}/\tan 2\theta$ . Assuming $\theta$ is a fairly small angle we get

$\displaystyle F_{B}$ $\displaystyle={F_{A}\over\tan{2\theta}}\approx{F_{A}\over 2\theta}\approx{1% \over 2}{F_{A}\over\tan\theta}$

$\displaystyle\approx{1\over 2}\cdot\mbox{(the value of $F_{B}$ had there been % no friction).}$

Thus the design guideline:

Non-back-driveable transmissions are generally 50% or less efficient, they transmit 50% or less of the force they would transmit if they were frictionless.

To use a wedge in this backwards way requires very low friction. A rare case where a narrow wedge is back drivable is this: a fresh wet watermelon seed squeezed between two pinched fingers.

Problems for 7.2 Levers, Wedges, Toggles, Gears and Pulleys

7.2.1 A suitcase with bad wheels has length $\ell=0.5\,\mathrm{m}$ and thickness (height) of $\ell=0.25\,\mathrm{m}$ is pulled along steadily with a force $F=100\,{\mbox{N}}$ as shown in the figure.

(a)

Find the weight $W$ of the suitcase.
(b)

Find the ground force on the wheel (both magnitude and direction).
(c)

What is force amplification if you consider $F$ as the input and $W$ as the output.

Filename:pfigure4-4-rp13

Figure 7.71

7.2.2 A wheelbarrow containing $100\,\mathrm{kg}$ of this-n-that is wheeled steadily with a force $F$ as shown in the figure. For the given geometry and $g\approx 10\,\mathrm{m}/\mathrm{s}^{2}$ , find the required force $F$ .

Filename:f92h7p1

Figure 7.72

7.2.3 A bottle-opener ABC contains a cut-out AB of approximate diameter $2\,\mathrm{cm}$ that clamps on the bottle cap. The arm BC is approximately $15\,\mathrm{cm}$ long. If the cap is opened by applying a vertical force $F=10\,{\mbox{N}}$ at C, find the force on the cap at B.

Filename:pfigure-blue-52-2

Figure 7.73

7.2.4 A cut-out view of a garlic press is shown in the figure. For an input force $F_{i}=10\,{\rm lbf}$ , find the output force $F_{o}$ at the site of the press. What is the force amplification?

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Figure 7.74

7.2.5 Assuming all frictionless contacts, find the force $F$ on the wedge required to lift the sphere weighing $500\,{\mbox{N}}$ if the wedge angle $\theta=10^{\circ}$ .

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Figure 7.75

7.2.6 A cutter, shown in the figure, uses a toggle mechanism BCD to get a big force amplification at the cutting edge. A partial free-body diagram of one of the arms of the cutter is shown in the figure. Assuming an input force of $F_{i}=20\,{\mbox{N}}$ at A, find the intermediate output force $F_{o}$ at C when

(a)

$\theta=30^{\circ}$ ,
(b)

$\theta=10^{\circ}$ .

Filename:pfigure4-4-rp15

Figure 7.76

7.2.7 A toggle-like mechanism is used in a folding chair shown in the pictures here. The metallic link DB gets almost parallel to the seating plank AC when the chair is open. Given the dimensions $d_{1}=30\,\mathrm{cm},d_{2}=10\,\mathrm{cm},\delta=2\,\mathrm{cm}$ and the force at A, $F_{i}=500\,{\mbox{N}}$ , find the force in the link DB. Why is this force so big or small? In practice why would you never see such a large force? Answer: The force is so big because of the toggle mechanism: because DB is nearly parallel to AC, to balance moments about C the force in DB has to be huge. Look at the leg whose top is at D. That leg could only have a big force on it at D if the bottom of the leg was restrained. So the big force would only occur if there was, say, a cable connecting the bottoms of the two legs.

Filename:tfigure4-spherical-rotaxis

Figure 7.77

7.2.8 A gear of radius $250\,{\rm mm}$ is meshed in with a rack that carries a horizontal load $F=50\,{\mbox{N}}$ . Find the torque $M$ on the gear that is required for equilibrium.

Filename:tfigure5-7

Figure 7.78

7.2.9 The input gear A of radius $r_{A}=10\,\mathrm{cm}$ drives gear B that is one and a half times bigger than gear A. Gear B, in turn, drives a rack. If the input torque on gear A is $M_{i}=30\,{\rm N}\mathord{\cdot}{\rm m}$ , find the load $F$ on the rack.

Filename:tfigure5-gen-rigid-body

Figure 7.79

7.2.10 In the gear arrangement shown, gears $G_{1}$ and $G_{2}$ are welded together. The output gear $G_{3}$ is one third the size of gear $G_{2}$ .

(a)

Is this gear train for torque amplification or for torque reduction?
(b)

If the input torque $M_{\rm in}$ on gear $G_{1}$ is $300\,{\rm N}\mathord{\cdot}{\rm m}$ , find the output torque $M_{\rm out}$ .

Filename:tfigure5-term1-a

Figure 7.80

7.2.11 At the input to a gear box, a $100\,{\rm lbf}$ force is applied to gear A. At the output, the machinery (not shown) applies a force of $F_{B}$ to the output gear. Assume the system of gears is at rest. What is $F_{B}$ ?

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Figure 7.81

7.2.12 A $100\,{\rm lbf}$ force is applied to one rack. At the output, the machinery (not shown) applies a force of $F_{B}$ to the other rack. Assume the gear-train is at rest. What is $F_{B}$ ?

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Figure 7.82

7.2.13 The gear train and spindle shown in the figure are used for hoisting heavy loads. For the dimensions given, if the load $F=2\,{\rm kN}$ , find the torque $M$ that the motor A must apply for equilibrium.

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Figure 7.83

7.2.14 The figure shows a brush gear (also called a crown wheel) where wheel $\cal C$ of radius $r_{o}$ , rolls on the surface of wheel $\cal D$ without slipping. In addition, the position $r$ of wheel $\cal C$ from the center of wheel $D$ can be varied. Let the input torque on wheel $\cal D$ be $M_{i}$ .

(a)

Find the output torque $M_{o}$ on wheel $\cal C$ as a function of $r$ .
(b)

Find the output torque $M_{o}$ when $r=r_{o}$ and when $r=r_{o}/4$ .
(c)

If the output torque were not to exceed 100 times the input torque, where will you put safety latches on the axle of wheel $\cal C$ ?

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Figure 7.84

7.2.15 For the gear train shown in the figure, find the torque amplification $M_{\rm out}/M_{\rm in}$ .

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Figure 7.85

7.2.16 A torque amplifying planetary gear is shown in the figure where the sun-gear is free to rotate but the ring-gear is fixed. The sun-gear drives five planet-gears that drive the spider-gear through their axles housed in bearings in the spider. The radius of the planet-gears $r_{p}=50\,{\rm mm}$ and the radius of the sun-gear is twice as big. If the input torque on the sun-gear is $2000\,{\rm N}\mathord{\cdot}{\rm m}$ , find the output torque on the spider.

Filename:sfig4-6-5c

Figure 7.86

7.2.17 Consider the high gear ratio planetary gear discussed on page 7.2 of the text. Let $r_{D_{i}}$ and $r_{D_{o}}$ be the inner and outer radii, respectively, of the drive gear, and $r_{R_{i}}$ and $r_{R_{o}}$ be the inner and outer radii, respectively, of the ring-gear. Let $r_{s}$ and $r_{p}$ denote the radii of the sun and the planet-gear respectively. Show that the ratio of the output torque $M_{\rm out}$ on the spider-gear to the input torque $M_{\rm in}$ on the drive gear is approximately given by

\frac{M_{\rm out}}{M_{\rm in}}\approx\frac{2}{\frac{R_{D}}{R_{R}}-1}

where $R_{D}=r_{D_{i}}/r_{D_{o}}$ and $R_{R}=r_{R_{i}}/r_{R_{o}}$ .

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Figure 7.87

7.2.18 A force $F=100\,{\mbox{N}}$ acts at A. The pulleys are frictionless. Find the force on the box applied by the pulley. Answer: $200\,{\mbox{N}}$

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Figure 7.88

7.2.19 A force $F$ is applied as shown in the pulley arrangements shown in (a) and (b). Which arrangement gives a bigger force amplification on the box?

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Figure 7.89

7.2.20 A weight $W$ is held in place with a force $F=100\,{\mbox{N}}$ applied through a massless pulley as shown in the figure. The pulley is attached to a rod AB which, in turn, is held horizontal with the help of a string CB. Find the tension (or compression) in rod AB.

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Figure 7.90

7.2.21 Given $W$ and the frictionless pulleys shown find the tension $T$ needed to lift the weight in the situations shown.

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Figure 7.91

7.2.22 In the two cases shown in (a) and (b), find the maximum force $F$ that can be applied before the box starts skidding on the ground. Take $m=50\,\mathrm{kg}$ and $g\approx 10\,\mathrm{m}/\mathrm{s}^{2}$ . Which arrangement requires smaller force and why?

Filename:sfig4-6-8b

Figure 7.92

7.2.23 The pulley arrangement shown in the figure uses a spring EG of stiffness $k=200\,{\rm N}/{\rm cm}$ . If the spring is stretched by $1.5\,\mathrm{cm}$ under the application of force $F$ for equilibrium, find $F$ .

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Figure 7.93

7.2.24 Find the force on the mass at A in terms of $F$ and thus find the force amplification provided by the pulley arrangement used.

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Figure 7.94

7.2.25 In the figure shown, there is no friction between block A and the vertical wall but there is friction ( $\mu=0.3$ ) between block B and the floor. If $m_{B}=30\,\mathrm{kg}$ , find the mass of block A for equilibrium.

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Figure 7.95

7.2.26 Find the ratio of the masses $m_{1}$ and $m_{2}$ so that the system is at rest.

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Figure 7.96

7.2.27 If the mass and pulley system shown in the figure is in equilibrium when the spring is stretched by $3\,\mathrm{cm}$ , find $m$ , given $k=500\,{\rm N}/{\rm m}$ and $g\approx 10\,\mathrm{m}/\mathrm{s}^{2}$ .

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Figure 7.97

7.3 Mechanisms

We would now like to analyze ways to connect pieces so that a force or moment is amplified, attenuated or redirected.

For completeness, we present the statics recipe for machines, although it is an exact repeat of the recipe used for frames.

•
Draw free-body diagrams of
- –
  
  the whole machine; and
- –
  
  the separate parts of the machine; and
- –
  
  collections of parts of the machine, if such seem likely to be fruitful;
- –
  
  Use the principle of action and reaction in the free-body diagrams so that there is only one unknown force at a point where two bodies contact;
•

for each free-body diagram write equilibrium conditions. These should yield three independent scalar equations for each non-point part (in 2D)
•

solve some or all of the equilibrium equations for the desired unknowns

Some useful tricks and shortcuts include:

•

for any two-force bodies, assign an equal-valued tension to each end (thus eliminating any need or use for equilibrium equations for that object)
•
To minimize calculation, look for a subset of the equilibrium equations that
- –
  
  contains your unknowns of interest, and
- –
  
  has as many unknowns as scalar equations, and
- –
  
  contains as few equations as possible.

Example: Stamp machine

Pulling on the handle (below) causes the stamp arm to press down with a force $N$ at D. We can find $N$ in terms of $F_{h}$ by drawing free-body diagrams of the handle and stamp arm, writing three equilibrium equations for each piece and then solving these 6 equations for the 6 unknowns ( $A_{x}$ , $A_{y}$ , $F_{C}$ , $N$ , $B_{x}$ , and $B_{y}$ ).

Filename:tfigure-stampmachine

For this problem, the answer can be found more quickly with a judicious choice of equilibrium equations.

\begin{array}[]{lccc}\mbox{For the handle},\left\{\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/B}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}\right\}\cdot\hat{\boldsymbol{k}}&\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}&-hF_{h}+dF_{c}&=0\\ \mbox{For the stamp arm},\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{M% }}}_{\mathrm{/A}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\right\}\cdot% \hat{\boldsymbol{k}}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&-(d+w)F_{c}+% \ell N&=0.\\ \end{array}

\mbox{Eliminating $F_{c}$}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}N={h(d+w)% \over d\ell}F_{h}.

Note that the stamp force $N$ can be made very large by making $d$ small and thus the handle nearly vertical. Often in structural or machine design, one or another force gets extremely large or small as the design is changed to put pieces in near alignment.

Example: Improved stamp machine

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Figure 7.98: An improved stamp machine.

Figure 7.98 shows a stamp machine with all the same components. The method of analysis is identical. However, the design represents an improvement in two ways:

•

The lever in the stamp arm amplifies rather than attenuates the stamp force.
•

In the previous design, it is harder and harder to generate a given stamp force as the stamped object compresses. In this design, the toggle mechanism associated with the lever arm and sliding pin is in compression. Thus, as the stamping progresses and the handle becomes more vertical, for a constant hand-force, the stamping force increases as the motion progresses.

Non-rigid structures are mechanisms

A non-rigid structure cannot carry all loads and, if not also redundant, has more equilibrium equations than unknown reaction or interaction force components. Such a structure is also called a mechanism. The stamp machine above is a mechanism if we assume there is no contact at D. In particular, the equilibrium equations cannot be satisfied unless $F_{h}=0$ . Mechanisms have variable configurations. That is, the constraints still allow relative motion.

An attempt to design a rigid structure that turns out to be a mechanism is a design failure. But for machine design, the mechanism aspect of a structure is essential. Even though mechanisms are called ‘statically indeterminate’ because they cannot carry all possible loads, the desired forces can often be determined using statics. For the stamp machine above, the equilibrium equations are made solvable by treating one of the applied forces, say $N$ , as an unknown, and the other, $F_{h}$ in this case, as a known. This is a common situation in machine design where you want to determine the loads at one part of a mechanism in terms of loads at another part. For the purposes of analysis, a trick is to make a mechanism determinate by putting a pin on the roller’s connection to the ground at the location of any forces with unknown magnitudes but known directions.

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Figure 7.99: Stamp machine with roller.

Example: Stamp machine with roller

Putting a roller at D, the location of the unknown stamp force, turns the stamp machine into a determinate structure.

Pulley and chain drives

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Figure 7.100: a) A chain or pulley drive involving two sprockets or pulleys and one chain or belt, b) free-body diagrams of each of the sprockets/pulleys.

Chain and pulley drives may be thought of as spread-out gears (fig. 7.100). The rotation of two shafts is coupled not by the contact of gear teeth but by a belt around a pulley or a chain around a sprocket. For a simple analysis, one draws free-body diagrams for each sprocket or pulley with a little bit of chain as in fig. 7.100b. Note that $T_{1}\neq T_{2}$ , unlike the case of an ideal undriven pulley. Applying moment balance we find,

\begin{array}[]{lccc}\mbox{For gear A},\left\{\sum{\accentset{\rightharpoonup}% {\boldsymbol{M}}}_{\mathrm{i/A}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}% \right\}\cdot\hat{\boldsymbol{k}}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&-R% _{\rm A}(T_{2}-T_{1})+M_{\rm A}&=0\\ \mbox{For gear B},\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{% \mathrm{i/B}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\right\}\cdot\hat{% \boldsymbol{k}}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&R_{\rm B}(T_{1}-T_{2% })-M_{\rm B}&=0.\\ \end{array}

\mbox{Eliminating $(T_{2}-T_{1})$}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}M_{% B}={R_{B}\over R_{A}}M_{A}\hbox to36.135pt{\hfil or\hfil}M_{A}={R_{A}\over R_{% B}}M_{B},

exactly as for a pair of gears. Note that we cannot find $T_{2}$ or $T_{1}$ but only their difference. Typically in design if, say, $M_{A}$ is positive, one would try to keep $T_{1}$ as small as possible without the belt slipping or the chain jumping teeth. If $T_{1}$ grows, then so must $T_{2}$ , to preserve their difference. This increase in tension increases the loads on the bearings as well as on the chain or belt itself.

Four-bar linkages

Four-bar linkages often, confusingly, have 3 bars, the fourth piece is something bigger that the linkage is attached to.

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Figure 7.101: Four bar linkages. a) A bicycle, thigh, calf, and crank, b) a door closer, c) a folding ladder, d) a generic mechanism, e) free-body diagrams of the parts of a generic mechanism.

A planar mechanism with four pieces connected in a loop by hinges is a four-bar linkage. Four-bar linkages are remarkably common. After a single object connected at a hinge (like a gear or lever), a four-bar linkage is one of the simplest mechanisms that can move in just one way (i.e., have just one degree of freedom).

A reasonable model of a person seated on a bicycle uses a 4-bar linkage (fig. 7.101a). The whole bicycle frame is one bar, the human thigh is the second, the calf is the third, and the bicycle crank is the fourth. The four hinges are the hip joint, the knee joint, the pedal axle, and the bearing at the bicycle crank axle. A more sophisticated model of the system would include the ankle joint and the foot would make up a fifth bar.

A standard door-closing mechanism is part of a 4-bar linkage (fig. 7.101b). The door jamb and door are two bars and the mechanism pieces make up the other two.

A standard folding ladder design is, until locked open, a 4-bar linkage (fig. 7.101c).

An abstracted 4-bar linkage with two loads is shown in fig. 7.101d with free-body diagrams in fig. 7.101e. If one of the applied loads is given, then the other applied load along with interaction and reaction forces compose nine unknown components (after using the principle of action and reaction). With three equilibrium equations for each of the three bars, all these unknowns can be found.

Slider crank

A mechanism closely related to a four-bar linkage is a slider crank (fig. 7.102a). An umbrella is one example (rotated $90^{{}^{\circ}}$ in fig. 7.102b). If the sliding part is replaced by a bar, as in fig. 7.102c, the point C moves in a circle instead of a straight line. If the height $h$ is very large, then the arc traversed by C is nearly a straight line so the motion of the four-bar linkage is almost the same as the slider crank. For this reason, slider cranks are sometimes regarded as a special case of a four-bar linkage in the limit as one of the bars gets infinitely long.

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Figure 7.102: a) a slider crank, b) an umbrella has a slider-crank mechanism, c) the equivalent four-bar linkage, at least when

h\rightarrow\infty

Box 7.6 Shears with gears

Many cutters, pliers and shears are essentially two levers pivoting against each other. For example these shears

Filename:tfigure-FiskarsBefore

consist of two levers, JAQ and KAP, pivoted at A. The hands squeeze the handles at J and K, causing a cutting force on an object between the blades at P and Q. The force at P, say, is $|KA|/|AP|$ times the force at K (from moment balance about A using a free-body diagram of KAP). Two possible deficiencies of this bi-lever design are that

•

One may want more mechanical advantage but not longer handles, and
•

For a given hand strength (available force at J and K) the force at the cutting edge gets less and less as the location of the cut force at P and Q moves farther out on the blade, away from A.

The Fiskars company, known mostly for scissors using the basic design above, has some designs that address these deficiencies. The loppers in problem 7.124 use a 3-piece mechanism to address these issues. Here, even more elaborately, are Fiskars shears using 4 moving parts.

Filename:tfigure-FiskarsShears

The two identical blades AP and AQ are hinged at A. The two identical handles JB and KC are hinged to the blades at B and C. Each handle also has gear teeth at the end that engage gear teeth on the opposite blade. Let’s take P and Q to be the point of contact of the object being cut.

Let’s try to understand the mechanism without detailed analysis (see homework problem 7.125).

To start, forget handle KC and assume that blade BAQ is held firmly by something outside. Blade CAP is attached at A about which it is free to spin. Handle JB is attached at B about which it is free to spin. But JB and CAP roll against each other with engaged gear teeth. So if handle JB rotates counter-clockwise about B, then CAP rotates clockwise about A.

Although the gear teeth are complex-looking, there is always an effective contact point G between handle JB and blade CAP on the line segment AB. G is effectively a hinge between JB and CAP. You can think of the handle as a lever with force points at J, B and G. Thus blade CAP is closed by the force on the gear teeth at G. The shorter BG, the bigger the forces at B and G.

Simultaneously you could think of blade CAP as fixed with blade BAQ and handle KC hinged to it and geared to each other at G’ (not shown). Thus blade CAP is also closed by an upwards force at C from handle KC. Similarly blade BAQ is closed by a downwards force at B from handle JB and a downwards force at G’ from handle KC.

The effective hinges G and G’ have locations which change as the blades close. When the blades are wide open, G and G’ are near A. When the blades are closed, G and G’ have moved to about the midpoint between B and A and C and A, respectively.

If G was at A, then this 4-piece design would be equivalent to a standard 2-piece cutter.

Because BG is shorter than BA, this design gives a bigger downwards force at B.

The shape of the geared curves makes the distance BG decrease, and the distance AG increase, as the blades close. Thus for given forces acting at J and Q, as the blades close the force at B increases, the force at G increases, and the lever-arm AG increases. These three effects partially compensate for the standard scissors problem, the decreasing mechanical advantage from the distance AP increasing as the blades close.

Another way to see the mechanical advantage of this design compared to the 2-piece design is to see that during a cut the handle angle decrease is greater than the blade angle decrease. Following the general rule for mechanisms, a motion attenuation is a force gain.

Filename:tfigure-FiskarsShears2

SAMPLE 7.13

Filename:pfigure-s94h6p4

Figure 7.103:

A slider crank: A torque $M=20\,{\rm N}\mathord{\cdot}{\rm m}$ is applied at the bearing end A of the crank AD of length $\ell=0.2\,\mathrm{m}$ . If the mechanism is in static equilibrium in the configuration shown, find the load $F$ on the piston.

Solution

Filename:pg84-3

Figure 7.104:

The free-body diagram of the whole mechanism is shown in Fig. 7.104. From the moment equilibrium about point A, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}\times({\accentset{% \rightharpoonup}{\boldsymbol{B}}}+{\accentset{\rightharpoonup}{\boldsymbol{F}}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle-M\hat{\boldsymbol{k}}+2\ell\cos\theta\hat{\boldsymbol{\imath}}% \times(B_{y}\hat{\boldsymbol{\jmath}}-F\hat{\boldsymbol{\imath}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle(-M+2B_{y}\ell\cos\theta)\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}B_{y}$	$\displaystyle=$	$\displaystyle\frac{M}{2\ell\cos\theta}.$

The force equilibrium, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ , gives

	$\displaystyle(A_{x}-F)\hat{\boldsymbol{\imath}}+(A_{y}+B_{y})\hat{\boldsymbol{% \jmath}}=0$
			$\displaystyle A_{x}=F$
			$\displaystyle A_{y}=-B_{y}$

Filename:pfigure-blue-36-1

Figure 7.105:

Note that we still need to find $F$ or $A_{x}$ . So far, we have had only three equations in four unknowns ( $A_{x},A_{y},B_{y},F$ ). To solve for the unknowns, we need one more equation. We now consider the free-body diagram of the mechanism without the crank, that is, the connecting rod DB and the piston BC together. See Fig. 7.105. Unfortunately, we introduce two more unknowns (the reactions) at D. However, we do not care about them. Therefore, we can write the moment equilibrium equation about point D, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{D}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ and get the required equation without involving $D_{x}$ and $D_{y}$ .

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/D}}% \times(-F\hat{\boldsymbol{\imath}}+B_{y}\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\ell(\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}})\times(-F\hat{\boldsymbol{\imath}}+B_{y}\hat{\boldsymbol{% \jmath}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle B_{y}\ell\cos\theta\hat{\boldsymbol{k}}-F\ell\sin\theta\hat{% \boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Dotting the last equation with ˆ k we get

$\displaystyle F$	$\displaystyle=$	$\displaystyle B_{y}\frac{\cos\theta}{\sin\theta}$
	$\displaystyle=$	$\displaystyle\frac{M}{2\ell\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}$
	$\displaystyle=$	$\displaystyle\frac{M}{2\ell\sin\theta}$
	$\displaystyle=$	$\displaystyle\frac{20\,{\rm N}\mathord{\cdot}{\rm m}}{2\cdot 0.2\,\mathrm{m}% \cdot\sqrt{3}/2}$
	$\displaystyle=$	$\displaystyle 57.74\,{\mbox{N}}.$

Answer: $F=57.74\,{\mbox{N}}$

Note that the force equilibrium carried out above is not really useful since we are not interested in finding the reactions at A. We did it above to show that just one free-body diagram of the whole mechanism was not sufficient to find $F$ . On the other hand, writing moment equations about A for the whole mechanism and about D for the connecting rod plus the piston is enough to determine $F$ .

SAMPLE 7.14

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Figure 7.106:

A flyball governor: A flyball governor, usually considered in a dynamics context, is shown in static equilibrium. The relaxed length of the spring is 0.15 $\,\mathrm{m}$ and its stiffness is 500 N/m.

1.

Find the static equilibrium position of the center collar.
2.

Find the force in the strut AB.
3.

How does the spring force required to hold the collar depend on $\theta$ ?

Solution Let $\ell_{0}(=0.15\,\mathrm{m})$ denote the relaxed length of the spring and let $\ell$ be the stretched length in the static equilibrium configuration of the flyball, i.e., the collar is at a distance $\ell$ from the fixed support EF. Then the net stretch in the spring is $\delta\equiv\Delta\ell=\ell-\ell_{0}$ . We need to determine $\ell$ , the spring force $k\delta$ , and its dependence on the angle $\theta$ of the ball-arm.

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Figure 7.107:

The free-body diagram of the collar is shown in fig. 7.107. Note that the struts AB and CD are two-force bodies (forces act only at the two end points on each strut). Therefore, the force at each end must act along the strut. From geometry (AB = BE = $d$ ), then, the strut force $F$ on the collar must act at angle $\theta$ from the vertical. Now, the force balance in the vertical direction, i.e., $[\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup% }{\boldsymbol{0}}}]\cdot\hat{\boldsymbol{\jmath}}$ , gives

-2F\cos\theta+k\delta=mg.

(7.17)

Filename:pfigure-blue-86-1

Figure 7.108:

Thus to find $\delta$ we need to find $F$ and $\theta$ . Now we draw the free-body diagram of arm EBG as shown in fig. 7.108. From the moment balance about point E, we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/E}}% \times(-2mg\hat{\boldsymbol{\jmath}})+{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{\!\mathrm{B/E}}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle 2d\hat{\boldsymbol{\lambda}}\times(-2mg\hat{\boldsymbol{\jmath}}% )+d\hat{\boldsymbol{\lambda}}\times F(-sin\theta\hat{\boldsymbol{\imath}}+\cos% \theta\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle-4mgd(\underbrace{\hat{\boldsymbol{\lambda}}\times\hat{% \boldsymbol{\jmath}}}_{-\sin\theta\hat{\boldsymbol{k}}})+Fd[-\sin\theta(% \underbrace{\hat{\boldsymbol{\lambda}}\times\hat{\boldsymbol{\imath}}}_{\cos% \theta\hat{\boldsymbol{k}}})+\cos\theta(\underbrace{\hat{\boldsymbol{\lambda}}% \times\hat{\boldsymbol{\jmath}}}_{-\sin\theta\hat{\boldsymbol{k}}})]$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle 4mgd\sin\theta\hat{\boldsymbol{k}}+Fd(-\sin\theta\cos\theta\hat{% \boldsymbol{k}}-\cos\theta\sin\theta\hat{\boldsymbol{k}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle(4mgd\sin\theta-2Fd\sin\theta\cos\theta)\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Dotting this equation with $\hat{\boldsymbol{k}}$ and assuming that $\theta\neq 0$ , we get

2F\cos\theta=4mg.

(7.18)

Substituting eqn. (7.18) in eqn. (7.17) we get

	$\displaystyle k\delta$	$\displaystyle=$	$\displaystyle mg+2F\cos\theta=mg+4mg=5mg$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\delta$	$\displaystyle=$	$\displaystyle\frac{5mg}{k}=\frac{5\cdot 2\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/% \mathrm{s}^{2}}{500\,{\rm N}/{\rm m}}=0.196\,\mathrm{m}.$

1.

The equilibrium configuration is specified by the stretched length $\ell$ of the spring (which specifies $\theta$ ). Thus,

$\ell=\ell_{0}+\delta=0.15\,\mathrm{m}+0.196\,\mathrm{m}=0.346\,\mathrm{m}.$

Now, from $\ell=2d\cos\theta$ , we find that $\theta=30.12^{\circ}$ .
2.

The force in strut AB is

$F=2mg/\cos\theta=45.36\,{\mbox{N}}.$
3.

The force in the spring $k\delta=5mg$ as shown above and thus, it does not depend on $\theta$ ! In fact, the angle $\theta$ is determined by the relaxed length of the spring.

Answer: $\mbox{(a) }\ell=0.346\,\mathrm{m},\quad\mbox{(b) }F=45.36\,{\mbox{N}},\quad% \mbox{(c) }k\delta\neq f(\theta)$

SAMPLE 7.15

Filename:pfigure-blue-88-1

Figure 7.109:

: A motor housing support: A slotted arm mechanism is used to support a motor housing that has a belt drive as shown in the figure. The motor housing is bolted to the arm at B and the arm is bolted to a solid support at A. The two bolts are tightened enough to be modeled as welded joints (i.e., they can also take some torque). Find the support reactions at A.

Solution

Filename:pfigure4-2-rp3

Figure 7.110:

Although the mechanism looks complicated, the problem is straightforward. We cut the bolt at A and draw the free-body diagram of the motor housing plus the slotted arm. Since the bolt, modeled as a welded joint, can take some torque, the unknowns at A are ${\accentset{\rightharpoonup}{\boldsymbol{A}}}(=A_{x}\hat{\boldsymbol{\imath}}+% A_{y}\hat{\boldsymbol{\jmath}})$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}$ . The free-body diagram is shown in fig. 7.110. Note that we have replaced the tension at the two belt ends by a single equivalent tension $2T$ acting at the center of the axle. Now taking moments about point A, we get

{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}\times 2{\accentset{% \rightharpoonup}{\boldsymbol{T}}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{\!\mathrm{G/A}}\times m{\accentset{\rightharpoonup}{\boldsymbol{g}}}={% \accentset{\rightharpoonup}{\boldsymbol{0}}}

where

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}% \times 2{\accentset{\rightharpoonup}{\boldsymbol{T}}}$	$\displaystyle=$	$\displaystyle(\ell\hat{\boldsymbol{\imath}}+h\hat{\boldsymbol{\jmath}})\times 2% T(-\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle 2T(\ell\sin\theta+h\cos\theta)\hat{\boldsymbol{k}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}% \times m{\accentset{\rightharpoonup}{\boldsymbol{g}}}$	$\displaystyle=$	$\displaystyle[(\ell+d)\hat{\boldsymbol{\imath}}+(\mbox{anything})\hat{% \boldsymbol{\jmath}}]\times(-mg\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-mg(\ell+d)\hat{\boldsymbol{k}}.$

Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}$	$\displaystyle=$	$\displaystyle-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}% \times 2{\accentset{\rightharpoonup}{\boldsymbol{T}}}-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}\times m{\accentset{% \rightharpoonup}{\boldsymbol{g}}}$
	$\displaystyle=$	$\displaystyle-2T(\ell\sin\theta+h\cos\theta)\hat{\boldsymbol{k}}+mg(\ell+d)% \hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle-2(5\,{\mbox{N}})(0.1\,\mathrm{m}\cdot\sin 60^{\circ}+0.04\,% \mathrm{m}\cdot\cos 60^{\circ})\hat{\boldsymbol{k}}$
		$\displaystyle{}+2\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot(0.1+0% .01)\,\mathrm{m}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle 1.092\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}.$

The reaction force ${\accentset{\rightharpoonup}{\boldsymbol{A}}}$ can be determined from the force balance, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ as follows.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{A}}}+2{\accentset{% \rightharpoonup}{\boldsymbol{T}}}+m{\accentset{\rightharpoonup}{\boldsymbol{g}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{A}}}$	$\displaystyle=$	$\displaystyle-2{\accentset{\rightharpoonup}{\boldsymbol{T}}}-m{\accentset{% \rightharpoonup}{\boldsymbol{g}}}$
	$\displaystyle=$	$\displaystyle-10\,{\mbox{N}}(-\frac{1}{2}\hat{\boldsymbol{\imath}}+\frac{\sqrt% {3}}{2}\hat{\boldsymbol{\jmath}})-(-19.62\,{\mbox{N}}\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle 5\,{\mbox{N}}\hat{\boldsymbol{\imath}}+10.96\,{\mbox{N}}\hat{% \boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}=1.092\,{\rm N}% \mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}\qquad\mbox{and}\qquad{\accentset{% \rightharpoonup}{\boldsymbol{A}}}=5\,{\mbox{N}}\hat{\boldsymbol{\imath}}+10.96% \,{\mbox{N}}\hat{\boldsymbol{\jmath}}$

SAMPLE 7.16

Filename:pfigure-s94q6p1

Figure 7.111:

Push-up mechanics: During push-ups, the body including the legs, usually moves as a single rigid unit; the ankle is almost locked, and the push-up is powered by the shoulder and the elbow muscles. A simple model of the body during push-ups is a four-bar linkage ABCDE shown in the figure. In this model, each link is a rigid rod, joint B is rigid (thus ABC can be taken as a single rigid rod), joints C, D, and E are hinges, but there is a motor at D that can supply torque. The weight of the person, $W=150\,{\rm lbf}$ , acts through G. Find the torque at D for $\theta_{1}=30^{\circ}$ and $\theta_{2}=45^{\circ}$ .

Solution

Filename:pfigure-spinningbrick

Figure 7.112:

The free-body diagram of part ABC of the mechanism is shown in fig. 7.112. Writing the moment-balance equation about point A, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , we get

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}\times{\accentset{% \rightharpoonup}{\boldsymbol{C}}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{\!\mathrm{G}}\times{\accentset{\rightharpoonup}{\boldsymbol{W}}}={% \accentset{\rightharpoonup}{\boldsymbol{0}}}.

Let ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}=r_{C_{x}}\hat{% \boldsymbol{\imath}}+r_{C_{y}}\hat{\boldsymbol{\jmath}}\qquad\mbox{and}\qquad{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G}}=r_{G_{x}}\hat{% \boldsymbol{\imath}}+r_{G_{y}}\hat{\boldsymbol{\jmath}}$ for now (we can figure it out later). Then, the moment equation becomes

	$\displaystyle(r_{C_{x}}\hat{\boldsymbol{\imath}}+r_{C_{y}}\hat{\boldsymbol{% \jmath}})\times(C_{x}\hat{\boldsymbol{\imath}}+C_{y}\hat{\boldsymbol{\jmath}})% +(r_{G_{x}}\hat{\boldsymbol{\imath}}+r_{G_{y}}\hat{\boldsymbol{\jmath}})\times% (-W\hat{\boldsymbol{\jmath}})={\accentset{\rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle{[}(C_{y}r_{C_{x}}-C_{x}r_{C_{y}})\hat{\boldsymbol{k}}-Wr_{G_{x}}% \hat{\boldsymbol{k}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}{]}$
	$\displaystyle\mbox{[~{}]}\cdot\hat{\boldsymbol{k}}\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}\qquad C_{y}r_{C_{x}}-C_{x}r_{C_{y}}=Wr_{G_{x}}.$		(7.19)

We now draw free-body diagrams of the links CD and DE separately (fig. 7.113) and write the moment- and force-balance equations for them.

For link CD, the force equilibrium $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ gives

Filename:pfigure-s94f1p2

Figure 7.113:

(-C_{x}+D_{x})\hat{\boldsymbol{\imath}}+(D_{y}-C_{y})\hat{\boldsymbol{\jmath}}% ={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

Dotting with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ gives
$\displaystyle\mbox{\hskip 108.405pt}D_{x}$ $\displaystyle=$ $\displaystyle C_{x}$ $\displaystyle D_{y}$ $\displaystyle=$ $\displaystyle C_{y}$
and the moment equilibrium about point D, gives

	$\displaystyle M\hat{\boldsymbol{k}}-a(\cos\theta_{2}\hat{\boldsymbol{\imath}}+% \sin\theta_{2}\hat{\boldsymbol{\jmath}})\times(-C_{x}\hat{\boldsymbol{\imath}}% -C_{y}\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle M\hat{\boldsymbol{k}}+(C_{y}a\cos\theta_{2}-C_{x}a\sin\theta_{2}% )\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$		(7.21)

Similarly, the force equilibrium for link DE requires that
$\displaystyle\mbox{\hskip 108.405pt}E_{x}$ $\displaystyle=$ $\displaystyle D_{x}$ $\displaystyle E_{y}$ $\displaystyle=$ $\displaystyle D_{y}$
and the moment equilibrium of link DE about point E gives

-M+D_{x}a\sin\theta_{1}+D_{y}a\cos\theta_{1}=0.

(7.23)

Now, from eqns. (LABEL:pushup.eq2) and (7.23)

-M+C_{x}a\sin\theta_{1}+C_{y}a\cos\theta_{1}=0.

(7.24)

Adding eqns. (7.21) and (7.24) and solving for $C_{x}$ , we get

C_{x}=\frac{\cos\theta_{1}+\cos\theta_{1}}{\sin\theta_{2}-\sin\theta_{1}}C_{y}.

For simplicity, let

f(\theta_{1},\theta_{2})=\frac{\cos\theta_{1}+\cos\theta_{1}}{\sin\theta_{2}-% \sin\theta_{1}}

so that

C_{x}=f(\theta_{1},\theta_{2})C_{y}.

(7.25)

Now substituting eqn. (7.25) in (7.19), we get

C_{y}=\frac{r_{G_{x}}}{r_{C_{x}}-r_{C_{y}}f}W.

Now substituting $C_{y}$ and $C_{x}$ into eqn. (7.24) we get

M=\frac{r_{G_{x}}a(\cos\theta_{1}+f\sin\theta_{1})}{r_{C_{x}}-r_{C_{y}}f}W

where

$\displaystyle r_{G_{x}}$	$\displaystyle=$	$\displaystyle(\ell/2)\cos\theta-h\sin\theta$
$\displaystyle r_{C_{x}}$	$\displaystyle=$	$\displaystyle\ell\cos\theta-h\sin\theta$
$\displaystyle r_{C_{y}}$	$\displaystyle=$	$\displaystyle\ell\sin\theta+h\cos\theta.$

Now plugging in all the given values: $W=160\,{\rm lbf},\theta_{1}=30^{\circ},\theta_{2}=45^{\circ},\ell=5\,\mathrm{% ft},h=1\,\mathrm{ft},a=1.5\,\mathrm{ft}$ , and, from simple geometry, $\theta=9.49^{\circ}$ ,

$\displaystyle f$	$\displaystyle=$	$\displaystyle 7.60$
$\displaystyle r_{C_{x}}$	$\displaystyle=$	$\displaystyle 4.77\,\mathrm{ft},$
$\displaystyle r_{C_{y}}$	$\displaystyle=$	$\displaystyle 1.81\,\mathrm{ft},$
$\displaystyle r_{G_{x}}$	$\displaystyle=$	$\displaystyle 2.30\,\mathrm{ft}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}M$	$\displaystyle=$	$\displaystyle-269\,{\rm lb}\mathord{\cdot}{\rm ft}.$

Answer: $M=-269\,{\rm lb}\mathord{\cdot}{\rm ft}$

SAMPLE 7.17

Filename:s97f2

Figure 7.114:

A spring and rod buckling model: A simple model of sideways buckling of a flexible (elastic) rod can be constructed with a spring and a rigid rod as shown in the figure. The solution comes out most beautifully if we assume that the rod is held in place with a zero-rest-length spring that has no tension when B is on top of A. If $P=0$ the system is in equilibrium with the rod vertical, $\theta=0$ . We find the so-called ‘buckling’ load by finding a load $P$ at which there are solutions with $\theta\neq 0$ . So, we assume the rod to be in static equilibrium at some angle $\theta\neq 0$ from the vertical. The vertical load is $P$ , spring stiffness $k$ , and bar length $\ell$ . For this problem, with a zero-rest-length spring we need not assume small angles.

Solution When the rod is displaced from its vertical position, the zero-rest-length spring gets stretched, i.e., , it is now under tension. The spring then exerts a force on the rod in the opposite direction of the tilt. The free-body diagram of the rod with a counterclockwise tilt $\theta$ is shown in Fig. 7.115.

Filename:pg92-2

Figure 7.115:

From the moment balance $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{O}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ (about the bottom support point O of the rod, which is also used as the origin for finding position vectors in the calculations below), we have

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{P}}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{\!\mathrm{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{s}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

Noting that

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}$	$\displaystyle=$	$\displaystyle\ell\hat{\boldsymbol{\lambda}},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{P}}}$	$\displaystyle=$	$\displaystyle-P\hat{\boldsymbol{\jmath}},$
$\displaystyle\qquad\mbox{and}\qquad{\accentset{\rightharpoonup}{\boldsymbol{F}% }}_{\!\!\mathrm{s}}$	$\displaystyle=$	$\displaystyle k({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}-{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}})\quad\mbox{(For a % zero-rest-length spring)}$
	$\displaystyle=$	$\displaystyle k(\ell\hat{\boldsymbol{\jmath}}-\ell\hat{\boldsymbol{\lambda}}),$

we get

	$\displaystyle\ell\hat{\boldsymbol{\lambda}}\times(P\hat{\boldsymbol{\jmath}})+% \ell\hat{\boldsymbol{\lambda}}\times k\ell(\hat{\boldsymbol{\jmath}}-\hat{% \boldsymbol{\lambda}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle-P\ell(\hat{\boldsymbol{\lambda}}\times\hat{\boldsymbol{\jmath}})% +k\ell^{2}(\hat{\boldsymbol{\lambda}}\times\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Dotting this equation with $(\hat{\boldsymbol{\lambda}}\times\hat{\boldsymbol{\jmath}})$ we get

	$\displaystyle-P\ell+k\ell^{2}$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}P$	$\displaystyle=$	$\displaystyle k\ell.$

Thus the equilibrium only requires that $P$ be equal to $k\ell$ and it is independent of $\theta$ ! That is, the system will be in static equilibrium at any $\theta$ as long as $P=k\ell$ .

Answer: If $P=k\ell$ , any $\theta$ is an equilibrium position. This is the ‘buckling’ load.

Problems for 7.3 Machines

7.3.1 A simply supported two bar mechanism supports a load of $200\,{\mbox{N}}$ at joint B with the help of a horizontal force $F$ applied at joint C. Find $F$ .

Filename:summer95p2-3

Figure 7.116

7.3.2 Pulling on the handle causes the stamp arm to press down at D. Neglect gravity and assume that the hinges at A and B, as well as the roller at C, are frictionless. Find the force $N$ that the stamp machine causes on the support at D in terms of some or all of $F_{h},w,d,\ell,h,$ and $s$ . Answer: $N=\left(h(w+d)/d\ell\right)F_{h}$

Filename:p-f96-p3-3

Figure 7.117

More-Involved Problems

7.3.3 See Problem 5.77 on page 5.77. A person who weighs $W$ stands on tiptoes on one foot. Assume the weight of the foot is negligible.

(a)

Draw a free-body diagram of the whole person and find the force of the ground on the foot front.
(b)

Draw a free-body diagram of the foot and find the force of the calf on the foot at the ankle and the tension in the Achilles tendon.

Filename:s97p3-3

Figure 7.118:

7.3.4 See Problem 5.77 on page 5.77. A person with weight $W=140\,{\rm lbf}$ has an upper body with weight $0.7W$ with center of mass at C. The back muscles are idealized as a single muscle with one end (the muscle origin also at C. Use the idealization and geometry shown.

(a)

Find the back-muscle tension and the force of the lower body on the upper body at the hips.
(b)

Repeat the problem but assume that the person is lifting a $30\,{\rm lbf}$ load at D.

Filename:pfigure-s94h8p1

Figure 7.119:

7.3.5 In the flyball governor shown, the mass of each ball is $m=5\,\mathrm{kg}$ , and the length of each link is $\ell=0.25$ m. There are frictionless hinges at points $A$ , $B$ , $C$ , $D$ , $E$ , $F$ where the links are connected. The central collar has mass $m/4$ . Assuming that the spring of constant $k=500$ N/m is uncompressed when $\theta=\pi\,{\rm radians}$ , what is the compression of the spring?

Filename:sfig4-7-DH1

Figure 7.120:

7.3.6

(a)

Find $F$ for equilibrium for the parallelogram structure shown assuming the rest length of the spring is zero.
(b)

Comment on how your answer above depends on $\theta$ .

Filename:sfig4-7-DH2

Figure 7.121:

7.3.7 A common lamp design is shown. In principle, the lamp should be in equilibrium in all positions. According to the original patent from the 1930s, it can be in equilibrium, even with no friction in the joints. Unfortunately, the recent manufacturers of this lamp seem to have lost the wisdom of the original patent. Show how to place the springs so this lamp is in equilibrium for all $\theta<\pi/2$ and $\phi<\pi/2$ . [Hint: use springs with zero rest length.]

Filename:sfig4-7-DH3

Figure 7.122:

Lamp.

7.3.8 Log carrier. This self-locking scissors-mechanism gadget is used to pick up logs and blocks of ice. The $5\,\mathrm{cm}$ wide and $3\,\mathrm{cm}$ high diamond arrangement of hinges ABCD makes up a 4-bar linkage. The grips E and H are $16\,\mathrm{cm}$ apart and $9\,\mathrm{cm}$ below D. The block weighs $250\,{\mbox{N}}$ . Neglect the weight of the mechanism.

(a)

What is the horizontal component of the force on the block at E?
(b)

What is the minimum coefficient of friction $\mu$ for which this device self locks?

Filename:sfig4-7-1

Figure 7.123:

7.3.9 Gear teeth on handle JB mesh with teeth on handle-and-blade KAP at point G midway between hinges A and B. Assume that in the configuration of interest J, B and A are co-linear, that K, A and Q are co-linear and that the cutting contact points Q and P are effectively coincident, that angle JAK = $20^{{}^{\circ}}$ , $JB=40\,\mathrm{cm}$ , $BA=6\,\mathrm{cm}$ , $KA=46\,\mathrm{cm}$ , $AP=AQ=3\,\mathrm{cm}$ , and that the co-linear squeezing forces at J and K are $100\,{\mbox{N}}$ .

(a)

Find the cutting force at Q and P.
(b)

Replace this design with one that has no tooth engagement at G. But instead handle JB and blade BAQ are welded together as one piece. Assuming the same geometry as before, what then is the cutting force at Q and P?
(c)

Without detailed calculations, explain the ratio of the two answers above. Answer: Either by looking at part KAP or at part BAQ, if we think of moment balance about A we see that the cutting force has to fight about twice the torque in the gear mechanism as in the ungeared mechanism. For example KAP is aided in its cutting by the torque from the force at G.

Filename:sfig4-7-1a

Figure 7.124:

7.3.10 These 4-piece shears use the mechanism in problem 7.124 twice over. Co-linear hand forces $F_{JK}$ are applied to handles JB and KC at J and K. Handle JB is hinged to blade BAQ at B. Handle KC is hinged to blade CAP at C. The blades are hinged to each other at A. Handle JB is effectively hinged to CAP, by means of gear teeth, at G, a point on the line segment AB. Similarly KC is effectively hinged to BAQ at point G’ on the segment AC. The cut object presses with colinear forces $F_{PQ}$ on the blades at P and Q. See box 7.3 on page 7.3 for more pictures of these shears.

Assume $F_{JK}=100\,{\mbox{N}}$ , $JB=KC=30\,\mathrm{cm}$ , $AB=AC=6\,\mathrm{cm}$ , $AG=AG^{\prime}=3\,\mathrm{cm}$ , and $AP=AQ=20\,\mathrm{cm}$ . Assume AB, BJ, AC,and CK all make angles of $\pm 20^{{}^{\circ}}$ with a horizontal line. Assume P and Q are coincident and on a horizontal line extending from A.

(a)

Find $F_{PQ}$ .
(b)

Replace this design with one where JB is welded to BAQ at B, KC is welded to CAP at C, and there are no contacting gears. In this same geometry what is $F_{PQ}$ ?
(c)

Give a quantitative estimate, but not a detailed calculation that tells you the ratio of the forces in the above two problems? Answer: The mechanism multiplies the force at B and C by a factor of 2 compared to having the handle hinged at A. The force at G also gets (a shade less than) this force but with half the lever arm. Together they give a force multiplication of (a shade less than) 2+1=3.

Filename:sfig4-7-1b

Figure 7.125:

7.3.11 The garden cutters shown are a 4-bar linkage. Estimate the locations of points, as needed, using the given dimensions as a scale (the drawn clippers are shrunk slightly from reality to simplify the numbers).

(a)

If the handle is squeezed with a pair of $50\,{\mbox{N}}$ forces at J and K what is the cutting force at P and Q? Answer: $F_{P}=125\,{\mbox{N}}$
(b)

If the handle is squeezed with a pair of $50\,{\mbox{N}}$ forces at I and H what is the cutting force at P and Q? Answer: $F_{P}=125\,{\mbox{N}}$
(c)

If this design was changed by eliminating link DB and welding handle JCDI to the blade CAQ, what would be the answers to the two questions above. Answer: For the load at I, $F_{P}=75\,{\mbox{N}}$ . For the load at J, $F_{P}=250N$ .
(d)

Describe in words, the reasons for the similarities and differences between the answers above. Answer: With the welded handle there is just a simple lever and the mechanical advantage comes from the horizontal distance between the load and hinge A. For the 4 bar mechanism the force at C is the applied vertical load, no matter where it is applied. So the lever arm is the horizontal distance from A to C.

Filename:sfig4-7-2

Figure 7.126:

7.3.12 The pliers shown are made of five pieces modeled as rigid: HEG and its mirror image, DCE and its mirror image, and link CC’. You may assume that the geometry is symmetric about a horizontal line (the top is a mirror image of the bottom). The load $F$ and dimensions shown are given. (See also similar problem 7.76)

(a)

Find the force squeezing the piece at D; Answer: $F_{D}=\ell_{EC}(\ell_{EH}-d)F/d\ell_{CD}$
(b)

Find the tension in CC’; Answer: $T_{CC^{\prime}}=\left(\ell_{EH}/d-1\right)\left(\ell_{EC}/\ell_{CD}+1\right)F$
(c)

What happens to the squeezing force if $d$ is made smaller, approaching zero? Why can’t this work in practice? Answer: As $d\rightarrow 0,F_{D}\rightarrow\infty$ . Two problems: the amount of motion goes to zero and the assumption of rigidity becomes non-negligibly inaccurate.

Filename:sfig4-7-2a

Figure 7.127:

7.3.13 For simplicity the vice grips shown in the photo are approximated as in the drawing. Round piece AA’ is gripped between the upper handle/jaw ABEG and the lower jaw A’BC. The upper handle ABEG is pinned to the lower jaw A’BC at B. Handle CDH is pinned to the lower jaw at C and to the bar DE at D. Bar DE is pinned to the upper handle ABEG at E. The $25\,{\rm lbf}$ forces act at G and H as shown. Dimensions are as shown. What is the magnitude of the force at A? Answer: $F_{A}=500\,{\rm lbf}$

Filename:sfig4-7-2b

Figure 7.128:

7.3.14 Pipe wrench. A wrench is used to turn a pipe as shown in the figure. Neglecting the weight of the pipe, find

(a)

the torque of the pipe wrench forces about the center of the pipe
(b)

the forces on the pipe at C and D
(c)

the needed friction coefficient between the wrench and pipe for the wrench not to slip.
(d)

what design change would reduce this needed coefficient of friction (what change of dimensions)? Answer: reduce the dimension marked “2 inches”. The smaller the less the friction needed.
(e)

given that the design change above is possible, why isn’t it used? [hint: implement the design change and calculate the forces on the pipe.] Answer: As the “2 inch” dimension is reduced to zero, the needed coefficient of friction goes to zero and the forces squeezing the pipe go to infinity. This is bad because it can damage the pipe. It is also bad because a small pipe deformation will cause the hinge on the wrench to snap through, like a so called “toggle mechanism” and thus not grab at all.

Filename:sfig4-7-2c

Figure 7.129:

7.3.15 The center of mass of 200 pound structure AEGBC is at G. It is held by rollers at A and B as well as with the rope which starts at E, wraps around the pulley at C, and ends at D. You can assume the pulley at C has negligible size.

(a)

Find the force of the ground on the structure at A. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{A}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$
(b)

Find the tension in the rope. Answer: $T=200\,{\rm lbf}$

Filename:pfigure-s94q7p1

Figure 7.130:

7.3.16 Consider a bike on level ground that is held from falling sideways with forces that don’t push it forward or back. Assume that all the bearings are ideal and that the wheels don’t slip.
$R_{r}=$ radius of rear wheel,
$R_{s}=$ radius of rear sprocket,
$R_{p}=$ crank length from crank-axle to pedal, and
$R_{c}=$ radius of chain wheel (front sprocket).
What backwards force $F$ on the seat is required to keep the bike from going forward (i.e., to maintain static equilibrium) if

(a)

A person sits on the bike and pushes back on the bottom pedal with a force $F_{p}$ ? (is $F>0$ ?)
(b)

A person standing next to the bike pushes back on the pedal with force $F_{p}$ ? (is $F>0$ ?)

Your answer should be in terms of some or all of $R_{r},R_{s},R_{p},R_{c},$ and $F_{p}$ . Of great interest is whether $F$ is bigger or less than zero. So pay close attention to signs.
To solve this problem you have to draw several free-body diagrams: 1) of the whole bike and rider (if the rider is on the bike), 2) of the crank-pedal-chain-wheel system, with a little bit of chain, 3) The rear wheel and rear sprocket, with a little bit of chain.

Filename:pfigure-s94h7p2

Figure 7.131:

7.3.17 The proposed nutcracker design consists of two moving parts: a lever hinged to the fixed base at B and a punch hinged to the fixed base at A. All joints and slots are assumed to have negligible friction.
Mechanism and geometry clarifications: The vertical lever has a pin at C and a horizontal force $F$ applied at D. The punch has a slot in which the lever pin slides at C. The slot is parallel to the line AC. The spherical nut is cracked by being squeezed between the vertical surface of the punch at N and the vertical surface attached to the base. Point N at the left edge of the nut is level with the sliding pin at C. The horizontal distance from C to N does not enter the solution, but assume it is $c$ if you need it for an intermediate calculation.
Quantities: $F=10\,$ lbf, $a=2\,$ in, $b=10\,$ in.

(a)

Find the force acting on the nut at N. A number is desired (i.e., so many lbf force). [Hint: Only substitute in numbers when you have a formula for your answer in terms of $a$ , $b$ and $F$ .] Answer: $F_{N}\left(b(a^{2}+b^{2})/a^{2})\right)F=130F=1300\,{\rm lbf}$
(b)

The answer to (a) is conspicuous in its being either much smaller than $F$ , very similar to $F$ , or much bigger than $F$ . Which is it? Explain, in words, why. The best possible answer will generate an approximate formula for the force at N using next-to-no equations. Answer: The mechanism uses three tricks to multiply the force: a lever, a wedge, and a toggle. Each of these multiplies by about 5. Thus the nut-force $F_{N}$ is on the order of $5^{3}=125$ times as big as $F$ .

Filename:ch4-7-c

Figure 7.132:

	$\displaystyle\frac{1}{k_{s}}$	$\displaystyle=$	$\displaystyle\frac{1}{k_{p}}+\frac{1}{2k}=\frac{1}{2k}+\frac{1}{2k}=\frac{1}{k}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}k_{s}$	$\displaystyle=$	$\displaystyle k=20\,{\rm kN}/\,\mathrm{m}.$

	$\displaystyle F_{B}$	$\displaystyle={F_{A}\over\tan{2\theta}}\approx{F_{A}\over 2\theta}\approx{1% \over 2}{F_{A}\over\tan\theta}$
		$\displaystyle\approx{1\over 2}\cdot\mbox{(the value of $F_{B}$ had there been % no friction).}$