(a)

(b)

Fig. 10–1 (a) Forces F and –F cause ameter stick to rotate. (b) Forces F and-F, applied now at the same place onthe meter stick, cause no rotation.

237

Static Equilibrium

I n the last chapter we saw how rotation can result from external forces producing

torque. In this chapter we shall see how external forces can be balanced to produce

no rotation of an extended body—for example, how the forces on a ladder must be

balanced to prevent it from tipping over. We shall also see how excessive internal

forces in a body can result in the body breaking, for example, the breaking of a bone.

Conditions for Static EquilibriumThe external forces acting on a body that stays at rest in an inertial reference framealways satisfy certain conditions. For a single particle, there is only one condition, andit is quite simple: the vector sum of the forces must equal zero. For an extendedbody, things are more complicated. A resultant force of zero ensures only that thecenter of mass of the body remains at rest, as in the following example.Suppose a meter stick is initially at rest on a table. If two equal-magnitude, hori-

zontal forces F and �F are applied as shown in Fig. 10–1a, the stick will begin torotate. It will not remain at rest, despite the fact that the resultant force acting on it iszero. Only its center of mass does not move. This one point remains at rest, and themeter stick is said to be in translational equilibrium.The meter stick remains at rest when F and �F are applied as shown in Fig. 10–1b.

In this case, the stick neither rotates nor translates. It is in rotational equilibrium aswell as translational equilibrium.The meter stick rotates when the forces do not act along the same line (Fig. 10–1a)

because in this case there is a net torque. Both forces produce negative (clockwise)torque. Therefore the stick will rotate with negative angular acceleration about itscenter of mass. On the other hand, when the two opposing forces have the same lineof action (Fig. 10–1b), the stick does not rotate because there is no net torque. (The twoforces F and �F have equal magnitudes and equal moment arms, and they producetorques of opposite sign, and so these torques cancel.)

10–1

C.HAPTER10Return to Table of Contents

Giant sequoias up to 100 m highstand in static equilibrium.

CHAPTER 10 Static Equilibrium238

The general condition for rotational equilibrium follows as a special case ofEq. 7–15 (� & I�). When the sum of the torques produced by external forcesequals zero (� & 0), a body will experience no angular acceleration (� 0). If thebody is initially not rotating ((0 0), it will continue in a state of nonrotation (( 0):

� & 0 (rotational equilibrium) (10–1)

The condition for translational equilibrium is that the sum of the external forcesmust equal zero. For forces acting in the xy plane, we express this condition:

� Fx 0 � Fy 0 (translational equilibrium) (10–2)

A body in both rotational and translational equilibrium is said to be in static equi-librium. Eqs. 10–1 and 10–2 give the conditions that must be satisfied by the externalforces acting on a body in order for that body to be in static equilibrium. These equa-tions are very useful in solving for unknown forces acting on a stationary body.

Choice of Rotation AxisIn applying the condition for rotational static equilibrium (� & 0), you may chooseany rotation axis to compute the torques. Although the values of the individual torqueswill depend on the axis chosen, the resultant torque will always be zero when the bodyis in rotational equilibrium, since there is no rotation about any axis. You shouldchoose an axis that makes the calculation as simple as possible. As we shall see in theexamples that follow, you can often choose the axis so that some of the unknownforces have zero moment arms and therefore zero torques with respect to that axis.Then these unknown forces will not appear in the torque equation, which reduces to asimple equation you can solve for a single unknown force.

EXAMPLE 1 Balancing on a Seesaw

Two children sit balanced on a seesaw (Fig. 10–2a). Find theweight of the smaller child, who sits 2.00 m from the center ofthe board, if the larger child, who sits 1.00 m from the center,weighs 648 N.

SOLUTION Fig. 10–2b shows the free-body diagram of theboard, which is subjected to three contact forces—two down-ward forces equal to the weights of the children, w1 and w2, andone upward force N provided by the support. We assume thatthe board has negligible weight.

We choose the pivot point O as the origin. Since the un -known force N acts directly at O, this force produces no torqueabout O. Therefore N will not appear in the torque equation. Weapply the torque equation (Eq. 10–1: � � � 0), expressing themagnitude of each torque as the product of its force and itsmoment arm. The torque �1 produced by force w1 is positivebecause this torque tends to rotate the board counterclockwise.The torque �2 produced by force w2 is negative because thistorque tends to rotate the board clockwise:

(a)

(b)

Fig. 10–2

� � � 0�1 � �2 � 0

�w1(2.00 m) � (648 N)(1.00 m) � 0w1 � 324 N

The board will balance if the smaller child has half the weightof the larger child and sits twice as far from the center.

10–1 Conditions for Static Equilibrium 239

EXAMPLE 2 Forces on a Leaning Ladder

An aluminum ladder of negligible weight and length � leansagainst a frictionless wall (Fig. 10–3a). A man weighing 756 N(170 lb) stands halfway up the ladder. The coefficient of frictionbetween the ground and the ladder is 0.500. (a) Find the forcesexerted on the ladder by the ground and the wall. (b) Howhigh can the man climb before the ladder begins to slip?

SOLUTION (a) First we draw a free-body diagram of theladder and man (Fig. 10–3b). We choose point O as our originso that both N and N� produce no torque.* Then these twounknowns do not appear in the torque equation, allowing us toeasily solve for f:

� � � 0f (� sin 60.0°) � w( � cos 60.0°) � 0

Solving for f, we find

f � �

� 218 N

Cancellation of the x components of force implies that this isalso the value of N�, the force exerted by the wall on the ladder:

� Fx � 0

f � N� � 0

N� � f � 218 N

Finally, we apply the condition for equilibrium in the y directionto calculate N, the normal force exerted by the ground on theladder:

� Fy � 0

N � w � 0

N � w � 756 N

To be sure that the ladder does not slip we need to check thatthe frictional force exerted by the ground on the ladder does notexceed the maximum possible value:

fmax � �sN � (0.500)(756 N) � 378 N

The actual frictional force is less than this maximum allowablevalue.

(b) If the man climbs higher on the ladder, the torque associatedwith his weight increases. If the ladder is not to slip, this in -creased torque must be balanced by an increased frictionaltorque. Thus, as the man climbs, the frictional force increases sothat equilibrium is maintained. However, f cannot exceed themaximum value of 378 N.

*The base of the ladder is another reasonable choice for the origin. In thiscase, N and f are eliminated from the torque equation, and N� can be easilyfound.

(a)

Fig. 10–3

We shall solve for the maximum dis tance x the man may safelyclimb, corresponding to f � 378 N. Applying the condition forrotational equilibrium to the free-body diagram of Fig. 10–3c,we obtain

f � sin 60.0° � wx cos 60.0° � 0

Next we solve for x:

x � � �(tan 60.0°)�� (tan 60.0°)�

� 0.866�

Thus with the ladder leaning at this angle, the man can climbonly about 90% of the total length. How could the ladder’sposition be changed so that the man can climb to the top?

(c)(b)

378N756N

fw

12

756 N2 tan 60.0°

w cos 60.0°2 sin 60.0°

MusclesThe human body moves by means of muscles and bones. A muscle consists of abundle of fibers (Fig. 10–4), the ends of which are attached by tendons to adjacentbones. When muscle fibers receive an electrical stimulus from nerve endings connectedto the brain, they shorten, or contract. Contracting muscles exert tension forces on thebones to which they are attached. If the force exerted by the muscle on a bone is notbalanced by other forces, the bone will rotate about a pivot point located between itselfand the bone adjacent to it.The primary muscle responsible for raising the forearm is the biceps, shown in Fig.

10–5a, where a heavy weight is held in the hand. A simplified model of Fig. 10–5a isshown in Fig. 10–5b, where a string under tension represents the biceps. The angle '1that the upper arm makes with the vertical is determined by the shoulder, and the angle'2 that the forearm makes with the horizontal is determined by the contraction of thebiceps.

(a)

(b)

Fig. 10–4 (a) Scanning electron micrograph of a muscle fiber and the (smaller) nerve fiber thatactivates it. (From Hubel DH: Sci Am 241[3]:53, Sept. 1979.) (b)The mechanism of muscle contrac-tion. Muscle fibers consist of overlapping protein filaments—myosin and actin. When the myosinis activated by a nerve impulse, the projections (myosin “heads”) pull it into the space betweenthe actin filaments, shortening the muscle fiber. (From Raven PH, Johnson GB: Understanding Biology,ed 3, St. Louis, 1992, Mosby.)

CHAPTER 10 Static Equilibrium240

10–1 Conditions for Static Equilibrium 241

(a) (b)

Fig. 10–5 (a) The bicepssupports the forearm as aweight is held in the hand.(b) A simple model in whicha string under tension Trepresents the biceps.

EXAMPLE 3 Biceps Tension Required to Lift a Weight

(a) Find the tension in the biceps, using the model in Fig.10–5b, when 1 � 0° and w � 134 N (30 lb). (b) Muscles canproduce a maximum tension of about 70.0 N per squarecentimeter of the muscle’s largest cross section. What is theminimum diameter of the biceps for this problem? (Assume acircular cross-sectional area.)

SOLUTION (a) We choose as a free body the forearm anddraw a free-body diagram in Fig. 10–6. We neglect the weightof the forearm because it is small compared with the otherforces acting on the forearm. Setting 1 � 0° in Fig. 10–5bmeans that the biceps exerts a tension T directed verticallyupward. Since both T and w are vertical, the force E on theelbow must also be vertical. Otherwise there would be aresultant force along the x-axis. And since w produces a nega-tive torque about the point P where T is applied to the forearm,E must be directed downward so that it produces an opposingpositive torque about P. (If the direction of E had been indicatedincorrectly in the free-body diagram, this could have beendiscovered by subsequent application of the conditions forstatic equilibrium. Here we shall choose the point O as ourorigin, so that there is no torque produced by E about O. Thedirection of E will therefore not affect the torque equation.)

We apply the condition for rotational equilibrium. Themoment arm of T is (4.00 cm)(cos 2) and the moment arm ofw is (40.0 cm)(cos 2):

� � � 0

�T(4.00 cm)(cos 2) � w(40.0 cm)(cos 2) � 0

Solving for T, we obtain

T �10.0w �10.0(134 N) �1340 N (or 300 lb)

Notice that this result is independent of 2.

Fig. 10–6

Because the biceps is connected to the forearm so close tothe elbow, a relatively large tension (1340 N) is produced by asmall applied force (134 N). If the muscle were attached, say,8 cm from the elbow, twice as much weight could be supportedwith the same tension of 1340 N. Having the attachment pointclose to the elbow does have an advantage, however. It permitsthe forearm to rotate through a large angle when the bicepscontracts only slightly, thereby allowing the forearm to moverapidly in the short time necessary for this small contraction.Evolution has favored a human anatomy designed for rapidlimb movement rather than weight lifting.

(b) The minimum cross-sectional area is

A � �19.1 cm2

We are assuming a circular cross section, and so this area isrelated to the diameter d by the formula

A � �r 2 �

Therefore

d � �4�A�/�� � 4.93 cm

This is a rather large biceps.

�d2

4

1340 N70.0 N/cm2

(a)

(b)

(c)

Fig. 10–7 (a) A body composed of nparticles, each with its own weight andmoment arm. (b) The body’s totalweight, the vector sum of the particles’weights, equals Mg, where M is the body’stotal mass. (c)We can compute thetotal gravitational torque on the bodyby treating the weight as though it actsat a single point, the center of gravity.

242 CHAPTER 10 Static Equilibrium

Center of GravityThe torque produced by the gravitational force on an extended body is due to theweight of each particle of the body. Fortunately, we do not have to compute and sumthe torques for each of the large number of particles that compose any extended body.Instead the total gravitational torque on an extended body may be computed as thoughthe weight were concentrated at a single point, called the center of gravity.The body shown in Fig. 10–7a consists of n particles of masses m1, m2, … , mn and

weights m1g, m 2g, … , mng. As shown in Fig. 10–7b, the resultant gravitational forceon the body is the vector sum of the particle weights m1g � m 2g � … � mng, orsimply Mg, where M m1 m 2 … mn. We shall show that, for purposes ofcalculating torque, the total weight Mg can be thought of as acting at a point, the centerof gravity, as illustrated in Fig. 10–7c.Referring to Fig. 10–7a, we see that the total torque produced by the weights of the

n particles can be expressed in terms of the x coordinates of the particles, x1, x 2, … , xn:

& �m1gx1 � m2gx2 �… � mngxn (10–3)

�g(m1x1 m2x2 … mnxn)

Recalling that the body’s center of mass has an x coordinate given by Eq. 5–5:

xcm

we may write

m1x1 m2x2 … mnxn Mxcm

Substituting this equation into Eq. 10–3, we obtain

& �Mgxcm (10–4)

This equation shows that the torque may be computed as though the body’s entireweight Mg is concentrated at its center of mass. This point serves as the “center ofgravity.” Thus the center of gravity illustrated in Fig. 10–7c is the same as the centerof mass.* As we saw in Chapter 5, the center of mass of a symmetric body is at itsgeometric center. So the geometric center is also the center of gravity.

*The center of mass and the center of gravity are logically distinct and in fact will not be the same point ifthe body is large enough so that g is not uniform over the entire body and could not therefore be factored inEq. 10–3.

10–2

m1x1 m2x2 … mnxn***

M

24310–2 Center of Gravity

EXAMPLE 4 Forces on a Book Balanced on a Table

The edge of a 9.00 N book extends over the edge of a table.Find the force exerted by the table on the book and the effectiveline of action of this force.

SOLUTION The book is acted upon by upward forces alongits surface of contact with the table. The vector sum of these forcesis the normal force N. Because the book is in translationalequil ibrium, this normal force must balance the book’s weight:

N � �wN � w � 9.00 N

Although N is distributed over the whole surface of contact,we can think of it as having a single effective line of action, thesame as for the book’s weight. Because the book is in rotational

equilibrium, this line of action must pass through the center ofgravity (Fig. 10–8). You can think of N as acting at a singlepoint directly beneath the book’s center of gravity. Of course,this is physically possible only so long as part of the table isbeneath the center of gravity.

Fig. 10–8

EXAMPLE 5 Forces Supporting a Car’s Front and Rear Wheels

A car weighing 1.80 �104 N is parked on level ground. Its centerof gravity is located 1.30 m in front of the rear axle and 2.00 mbehind the front axle. Find the force exerted by the ground(a) on the front wheels and (b) on the back wheels.

SOLUTION The free-body diagram of the car is given inFig. 10–9; NR is the net force on the rear wheels, and NF is thenet force on the front wheels.

(a) Applying the condition for rotational equilibrium with theorigin at the base of NR, we can eliminate NR from the torqueequation and solve for NF:

� � � 0

�NF(3.30 m) � w(1.30 m) � 0

NF � (1.30/3.30)w

� (1.30/3.30)(1.80 �104 N)

� 7.09 �103 N

(b) We apply the condition for translational equilibrium tofind NR:

� Fy � 0

NR � NF � w � 0

NR � w � NF �1.80 �104 N � 7.09 �103 N

�1.09 �104 N

Fig. 10–9

244 CHAPTER 10 Static Equilibrium

Stress and StrainWhen external forces are applied to a rigid body, its particles shift their relative posi-tions, thereby changing the internal forces between these particles. When a body is instatic equilibrium, the vector sum of the internal and external forces on each particleequals zero.To gain some idea of the nature of these internal forces, we shall analyze the

forces acting within a board being used as a child’s seesaw (Fig. 10–10a). We choosea section of the board to the left of point O as our free body. The external forces actingon this section are the smaller child’s weight w1 and the forces exerted by the adjacentsection of board to the right. The forces exerted on the free body along the surface ofcontact between the free body and the adjacent section must balance the force w1 andthe torque produced by w1. It is clear that the net force acting on this surface ofcontact is �w1; otherwise our free body would not be in translational equilibrium. Butthe force �w1 alone would produce a net positive torque and counterclockwise rotation.So there must be a balancing clockwise torque produced by other forces.A simple combination that would produce equilibrium is shown in Fig. 10–10b. The

large opposing horizontal forces can be accounted for if we understand that the boardbends under its load. In a thin plank, the bending is large enough to be obvious (Fig.10–10c), whereas in a thicker one it is not. But in either case, there is bending. Thisbending means that the particles along the top of the board are stretched apart,producing a tension (as in a stretched rope) and that the particles along the bottom ofthe board are pushed together, producing a compression (as in a compressed spring).

(a)

(b)

(c)

Fig. 10–10 (a) Two children balance on a seesaw. (b) The forces exerted on the section ofthe board on the left by the adjacent section must balance both the force w1 and the torqueproduced by w1. (c) The board’s upper surface is stretched and is under tension. The board’slower surface is compressed.

*10–3

(a)

(b)

Fig. 10–11 (a) A normal stress eitherstretches or compresses a rod. (b) Ashear stress causes an angular displace-ment of a book.

Fig. 10–12 A stress-strain curve for atypical rigid material.

StressInternal forces acting on surfaces within a body are said to produce stress in thebody. Stress, denoted by the Greek letter % (sigma), is defined as the force per unit areaon the surface on which it acts:

% (10–5)

When the force is perpendicular to the surface, it is called normal stress.When theforce is tangent to the surface, it is called shear stress.A normal stress can be eithertensile or compressive. In Fig. 10–10c the normal stress along the upper part of theboard is tensile stress and the normal stress along the lower part is compressive stress.The stress produced by the vertical force �w1 is shear, since this force is tangent to thevertical cross section on which it acts (Fig. 10–10c).

StrainThe change in shape of a body subjected to stress is called strain. The strain in theweight-bearing board of Fig. 10–10 is rather complex, since both shear and normalstresses are present. In Fig. 10–11, however, we see bodies subject to pure normalstress and pure shear stress.The Greek letter � (epsilon) is used to denote strain. Normal strain is denoted by �N

and defined to be the fractional change in length of the body that results when anormal stress is applied:

�N (10–6)

Shear strain is denoted by �S and is defined to be the tangent of the angular shift ' thatresults when a shear stress is applied:

�S tan ' (10–7)

Stress is a function of strain. In Fig. 10–12 a typical stress-strain curve is shown fora rigid material, such as metal, wood, or bone, under either compression or tension. Forsmall strains the relationship is approximately linear. The slope of the linear section iscalled Young’s modulus, denoted by Y:

Y (linear region) (10–8)

Young’s modulus may be used to compute the stress required to produce a givenstrain. For example, it follows from Eq. 10–8 that to produce a strain of 0.010, thestress required is % �Y 0.010Y. For steel Y 2.0 � 1011 N/m2, and so a stress of2.0 � 109 N/m2 must be applied to produce this 1% strain. For rubber Y 1.0 � 106

N/m2, and so a stress of only 1.0 � 104 N/m2 must be applied to produce the samestrain. Rubber is much easier to stretch or compress than steel, and this is reflected inrubber’s lower value of Y.The point where the stress-strain curve ends corresponds to fracture of the material.

The value of the stress at this point is typically on the order of 107 to 108 N/m2.

F*A

��*�

%*�

24510–3 Stress and Strain

When a rigid body is subject to a small external torque, large stress may beproduced by the body’s internal forces. This can be seen in Fig. 10–10c, where thehorizontal internal forces acting on the left section of the board are much larger thanthe weight w1—an external force. The large magnitude of these internal forces isnecessary to produce sufficient torque to balance the positive torque produced by w1

and �w1, since the horizontal forces have much smaller moment arms than w1 and �w1.A detailed analysis of the internal forces in a rigid rod shows that the rod will break

when the torque acting on it reaches a certain maximum value &max. This torque isrelated to the maximum stress at fracture %max by the equation

&max r3%max (10–9)

where r is the rod’s radius.

#*4

CHAPTER 10 Static Equilibrium246

EXAMPLE 6 Strained Spinal Disks

A weight lifter holds 1.0 �103 N (230 lb) directly overhead. Inthe midsection of his body this weight is supported by hisspine. Spinal vertebrae are separated by easily compressibledisks, each disk having an average cross-sectional area of 1.0 �

10�3 m2. Young’s modulus for a disk is 7.0 �106 N/m2. The totalthickness of all the intervertebral disks is 15 cm. Compute theweight lifter’s decrease in height resulting from the compres-sion of the disks. For simplicity, ignore the effect of the upperbody’s weight on the spine.

SOLUTION Consider a free-body diagram of the 1000 Nweight and the part of the weight lifter’s body directly aboveone of the disks (Fig. 10–13). The force F exerted by the disk onthe upper body must balance the 1000 N weight. The upperbody exerts a downward reaction force of 1000 N on the disk.The disk also experiences a balancing upward force of 1000 N,exerted by contact with the body below. Each disk experiencesthe same opposing forces of 1000 N, produced by contact withthe vertebrae above and below it. The stress on each disk is thisforce divided by the area:

� �

� �1.0 �106 N/m2

The strain in each disk may be computed by use of Young'smodulus. From Eq. 10–8, we find

� �

� � 0.14

Fig. 10–13

Each disk will be compressed by 14%, and hence the totalthickness of the disks is reduced by 14%. From the definition ofstrain (�N � ��/�), it follows that the change in thickness �� isthe product of the strain and the total length � of the disks:

�� � �� � (0.14)(15 cm) � 2.1 cm

This reduction in thickness should be approximately equal tothe weight lifter’s decrease in height.

1.0 �106 N/m2

7.0 �106 N/m2

�Y

1.0 �103 N1.0 �10–3 m2

FA

ScalingStructures cannot be increased arbitrarily in size while maintaining the same propor-tions. The weight of a body is proportional to its volume, but the strength of thesupporting members is proportional to their cross-sectional area (F %A). If alldimensions of a body are proportionately increased, the ratio of strength to weightdecreases. Thus a structure that is strong at one size may not be greatly scaled up insize without greatly diminishing its strength. To maintain sufficient strength, largerstructures must be proportionately thicker (Fig. 10–15).Galileo recognized this effect. In Two New Sciences he wrote:

…you can plainly see the impossibility of increasing the size of structures to vast dimensionseither in art or in nature; likewise the impossibility of building ships, palaces, or temples ofenormous size in such a way that their oars, yards, beams, iron-bolts, and, in short, alltheir other parts will hold together; nor can nature produce trees of extraordinary sizebecause the branches would break down under their own weight; so also it would be impos-sible to build up the bony structures of men, horses, or other animals so as to hold togetherand perform their normal functions if these animals were to be increased enormously inheight; for this increase in height can be accomplished only by employing a material whichis harder and stronger than usual, or by enlarging the size of the bones, thus changing theirshape until the form and appearance of the animals suggest a monstrosity… . if one wishesto maintain in a great giant the same proportion of limb as that found in an ordinary man hemust either find a harder and stronger material for making bones, or he must admit adiminution of strength in comparison with men of medium stature; for if his height beincreased inordinately he will fall and be crushed under his own weight.

10–3 Stress and Strain 247

EXAMPLE 7 Breaking a Leg

(a) Compute the torque required to break a tibia (leg bone) thathas a radius of 1.00 cm at its thinnest point, if the maximumstress at fracture is 2.00 � 108 N/m2. (b) Suppose a footballplayer is being tackled by one opponent who holds his ankle ina fixed position while another opponent applies a horizontalforce to his knee, 0.500 m above the ankle. What is the max -imum value of the force that can be applied before the leg willbreak?

SOLUTION (a) Applying Eq. 10–9, we find

�max � r 3�max � (1.00 �10�2 m)3(2.00 �108 N/m2)

�157 N-m

(b) From Fig. 10–14, we see that the external torque applied to theleg is related to the applied force’s magnitude F by the equation

� � Fd

Breaking occurs when � � �max, or

F � �

� 314 N (or 71 lb)

Fig. 10–14

This force is surprisingly small. If the leg were free to move, amuch larger force could be applied without serious injury. It isthe fact that the ankle is pinned in place that causes the prob -lem. To hold it in place, other external forces must be acting onthe ankle. It is the combination of all the forces that producesthe large internal stress within the leg.

157 N-m0.500 m

�maxd

�4

�4

Fig. 10–15 Giant redwood trees havedisproportionately large trunks andsmall branches.

1 A tightrope walker carries a long pole, which helps hermaintain her balance (Fig. 10–16). Suppose that at someinstant her center of gravity is displaced to the side sothat there is a torque tending to rotate her body off therope.(a) What quantity does the pole serve to increase?(b) How does this affect the time it takes her body to

rotate appreciably?

Fig. 10–16

248

HAPTER SUMMARY10C For a rigid body to be in static equilibrium, the followingequa tions must be satisfied by the external forces acting on it:

� Fx 0� Fy 0� & 0

Although the torque produced by a given force dependson the choice of rotation axis, the resultant torque withrespect to any axis equals zero if the body is in static equi-librium. An axis should be chosen so as to minimizecomputation.The weight of any body is distributed over the entire

body. When calculating torque, however, that weight can betreated as though it acts at a single point, called the “centerof gravity”. When g is uniform over a body, the center ofgravity is the same as the center of mass.The stress % in a body is the force per unit area acting on

a surface within the body:

%

Normal stress is perpendicular to the surface along which itacts, and shear stress is tangent to the surface. The strain �in a body is a measure of the change in body shape that

results from stress. Normal strain �N is defined as the frac-tional change in length of a body:

�N

Shear strain �S is defined as

�S tan '

where ' is the angular shift resulting from a shear stress.Stress is a function of strain. For many materials, stress

is proportional to strain when the strain is small. Theconstant of proportionality is called Young’s modulus:

Y *%

�*

Fracture of a material generally occurs when % isincreased to about 107 or 108 N/m2.If a rigid rod is subject to a torque that tends to make it

bend, stress is produced within the rod. If a torque andstress are great enough, the rod breaks. The maximumapplied torque &max is related to the maximum stress %max bythe equation

&max r 3%max

where r is the rod’s radius.

#*4

��*�

F*A

Questions

249

2 The wheels in Fig. 10–17 are subject to various forces,all of the same magnitude. Which of the wheels under-goes: (a) translational acceleration but not rotationalacceleration; (b) rotational acceleration but not transla-tional acceleration; (c) both translational and rotationalacceleration?

Fig. 10–17

3 Suppose the force F applied to the bicycle wheel inFig. 9–7 has a magnitude of 100 N. What two forcesapplied at R and Q could produce static equilibrium ofthe wheel?

4 Forces F1 and F2 act on a flat disk, as shown in Fig.10–18. The vector sum of the two forces equals 10 N,directed east. A third force, of magnitude 10 N anddirected west, is applied to the perimeter of the disk. Toproduce static equilibrium, this force must be applied at(a) P; (b) Q; (c) R; (d) S; (e) P or R; (f ) Q or S.

Fig. 10–18

5 Is the free body in Fig. 10–19 in static equilibrium?

Fig. 10–19

6 A square block is subject to the three forces shown inFig. 10–20. A fourth force is applied to the block so asto produce static equilibrium.(a) Find the magnitude and direction of the fourth force.(b) Where is the fourth force applied to the block?

Fig. 10–20

7 If the car in Fig. 10–9 is parked headed downhill ratherthan on level ground, would the following forces in crease,decrease, or remain the same: (a) Force of the road onthe rear wheels; (b) Force of the road on the front wheels;(c) Total force of the road on all the wheels?

8 A long board is held horizontally with two hands. Ifthe board is held with both hands near one end, themagnitude of the force exerted by each hand will be:(a) greater than the weight of the board; (b) less than theweight of the board.

9 (a) Is it possible for the center of gravity of a body to beat a point where there is no matter?

(b) Where is the center of gravity of a doughnut?10 The center of gravity of a standing, stationary personmust be (a) directly above a point halfway between thetwo feet; (b) directly above a point of contact betweenthe feet and the floor; (c) either directly above a point ofcontact between the feet and the floor or between twopoints of contact.

11 A waiter carries a heavy tray of food in front of him.(a) In what direction will he have to lean to maintain his

balance?(b) Will he lean less if he holds the tray close to his

body or at arm’s length?

Questions

Conditions for Static Equilibrium

1 In Fig. 10–21 a crowbar is used to lift a weight. If adown ward force of 225 N is applied to the end of thebar, how much weight does the other end bear? Thecrowbar itself has negligible weight.

Fig. 10–21

2 An artist designs a mobile of light horizontal rods con -nected by vertical strings and supporting various shapedweights (Fig. 10–22). Find the magnitudes w2, w3, and w4

if w1 1.00 unit of weight. The numerical values givenin the figure all have units of length.

Fig. 10–22

10–1

250

12 A physical therapist treating a patient with a damageddisk in the lower back emphasizes the importance ofcorrect posture in reducing irritation of the disk. In tryingto explain to the patient the physical effect of goodposture, the therapist claims that if the patient stands andsits correctly, the upper part of the spinal column willabsorb more of the body’s weight and less will be trans-mitted to the lower spinal column.(a) Is the therapist correct?(b) When your body is in a vertical position, how can

you prevent some of its weight from being supportedby the lower back?

(c) How is a very curved spine likely to cause greaterirritation to a disk?

13 Ski bindings rigidly attach a boot to the ski and are de -signed to automatically release the boot when the forceon the binding exceeds a certain value—usually about90 N (20 lb). Why is this an important safety feature?

Answers to Odd-Numbered Questions1 (a) moment of inertia; (b) reduces angular accelerationand therefore increases the time needed to rotate through agiven angle; 3At R, a 100 N force in the same direction asF; at Q, a 200 N force in the direction opposite F; 5 no;7 (a) decrease; (b) increase; (c) same; 9 (a) yes; (b) inthe hole; 11 (a) backward; (b) close; 13 The release featureprevents the foot from being trapped in a fixed position bythe ski. If the foot were trapped in a fixed position, a smallsideways force could break the leg.

CHAPTER 10 Static Equilibrium

Problems (listed by section)

251

3 A newly planted tree is supported against the wind by arope tied to a stake in the ground, as shown in Fig.10–23. The force of the wind, though distributed overthe tree, is equivalent to a single force F, as shown in thefigure. Find the tension in the rope. Assume that neitherthe tree’s weight nor the force of the ground on the tree’sroots produces any significant torque about point O.

Fig. 10–23

4 A deep-sea fisherman struggles to bring in a big salmon(Fig. 10–24). Find the force each hand must exert on thepole if the tension in the line is 175 N. Neglect theweight of the pole and assume that the force exerted bythe left hand is directed along the horizontal.

Fig. 10–24

5 A 90.0 N weight is held in the hand. The upper armmakes an angle of 30.0� with the vertical, and the lowerarm is 15.0� above the horizontal. Find the tension in thebiceps tendon. Refer to Fig. 10–5.

6 The swimmer in Fig. 10–25 propels himself forwardby exerting a 125 N backward force on the water withhis hand. He experiences a forward reaction force F ofmagnitude 125 N. Find the magnitude and direction ofthe torque exerted on the arm by F about an axis throughpoint O.

Fig. 10–25

7 As a swimmer pulls his arm through the water, variousmuscles exert forces on the upper arm. Fig. 10–26shows a force F exerted on the humerus (upper armbone) by the pectoral muscle. The muscle is connectedto the bone 8.0 cm from the center point O of theshoulder joint. Find the magnitude of F, if this force’storque on the arm provides half of the total torquebalancing the torque produced by the water pushingagainst the hand.

Fig. 10–26

Problems

252

*8 An athlete strengthens her leg and knee by doing “legcurls.” While sitting on a bench with the leg initiallybent at the knee, she supports a weight with the foot.She then slowly raises the foot and lower leg, with theupper leg held stationary. Find the tension in the quadri-ceps tendon for the position shown in Fig. 10–27.

Fig. 10–27

*9 In Fig. 10–28 the foot of a walking person is momen-tarily in contact with the ground and approximately instatic equilibrium. The foot bears the person’s full weightof 535 N (N w 535 N). Compute (a) the tension Tin the Achilles’ tendon; (b) the horizontal and verticalcomponents of the force P at the ankle.

Fig. 10–28

*10 The weightless strut in Fig. 10–29 is not attached to thewall; it is prevented from falling only by friction. Find(a) the magnitude of the force of friction between the walland the strut; (b) the normal force exerted by the wall onthe strut; (c) the minimum coefficient of static friction.

Fig. 10–29

Center of Gravity

11 A man doing a slow push-up is approximately in staticequilibrium. His body is horizontal, with his weight of756 N supported by his hands and feet, which are 1.35m apart. One hand rests on a spring scale, which reads267 N. If each hand bears an equal weight, how farfrom the shoulders is the man’s center of gravity?

12 A man and a boy carry a canoe 6.00 m long and weigh -ing 375 N. The canoe is held overhead in a horizontalposition. The boy holds one end of the canoe. How farfrom the opposite end should the man stand so that theboy exerts a force of only 125 N?

*13 A piece of furniture weighing 1.00 � 103 N is carried upstairs, as shown in Fig. 10–30. Find the vertical forcesexerted by the two people at P and Q.

Fig. 10–30

10–2

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253

*14 The two planks in Fig. 10–31 have a combined weightof 200.0 N. Assuming that the floor is frictionless, findthe forces the floor exerts at the two points of contact.

Fig. 10–31

*15 A raccoon wants to tip over a garbage can by pulling onits top edge with a force P acting 60.0� below the hori-zontal. The can is a cylinder of height 80.0 cm anddiameter 50.0 cm, weighing 40.0 N. The can’s center ofgravity is directly beneath its geometric center. What isthe minimum value of P that will cause the can to tip?

*16 A bench seat is attached to a wall by hinges and sup -ported from above by two ropes, as shown in Fig. 10–32.The seat weighs 135 N and is designed to support aperson weighing up to 1350 N, sitting on the edge. Howmuch tension must each rope be able to withstand?

Fig. 10–32

*17 A horizontal board weighing 200.0 N is supported atpoints A and B in Fig. 10–33. The plank serves as a plat-form for a painter, who weighs 600.0 N. Find the max -imum distance D, such that the plank will not tip, nomatter where the painter stands.

Fig. 10–33

18 What is the minimum horizontal force P necessary to tipover the 5.00 kg block shown in Fig. 10–34?

Fig. 10–34

*19 The deltoid muscle connects the shoulder blade to theupper arm (humerus), and so contraction of the deltoidraises the arm (Fig. 10–35). The deltoid tendon connectsthe humerus 15.0 cm from the shoulder joint at an angleof 20.0� with the horizontal. Suppose the arm weighs35.0 N and its center of gravity is 38.0 cm from theshoulder joint. A weight W is held in the outstretchedhand 90.0 cm from the shoulder joint. Find the tensionin the deltoid tendon in terms of W. Find the force theshoulder joint exerts on the humerus if W 40.0 N.Why is this force so large even for such a small weight?

Fig. 10–35

20 Find the magnitude of the vertical force F necessary tolift the end of a loaded wheelbarrow of weight 1400 N,shown in Fig. 10–36.

Fig. 10–36

Problems

254

*21 Find the minimum horizontal force that must be appliedto the handles of the wheelbarrow in the last problem, inorder that the wheel of diameter 20.0 cm will slowly goover a brick 6.00 cm high, as shown in Fig. 10–37. (Novertical component of force is applied to the handles.)

Fig. 10–37

*22 A rigid lean-to shelter of weight 90.0 N is supported bya horizontal rope, as shown in Fig. 10–38. The center ofgravity of the lean-to is at its center. What is the tensionin the rope? The answer is independent of d.

Fig. 10–38

Stress and Strain

*23 An old camping trick for breaking thick sticks for fire-wood is to wedge one end of a stick between two smalltrees that are very close together, and then apply a forceto the other end. Determine the largest radius a stick mayhave if it is to be broken in this way by a 4.00 � 102 Nforce applied perpendicular to the stick 1.00 m from wherethe stick is wedged. The breaking stress is 2.00 � 108

N/m2.

24 (a) Compute the maximum longitudinal force that maybe supported by a bone before breaking, given thatthe compressive stress at which bone breaks is 2.00� 108 N/m2. Treat the bone as a solid cylinder ofradius 1.00 cm.

(b) Young’s modulus for bone is 2.00 � 1010 N/m2. Esti-mate the strain in the bone just before breaking.

25 A certain concrete will support a maximum compressivestress of 1.20 � 107 N/m2 before being crushed. Acolumn of uniform width is to be made with thisconcrete. How tall can the column be if the stress in it atany point is not to exceed one fourth the maximumvalue? The weight density of the concrete is 2.40 � 104

N/m3.*26 A girl accidentally steps into a gopher hole, causing her

foot and ankle to be stuck. If all her weight of 525 N isborne by the one stuck foot, what is the maximum dis -tance her center of gravity can be displaced horizontallyfrom the point of support before her leg will break?

Additional Problems*27 An irregularly shaped, flat object is suspended from a

string, first at point P and then at point Q (Fig. 10–39).In each case, lines are marked on the object, extendingdown ward from the point of support. Prove that theobject’s center of gravity lies at the point of intersectionof the lines.

Fig. 10–39

*10–3

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255

*28 Three identical blocks are stacked with each extendingas far as possible over the block beneath it (Fig. 10–40).Find the distance D that the top block overhangs thebottom block if the length of each block is 20.0 cm.

Fig. 10–40

*29 A block of height 30.0 cm and width 10.0 cm is initiallyat rest on a table. One end of the table is then slowlyraised so that there is an increasingly large angle 'between the table’s surface and the horizontal. If thecoefficient of friction between block and table is 0.500,will the block tip over or slide down the incline?

*30 A symmetrical table of height 0.800 m, length 1.50 m,and weight 445 N is dragged across the floor by aforce applied to its front edge. The force is directed tothe right and upward and makes an angle of 30.0� withthe horizontal.(a) Find the minimum force necessary to drag the table

across the floor. The coefficient of sliding frictionbetween table and floor is 0.400.

(b) Calculate the normal force and the frictional forceon each leg, and show that the table will not tip.

*31 A sprinter of mass 80.0 kg has one foot on the ground,as shown in Fig. 10–41. The force F exerted by theground at point P is directed so that the torque about thebody’s center of mass is approximately zero.(a) Find the angle ' this force makes with the horizontal.(b) Find the instantaneous acceleration of the center of

mass, assuming that it is directed horizontally.

Fig. 10–41

*32 A car weighing 1.00 � 104 N is at rest on a 30.0�incline. Each wheel bears an equal part of the car ’sweight. Thus the axles apply a vertical downward forceof 2.50 � 103 N to the center of each wheel.(a) Find the frictional force exerted by the road on each

wheel.(b) Find the sum of the magnitudes of the frictional forcesfB applied by the brakes to the wheel, as shown inFig. 10–42.

Fig. 10–42

**33 An aluminum ladder of negligible weight leans againsta vertical wall at an angle of 60.0� with the horizontal.The coefficient of static friction between both ladderand wall and ladder and ground is 0.300. A sculptorclimbs the ladder and reaches a point at which the ladderbegins to slip. What fraction of the length of the ladderhas the sculptor climbed?

Problems

256

**34 When a person bends the body at an angle ' while lift -ing a weight w, large forces are produced near the baseof the spine. The back is supported by a system of liga-ments and muscles connecting the spine to the pelvis(Fig. 10–43a). The effect of these muscles is equivalentto having a cord connected at a point P two thirds upfrom the lower end of the spine and making an angle of12� with the spine (Fig. 10–43b). The weight of theupper body supported by the spine is typically about60% of body weight W, with a center of gravity at P.Additional weight w supported by the arms and shoul-ders acts at the top end of the spine.

(a)

(b)

Fig. 10–43

(a) Find the tension in the “cord” and the force on thebase of the spine if ' 30�, W 800 N, and w 0.

(b) Repeat the calculation with w 400 N.(c) The disk at the base of the spine is subject to a

compressive force. If the disk has an area of 1.0 �10�3 m2, what stress does it experience for the situa-tion described in (b)? By how much does the diskcompress if its original thickness is 0.80 cm?Although the relationship between stress and strainin this region is not linear, we may use an effectivevalue of Young’s modulus equal to 2.0 � 107 N/m2.

(d) How great would the weight w have to be to causea stress of 1.5 � 107 N/m2—enough to rupture thedisk?

This problem illustrates why most people are subject tolower back pain at some time in their lives. It is rathereasy to produce a large stress on the lower disks. Aheavily strained disk will bulge and press on the spinalnerve, causing pain in the lower back and the legs.

35 The weight in Fig. 10–44 hangs from a long board ofnegligible weight. Find the magnitude of the tension T inthe horizontal rope and the horizontal and verticalcomponents of force exerted on the board at P.

Fig. 10–44

36 Suppose the hand-held weight in Ex. 3 is raised becausethe biceps muscle contracts. Show that, as the forearmis raised to a horizontal position, the work done by thetension force in the muscle equals the work done onthe weight by the force the hand exerts on the weight.

CHAPTER 10 Static Equilibrium