8-1

EQUILIBRIUM AND ELASTICITY 8 Q8.1. Reason: Because the definition of equilibrium includes !! = 0 as well as

!!F =

!0, an object that

experiences exactly two forces that are equal in magnitude and opposite in direction may still not be in equilibrium because the forces could cause a net torque even though the sum of the forces is zero. See Figure 8.1.

Furthermore, an object could even have !" = 0 as well as !!F =

!0 (and therefore be in equilibrium) but still not be

in static equilibrium if it is moving at a constant velocity. Assess: Think carefully about the definition of equilibrium, especially what it doesn’t say.

Q8.2. Reason: See the figure below for a few possible forces.

Calculate torques about the pivot point. The moment arm for forces exerted at point P is smaller than that for the force at the end of the rod, so the component of the force at point P perpendicular to the radial line must be larger than that of the force at the tip of the rod. As long as the force has the appropriate magnitude component perpendicular to the radial line, the net torque will be zero. Assess: Note that the pivot provides an additional force that keeps the rod in static equilibrium in the horizontal and vertical direction.

Q8.3. Reason: The ladder could not be in static equilibrium. Consider the forces in the horizontal direction. There is a normal force exerted by the wall on the top of the ladder, but no other object (in the absence of friction) exerts a counterbalancing force on the ladder in the opposite direction. Examine Figure 8.9. Assess: This makes sense from a common sense standpoint. If a ladder is about to slip out one tries to increase the friction at the point of contact with the floor, or to produce a horizontal component of a normal force by wedging it.

Q8.4. Reason: As you move out farther on the branch the torque on the branch increases. The support torque provided by the branch must increase for the branch to remain in equilibrium. When the force that provides this torque exceeds the bending strength of the material making up the branch, the branch will break. Assess: The thickness of the branch decreases as you move farther out. This also increases the chance of the branch breaking.

Q8.5. Reason: For divers to be stable on the board before the dive their center of gravity must be over an area of support, that is, over the board. Extending their arms moves the center of gravity over the board. Assess: If the arms are not extended, then the center of gravity would be over the edge of the diving board when they stand on their toes with heels extended out. They would not be in static equilibrium and would topple over before getting off a good clean dive. The other option to get the center of gravity over the board (besides extending arms) is to lean forward slightly toward the board.

8-2 Chapter 8

Q8.6. Reason: In order to be stable, the centers of gravity of the people must lie over their base of support. The center of gravity of the person on the left must lie over his feet. For the person on the right, the center of gravity must be somewhere over her feet. Since she is leaning sideways, the center of gravity is shifted to the left. See the diagram below.

Assess: This result makes sense.

Q8.7. Reason: Your center of gravity must remain over a region of support (your feet). As you rise your rear end is behind your feet so you must lean forward to compensate. Assess: This seems to jibe with everyday experience.

Q8.8. Reason: We will assume that a stretch of 5 cm is still in the linear region of the spring. The given data tells us that it takes 10 N to stretch the spring 1 cm from equilibrium length. It would take another 10 N to stretch the spring an additional centimeter, and so on. Each additional stretch of 1 cm is produced by 10 N of additional force. Carefully examine Equation 8.3 and Figure 8.15; it is

!x, not x in Hooke’s law. So the spring force would

increase by 10 N if the spring is stretched from 4 cm away from equilibrium to 5 cm away from equilibrium. Assess: As long as the spring is still in the linear region, then the answer is always 10 N of additional force for each additional 1 cm of stretch.

Q8.9. Reason: Before Carlos came along the wall also pulled on the spring with a 200 N force when Bob did, that is, there was a 200 N tension force all along the spring. When Carlos arrives he takes the place of the wall but the spring must still stretch 20 cm. The only difference is that now Carlos also moves whereas the wall didn’t. (a) 10 cm. Though the spring stretched 20 cm originally, its center moved by 10 cm and so Bob’s end moved 10 cm away from (farther than) the center. In the tug-of-war the center stays still so Bob’s end only moves 10 cm. (b) 10 cm in the other direction. The total stretch under a 200 N tension must still be 20 cm. Assess: These answers fit well with Hooke’s law. In either case the 200 N tension produced a total stretch of 20 cm.

Q8.10. Reason: (a) The spring follows Hooke’s Law, since the graph of force applied to the spring versus extension of the spring is a straight line. (b) The displacement in Hooke’s Law is measured from the equilibrium length of the spring, which is when the force on the spring is zero. In this case, the equilibrium length of the spring is 10 cm. Measuring displacement from this point, the spring constant is

k = !(F

sp)

y

"y= !

(!10 N)

(0.10 m)= 100 N/m

Assess: Note that the force used in Hooke’s Law is the force that the spring exerts. In this case, the tension exerts an upward force on the spring. The force exerted by the spring is the reaction force to this and is downward. This is the reason for the double negative sign in the equation above.

Q8.11. Reason: Since both halves of the spring are made of the same material and constructed the same way, the spring constant of each half will be twice the spring constant of the original long spring. Assess: Hooke’s Law does not depend on the length of a spring.

Q8.12. Reason: The longer wire will also break at 5000 N. Because the two wires have the same diameter, the force per cross-sectional area is the same in both cases.

Equilibrium and Elasticity 8-3

Assess: You see that the length of the cable is not mentioned in Table 8.2 nor in Example 8.9; this means the length doesn’t matter. And the length doesn’t matter if the diameter is truly uniform and there are no defects in the cable. In practice, however, the longer a wire or cable is the more likely it is that it will contain a defect (impurity or a chance region of slightly smaller diameter) and it is therefore more likely to fail at a slightly lower tension (at the defect) than a shorter cable or wire would.

Q8.13. Reason: The force needed to bend a “beam,” whether it’s a nail or a steel wool fiber, depends on the thickness-to-length ratio. The diameter (thickness) of a steel wool fiber is much less, relative to its length, than that of a steel nail. Thus it takes only a very small force to bend and flex the thin fibers of steel wool, but a very large force to bend a steel nail. Assess: Fiberglass is also flexible while a thicker glass rod is not, for the same reasons. The extreme case is carbon nanotubes that are so thin that they bend easily, but if made into a solid bulk substance as thick as nails would be more resistant to bending.

Q8.14. Reason: The mass of the pole is negligible. Since the older child supports twice as much weight then the bucket must be half as far from the older child. If the bucket is 0.6 m from the older child and 1.2 m from the younger, then this condition is fulfilled and the sum is 1.8 m. So the answer is B. Assess: This makes intuitive sense, but it can be checked with equilibrium calculations !" = 0 as well.

Q8.15. Reason: Use equilibrium calculations !" = 0 around the suspended end. The weight of the rod acts at its center ( L / 2 away from the right end) and the normal force acts at L from the right end. For !" = 0 the normal force must be half the force at twice the distance, so the answer is A. Assess: For !F = 0 the tension in the suspension string must also be

7.0 N. This also makes sense when

computing the torques around the center of the rod.

Q8.16. Reason: Assume the weight of the board can be neglected. The student’s weight exerts a downward force on the board, and the two scales exert an upward force on the board. See the diagram.

There is no net force on the board in the vertical direction. Solving the second equation in Equation 8.1 gives

w = F

1+ F

2= (65 lb) + (100 lb) = 165 lb

The correct choice is D. Assess: Note that the scale on the right exerts a larger force since the center of mass of the student is closer to the right scale.

Q8.17. Reason: The fact that the board is “very light” means we will neglect its mass (which we weren’t given anyway). We know that the student weighs 165 lbs because the downward force of gravity on the student must equal the upward sum of the two scale readings for the student to be in equilibrium. Also required for equilibrium is !" = 0 and we are free to choose the axis around which we compute the torques anywhere we want. It would be most convenient to select a point above one of the scales so that the upward normal force due to that scale will not produce a torque. Furthermore, since we want to know the distance from the right hand scale, choose it as the pivot.

!" = d(165 lb) # (2.0 m)(65 lb) = 0

8-4 Chapter 8

where the counterclockwise torque is positive and the clockwise torque is negative:

d =(2.0 m)(65 lb)

165 lb= 0.79 m ! 0.8 m

So the correct choice is B. Assess: Not only could we have chosen the pivot point at the left scale and produced the same answer (using L – d as the lever arm), but we should do so as a check.

!" = (2.0 m)(100 lb) # (L # d)(165 lb) = 0

L # d =(2.0 m)(100 lb)

165 lb

d = 2.0 0m #(2.0 m)(100 lb)

165 lb= 0.79 m $ 0.8 m

Q8.18. Reason: The leg is in static equilibrium. Equation 8.1 applies. See the figure below.

The net torque on the leg must be zero. Calculating torques relative to the pivot point, the torque equation gives

wd

2!Td

1= 0

Solving for T ,

T =wd

2

d1

=(4.0 kg)(9.80 m/s2 )(0.25 m)

(0.05 m)= 200 N

The correct choice is B. Assess: This result makes sense. The tendon has a small moment arm compared to the weight of the leg.

Q8.19. Reason: As is quite clear from the figure, the force of the tendon on the lower leg is up. However, if we compute torques around a pivot point where the tendon attaches to the lower leg we see that there is a counterclockwise torque produced by the weight of the lower leg acting at the center of gravity 25 cm from the knee joint. There must be a clockwise torque to counterbalance if the lower leg is to be in equilibrium. That clockwise torque is provided by the force of the upper leg down on the lower leg at the knee joint. The correct answer is B. Assess: This may be surprising at first. It may help to visualize it as a lever with an upside down fulcrum (suspension point) where the tendon attaches to the lower leg.

Q8.20. Reason: The leg is in static equilibrium. Equation 8.1 applies. See the figure below.

The total force in the y-direction must be zero. The second equation in Equation 8.1 gives

F

y= w!T = mg !T = (4.0 kg)(9.80 m/s2 ) ! 200 N = !160 N

The correct choice is B.

Equilibrium and Elasticity 8-5

Assess: Note that the force is negative. This means that there is a downward force on the leg at the pivot.

Q8.21. Reason: Because !L = FL / AY if L is doubled then A must be doubled too to keep !L the same.

To double A requires an increase in the diameter by a factor of 2. So the answer is B. Assess: Doubling the diameter (an incorrect answer) would actually make the twice-as-long wire stretch less.

Q8.22. Reason: The center of gravity of the bottom board is 15.0 cm from its left end, or 7.0 cm to the left of the edge of the table. We can push the upper board to the right until its center of gravity is 7.0 cm to the right of the edge of the table, or 1.0 cm left of the right edge of the bottom board, which is 29.0 cm from the left edge of the bottom board. Therefore, for the top board,

x +15.0 cm = 29.0 cm ! x = 29.0 cm "15.0 cm = 14.0 cm.

The answer is C. Assess: The answer can be checked by computing the torques around any other point.

Q8.23. Reason: The tension in the rope is the same as if there were no pulley and one 20 kg block were

suspended, that is, T = mg = (20 kg)(9.8 m/s2) = 196 N. Use this tension as the F in the equation below.

!L =FL

YA=

(196 N)(2.0 m)

(1.5"109 N/m2 )# (0.0025 m)2= 1.3 cm

The answer is then D. Assess: A 1.3 cm stretch of 2.0 m of rope seems reasonable.

Problems

P8.1. Prepare: Because the board is “very light” we will assume that it is massless and does not contribute to the scale reading, nor does it contribute any torques. The sum of the two scale readings must equal the woman’s weight: w = mg = (64 kg)(9.8 m/s2) = 627 N " 630 N. Solve: Compute the torques around the point the board rests on the left scale. The woman’s weight creates a clockwise (negative) torque; and the normal force nright of the right scale creates a counterclockwise (positive) torque.

!" = (2.0 m)(n

right) # (1.5 m)(627 N) = 0 N $m

The right scale reads n

right:

nright

=(1.5 m)(627 N)

2.0 m= 470 N

By simple subtraction the left scale reads

n

left= 627 N ! 470 N = 160 N

Assess: The answer is reasonable. Since the woman is three times farther from the left scale than the right one, it (the left one) reads three times less. And the two scale readings sum to the woman’s weight, as required. Not only could we have chosen the pivot point at the right scale and produced the same answer, but we should do so as a check.

!" = (0.5 m)(627 N) # (2.0 m)(n

left) = 0 N $m

nleft

=(0.5 m)(627 N)

2.0 m= 160 N

And so

n

right= 627 N !157 N = 470 N

It is true that !" = 0 around any point (for equilibrium), but we picked the two we did (the second as a check) because then the resulting torque equations each had only one unknown in them.

P8.2. Prepare: The board is in static equilibrium. Equation 8.1 applies.

8-6 Chapter 8

Solve: See the following diagram.

The net torque must be zero. Calculating torques relative to the left scale, we have

!w

Bd

B! w

Wd

W+ F

2d

2= 0

Solving for F

2,

F2

=m

Bgd

B+ m

Wgd

W

d2

=(10 kg)(9.80 m/s2 )(1.0 m) + (54 kg)(9.80 m/s2 )(1.5 m)

2.0 m= 450 N

Applying the second equation in Equation 8.1 gives

F

1= w

B+ w

W! F

2= (10 kg)(9.80 m/s2 ) + (54 kg)(9.80 m/s2 ) ! 450 N = 180 N

Assess: Note that the axis for calculating torques can be chosen anywhere. Calculating the torques relative to the left scale eliminated the unknown force at that point from the torque equation, allowing a direct solution for the force at the right scale.

P8.3. Prepare: Compute the torques around the bottom of the right leg of the table. The horizontal distance

from there to the center of gravity of the table is

2.10 m

2= 0.55 m = 0.50 m.

Solve: Call the horizontal distance from the bottom of the right leg to the center of gravity of the man x.

!" = (56 kg)(9.8 m/s2 )(0.50 m) # (70 kg)(9.8 m/s2 )x = 0 $ x = 0.40 m

The distance from the right edge of the table is now 0.55 m ! 0.40 m = 0.15 m = 15 cm.

Assess: It seems likely that the table would tip if the man were closer than 15 cm to the edge.

P8.4. Prepare: When the beam just starts to tip, the left support will not exert any upward force on the beam. Simply apply the !F = 0 equilibrium condition. Solve: The net force must be zero, so the right support must exert an upward force equal to the combined weight.

(10 kg + 70 kg)(9.8 m/s2 ) = 780 N

Assess: Because the force of the left support is zero we did not need to use the !" = 0 condition.

P8.5. Prepare: Assume the pole is uniform in diameter and density. We will use the equilibrium

equation !! = 0. The weight of the pole is w = mg = (25 kg)(9.8 m/s2 ) = 245 N.

Equilibrium and Elasticity 8-7

Solve: Compute the torques around the left end, where the pole rests on the fence. The weight of the pole (acting at its center of gravity) will produce a clockwise (negative) torque, and force F near the right end of the pole will produce a counterclockwise (positive) torque.

! = (3.6 m " 0.35 m)F " (1.8 m)(245 N)#= (3.25 m)F " (1.8 m)(245 N)

= 0 N $m

Now solve for F:

F =(1.8 m)(245 N)

3.25 m= 140 N

Assess: This really is a rest because you only have to exert a force of just over half the pole’s weight, instead of the whole weight when you were carrying it. The fence is helping hold up the pole. Think carefully about the figure and imagine moving your hands toward the fence. The upward force you would have to exert to keep !! = 0 would increase, and when you support the pole at its center of gravity the torque equation says your force is equal to the weight of the pole. At that point the fence is no longer helping (and you aren’t resting) as it exerts no upward force. If you tried moving even farther toward the fence, past the center of gravity, there would be no way to keep the pole in equilibrium and it would rotate, fall, and hit the ground.

P8.6. Prepare: The rod is in rotational equilibrium, which means that !net = 0. As the weight of the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the figure below), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with your thumb on top (pushing down) and your forefinger underneath (pushing up). We will calculate the torque about the left end of the rod. The downward force exerted by the pin acts through

this point, so it exerts no torque. To prevent rotation, the pin’s normal force

!n

pinexerts a positive torque

(counterclockwise about the left end) to balance the negative torques (clockwise) of the weight of the mass and rod. The weight of the rod acts at the center of gravity.

Solve:

!

net= 0 N m = !

pin" (0.40 m)(2.0 kg)(9.8 m/s2) " (0.80 m)(0.50 kg)(9.8 m/s2) #!

pin= 12 N $m

Assess: We carefully chose our pivot at the left end of the rod where an unknown force acts, which contributes nothing to the torque.

P8.7. Prepare: The massless rod is a rigid body. To be in equilibrium, the object must be in both translational

equilibrium (!F

net= 0 N) and rotational equilibrium (!net = 0). We have

(F

net)

y= (40 N) ! (100 N) + (60 N) = 0 N, so

the object is in translational equilibrium. Our task now is to calculate the net torque.

8-8 Chapter 8

Solve: Measuring !net about the left end,

!net = (60 N)(3.0 m) sin (+90°) + (100 N)(2.0 m) sin (!90°) = !20 N !m

The object is not in equilibrium.

P8.8. Prepare: The object balanced on the pivot is a rigid body. Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium. There are three forces acting on the object: the

weight

!w

1 acting through the center of gravity of the long rod, the weight

!w

2acting through the center of gravity

of the short rod, and the normal force !P on the object applied by the pivot.

Solve: The translational equilibrium equation (F

net)

y= 0 N is

!w

1! w

2+ P = 0 N " P = w

1+ w

2= (1.0 kg)(9.8 m/s2 ) + (4.0 kg)(9.8 m/s2 ) = 49.0 N

Measuring torques about the left end, the equation for rotational equilibrium !net = 0 is

Pd ! w1(1.0 m) ! w

2(1.5 m) = 0 N "m # (49.0 N)d ! (1.0 kg) (9.8 m/s2 )(1.0 m) ! (4.0 kg) (9.8 m/s2 )(1.5 m)

= 0 N # d = 1.4 m

Thus, the pivot is 1.4 m from the left end. Assess: From geometry, the distance d is expected to be more than 1.0 m but less than 1.5 m. A value of 1.4 m is reasonable.

P8.9. Prepare: In this problem we are given the mass of the board, but because of the symmetry of the distances (the fulcrum, or right support, is right in the center) we don’t need it. We’ll compute the torques around the right support, which is under the center of gravity of the board, so the weight of the board won’t produce a torque.

The weight of the diver is w = mg = (60 kg)(9.8 m/s2 ) = 590 N.

Solve: Because of the symmetry of the situation (the two lever arms are equal in length), we can examine the torque equation in our heads and realize that the force exerted by the hinge must have the same magnitude as the weight of the diver. Therefore, the force the hinge exerts on the board is 590 N. Assess: We computed the torques around the right support because of the symmetry, but also because had we computed them around the hinge on the left, we would be eliminating from the equation the very torque we need to answer the problem. Since both the weight of the diver and the force of the hinge act in the downward direction, it is also clear that the right support must exert a force of twice the diver’s weight in the upward direction.

P8.10. Prepare: The beam is in static equilibrium. Equation 8.1 applies. Solve: Refer to the diagram below.

Equilibrium and Elasticity 8-9

The net torque on the beam is zero. Calculating torques about the pivot, the torque equation in Equation 8.1 gives

rTT!" (r

w)!

w = rTT sin# " r

ww = 0

Solving for the tension,

T =r

ww

rT

sin!=

(0.5 m)(10 kg)(9.80 m/s2 )

(1.0 m)sin30°= 100 N

Assess: Note that using different forms of the definition of torque led to simpler calculations.

P8.11. Prepare: The pole is in equilibrium, which means !" = 0; it is convenient to compute the torques

around the bottom of the pole (as suggested by the hinge there). It is definitely worth noting that the triangles made by the wires, the ground, and the pole are congruent (rotate one triangle 90° around corner at the bottom of the pole to see this). Solve: For !" = 0 we want the magnitude of the clockwise torque to equal the magnitude of the counterclockwise torque. The magnitude of each torque is the force (the tension in the guy wire) multiplied by the lever arm (the perpendicular distance from the line in the direction of the force to the pivot, see Equation 7.4). Examination of the figure, and realization that the triangles are congruent, leads to the realization that the lever arms are equal for the two wires. Since the magnitudes of the torques must be equal, and the lever arms are equal, then the magnitudes of the forces must be equal. Hence the ratio of the tension in the left wire to the tension in the right wire is 1

(T

2/T

1= 1).

Assess: Theoretically, the pole could be at rest in equilibrium if there were no wires, but any slight perturbation would make it fall over (an example of unstable equilibrium), and so guy wires are used which can keep the pole in equilibrium even if perturbed. Whether it is preferable to use guy wires like the one on the left or like the one on the right may depend on how much room around the base of the pole is available, and where one would rather attach the wires to the pole.

P8.12. Prepare: As long as the center of gravity of the board is over one of the tables, the board will not tilt. Solve: See the figure below.

When the end of the board reaches the other table its center of gravity must still be over the first table, otherwise it will tilt. The board’s center of gravity should be 20 cm from one of its ends. Assuming the board is uniform, the board must be at least 40 cm long. Assess: For an object to be stable, the center of gravity of the object must lie over its base of support.

P8.13. Prepare: The center of gravity of the magazine rack must be over the base of support to be stable. In this case the rule of thumb given in the text that “a wider base of support and/or a lower center of gravity improve stability” indicates that we expect the tipping angle to be small. The center of gravity need only move 2.5 cm horizontally for the rack to be on the verge of tipping.

8-10 Chapter 8

On the diagram construct a right triangle by first dropping a vertical from the center of gravity to the middle of the base and drawing the hypotenuse from the center of gravity to the edge of the base. Now we have a right triangle with legs of 16 cm and 2.5 cm. The angle we desire is the small angle at the top of the triangle.

If this is not yet clear, draw a second diagram with the rack tipped just so the center of gravity is over the edge of the base of support and see that the tipping angle is the same! that we labeled in the triangle. Solve: The angle will be given by the arctangent of the opposite side over the adjacent side.

! = tan"1 2.5 cm

16 cm

#

$%&

'(= 8.9°

Assess: The angle of 8.9° is fairly small, as expected. The precise shape of the cross section of the rack is unimportant as long as the base of support and the center of gravity are where they are.

P8.14. Prepare: In order not to tip, the vehicle’s center of gravity must be over its base of support. Solve: See the following figure.

The figure shows the vehicle just as it is about to tip. The center of gravity is at the edge of the base of support. Solving for the height,

h =1.0 m

tan(47°)= 0.93 m

Assess: Note that as the angle decreases the height required decreases.

P8.15. Prepare: Review Example 8.1, as this problem is very similar. We must find where the box will be when the combined center of gravity of the box and board is exactly over the edge of the table. We’ll use the edge of the table as the pivot point to calculate the torques. We will assume the board is uniform so that its center of gravity is at its geometrical center, which is 5.00 cm (which is 0.0500 m) left of the table edge. We will also assume that the box is uniform and that its center of gravity is at its geometrical center. The counterclockwise torque due to the board will be

!

counterclockwise= (r

board)(w

board) = (r

board)(m

boardg) = (0.0500 m)(6.00 kg)(9.80 m/s2 ) = 2.94 N "m

Solve: In equilibrium !

net= 0:

Equilibrium and Elasticity 8-11

!

net= !

counterclockwise" !

clockwise= (2.94 N #m) " (r

box)w

box= (2.94 N #m) " (r

box)m

boxg = 0.00 N #m

Solve for r

box, which is where the center of the box will be to just balance. This is the answer to the problem.

rbox

=2.94 N !m

mbox

g=

2.94 N !m

(2.00 kg)(9.80 m/s2 )= 0.150 m = 15 cm

Assess: The answer of 15.0 cm is the position of the center of the box, to the right of the edge of the table. It is still on the board (not off the edge), and three times farther from the pivot than the center of gravity of the board because the box is three times less massive than the board.

P8.16. Prepare: In order for the object not to tip over, the center of gravity of the object must be over its base of support. Solve: See the figure below.

Choose the origin at the bottom right corner of left segment delineated above. The mass of each segment is equal to the density of the material times the volume of the object. Assume the object has a uniform thickness T. The

mass of the left segment is M

L= !H

LW

LT . The mass of the right piece is

M

R= !H

RW

RT = !H

RxT .

The center of gravity of the right segment is located at the center of the right segment, a distance of x/2 from the origin. The center of gravity of the object is

xcg

=M

L(!0.01 m) + M

R(x/2)

ML

+ MR

The object will tip over if the x-position of its center of gravity passes the base of support, which is at the origin. Setting the x-position of the center of gravity is equal to zero in the equation above we have

M

L(!0.01 m) + M

R(x/2) = !"H

LW

LT (!0.01 m) + "H

RT (x

2/2) = 0

Solving for x,

x =2H

LW

L(0.01 m)

HR

=2(0.01 m)(0.04 m)(0.02 m)

(0.02 m)= 0.028 m

The largest x can be is 2.8 cm. Assess: This choice of origin makes solving the resulting equation much simpler.

P8.17. Prepare: Hooke’s law is given in Equation 8.3, (F

sp)

x= !k"x. It relates the force on a spring to the

stretch; the constant of proportionality is k, the spring constant that we are asked to find. Solve: The minus sign in Equation 8.3 simply indicates that the force and the stretch are in opposite directions (that the force is a restoring force); k is always positive, so we’ll drop the minus sign and just use magnitudes of (Fsp)x and !x since we would otherwise have to set up a more explicit coordinate system. See Equation 8.1.

k =(F

sp)

x

!x=

25 N

0.030 m= 830 N/m

Assess: This result indicates a fairly stiff spring, but certainly within reason.

8-12 Chapter 8

P8.18. Prepare: For small displacements, the DNA obeys Hooke’s law, Equation 8.3. Solve: The force the DNA exerts is equal and opposite to the force the tweezers exert. See the figure below.

Applying Equation 8.3,

k = !

Fsp

"x= !

!1.5#10!9 N

5.0 #10!9 m= 0.30 N/m

Assess: Note that Equation 8.3 refers to the force the spring exerts, not the force applied to the spring.

P8.19. Prepare: A visual overview below shows the details, including a free-body diagram, of the problem. We will assume an ideal spring that obeys Hooke’s law.

Solve: (a) The spring force or the restoring force is F

sp= !k"y. For

!y = –1.0 cm and the force in Newtons,

F

sp= F = !k"y # k = !F/"y = !F/(!0.01 m) = 100F N/m

Notice that !y is negative, so

F

spis positive.

We can now calculate the new length for a restoring force of 3F:

F

sp= 3F = !k"y = (!100 F )"y # "y = !0.03 m

From !y = y – ye = –0.03 m, or y = –0.03 m + ye, or y = –0.03 m + (–0.10 m) = –0.13 m, the length of the spring is 0.13 m. (b) The new compressed length for a restoring force of 2F can be calculated as:

F

sp= 2F = !k"y = (!100 F )"y " !y = –0.02 m

Using !y = ye – y = –0.02 m, or y = 0.02 m + ye, or y = 0.02 m + (– 0.10 m) = – 0.08 m, the length of the compressed spring is 0.08 m. Assess: The stretch !x is proportional to the applied force, as both parts of this problem demonstrate. Of course, this bet is off if the spring is stretched or compressed far enough to take it out of the linear region.

Equilibrium and Elasticity 8-13

P8.20. Prepare: A visual overview below shows the details, including a free-body diagram, of the problem. We will assume an ideal spring that obeys Hooke’s law.

Solve: (a) The spring force on the 2.0 kg mass is F

sp= !k"y. Notice that

!y is negative, so

F

spis positive. This

force is equal to mg, because the 2.0 kg mass is at rest. We have !k"y = mg. Solving for k:

k = !(mg/"y) = !(2.0 kg)(9.8 m/s2 )/(!0.15 m ! (!0.10 m)) = 392 N/m = 390 N/m

(b) Again using !k"y = mg:

!y = "mg/k = "(3.0 kg)(9.8 m/s2 )/(392 N/m) = "0.075 m

#y " ye

= "0.075 m $ #y = ye" 0.075 m = "0.10 m " 0.075 m = "0.175 m = "18 cm

The length of the spring is 18 cm when a mass of 3.0 kg is attached to the spring. Assess: The position of the end of the spring is negative because it is below the origin, but length must be a positive number. We expected the length to be a little more than 15 cm.

P8.21. Prepare: We do not know the original unstretched length of the spring, but that’s okay, since what Hooke’s law tells us is the stretch, or

!x, and that is given. We will assume that this problem is in the linear

region of the spring. We’ll use the initial data to find k and then compute the new stretch. The restoring force is equal in magnitude to

the weight of the hanging mass; in the initial data this is w = mg = (0.20 kg)(9.8 m/s2 ) = 1.96 N.

Solve: Solve Equation 8.1 for k.

k =F

sp

!x=

1.96 N

0.050 m= 39.2 N/m

If the 0.20 kg block is replaced with a 0.70 kg block, the new stretch (measured from the unstretched length) is

!x =F

sp

k=

mg

k=

(0.70 kg)(9.8 m/s2 )

39.2 N/m= 0.175 m = 17.5 cm " 18 cm

Assess: The above method is actually the long way. The short way is to realize that since we are in the linear region of the spring, the stretch is directly proportional to the force, so the new stretch must be 7/2 (that is, 0.70 kg/ 0.20 kg) of the old one.

7

2(5.0 cm) = 17.5 cm ! 18 cm

P8.22. Prepare: Assume the spring is massless and obeys Hooke’s law.

8-14 Chapter 8

Solve: The spring constant of the spring can be determined from the original stretch. The spring is at its equilibrium length before the first mass is added. Once the mass is added, assume it is in equilibrium. The net force on the mass is zero.

From Equation 8.1, F

sp= mg. Using Equation 8.3,

k = !F

sp

"y= !

mg

"y= !

(1.2 kg)(9.80 m/s2 )

!(0.024 m)= 490 N/m

When the new mass is added the spring stretches by

!y = "F

sp

k= "

mg

k= "

(1.8 kg)(9.80 m/s2 )

490 N/m= "0.036 m

With both masses, the spring is stretched 3.6 cm. It has stretched an additional 3.6 cm ! 2.4 cm = 1.2 cm.

Assess: The stretch in the spring is always measured from its equilibrium length.

P8.23. Prepare: Assume that the spring is ideal and obeys Hooke’s law. Also we will model the sled as a

particle. The only horizontal force acting on the sled is

!F

sp. A pictorial representation and a free-body diagram are

shown below.

Solve: Applying Newton’s second law to the sled gives

!(F

on sled)

x= F

sp= ma

x" k#x = ma

x" a

x= k#x/m = (150 N/m)(0.20 m)/20 kg = 1.5 m/s2

P8.24. Prepare: Assume that the spring is ideal and obeys Hooke’s law. According to Hooke’s law, the

spring force acting on a mass (m) attached to the end of a spring is given as F

sp= k!y , where !y is the change in

length of the spring. If the mass m is at rest, then Fsp is also equal to the weight w = mg.

Solve: We have F

sp= k!y = mg. We want a 0.10 kg mass to give

!y = 0.010 m. This means

k = mg/!y = (0.10 kg)(9.8 N/m)/(0.010 m) = 98 N/m

Assess: If you double the mass and hence the weight, the displacement of the end of the spring will double as well.

P8.25. Prepare: Rearrange Equation 8.6 to see that the stretch is proportional to the length (for part (a)) and inversely proportional to the area (for part (b)).

!L =LF

AY

Solve: (a) In this part everything on the right side of the equation stays constant except the length L. Since the length of the second wire is twice the length of the first wire, then the second wire will stretch twice as much by the same force. So the answer is 2 mm. (b) In this part everything on the right side of the equation stays constant except the cross-sectional area

A. The

cross-sectional area of the third wire is four times the area of the first wire, since A = !r

2= ! (D/2)2 and the

diameter of the third wire is twice the diameter of the first wire, so the third wire will stretch one-quarter as much by the same force. The answer is 0.25 mm.

Equilibrium and Elasticity 8-15

Assess: This problem is worth mentally reviewing to make sure the explanation given makes sense, and to tuck the results away as tidbits of practical knowledge.

P8.26. Prepare: The hanging mass creates tensile stress in the wire. The force (F) pulling on the wire, which is simply the weight (mg) of the hanging mass, produces tensile stress given by F/A, where A is the cross-sectional area of the wire. We will use Equation 8.6 to find the hanging mass. Solve: From the definition of Young’s modulus, we have

Y =mg/A

!L/L" m =

(#r 2 )Y!L

gL=# (2.50 $10%4 m)2(20 $1010 N/m2 )(1.0 $10%3 m)

(9.80 m/s2 )(2.0 m)= 2.0 kg

Assess: A 1.0 mm stretch of a 2.0 m wire by 2.0 kg hanging mass is reasonable.

P8.27. Prepare: Equation 8.6 relates the quantities in question.

Look up Young’s modulus for steel in Table 8.1: Y

steel= 20 !10

10N/m

2.

Convert all length data to meters: D = 1.0 cm = 0.010 m, !L = 5.0 mm = 0.0050 m.

Assume a circular cross section: A = !r

2= ! ( D

2)2

= ! (0.0050 m)2= 7.85"10#5 m2.

Solve:

F =YA

L!L =

(20 "1010 N/m2 )(7.85"10#5 m2 )

10 m0.0050 m = 7900 N

This is the force required to stretch a steel cable of the given length and diameter by 5.0 mm. Assess: A 1-cm-diameter cable is fairly substantial, so it ought to take a few thousand newtons to stretch it 5.0 mm. Notice the m2 cancel in the numerator and so do the other m, leaving only N.

P8.28. Prepare: Turning the tuning screws on a guitar string creates tensile stress in the string. The tensile stress in the string is given by T/A, where T is the tension in the string and A is the cross-sectional area of the string. Solve: From the definition of Young’s modulus,

Y =T/A

!L/L" !L =

T

A

L

Y

#

$%&

'(

Using T = 2000 N, L = 0.80 m, A = ! (0.0005 m)2, and Y = 20 ! 1010 N/m2 (from Table 8.1), we obtain "L = 0.010 m = 1.0 cm. Assess: 1.0 cm is a large stretch for a length of 80 cm, but 2000 N is a large tension.

P8.29. Prepare: Equation 8.6 relates the quantities in question. Since we know neither the material (and therefore Y ) nor the length nor the cross-sectional area, the best way to proceed is by ratios.

Known

F1

= 2000 N

!L1

= 0.001 m

L2

= 2L1

D2

= 2D1; A

2= 4A

1A scales as the square of r or D

!L2

= 0.001 m = !L

L3

= 2L1

D3

= D1A

3= A

1

F3

= 4000 N = 2F1

Find

F

2

!L

3

8-16 Chapter 8

Solve: (a)

F2

F1

=(YA2 /L

2)!L

2

(YA1/L1)!L

1

=Y (4A

1)/(2L

1)!L

1

(YA1/L

1)AL

1!L

1

=4

2= 2

So F

2= 2F

1= 2(2000 N) = 4000 N.

(b) Assume the third wire is made of the same material as the first and second wires. Solve Equation 8.6 for !L.

!L3

!L1

==L

3F

3/A

3Y

L1F

1/A

1Y

=(2L

1)(2F

1)/A

1Y

L1F

1/A

1Y

= 4

So !L

3= 4!L

1= 4(1 mm) = 4 mm.

Assess: This ratio technique is very powerful, especially when unknown quantities cancel out.

P8.30. Prepare: Equation 8.6 applies. Solve: The engine is in equilibrium so the weight of the engine must equal the restoring force exerted by the rod. See the figure below.

From Equation 8.1, F = w. Young’s modulus for steel is given in Table 8.1. Solving Equation 8.6 for the stretch gives

!L =FL

YA=

FL

Y"r2

=(4700 N)(1.2 m)

(20 #1010 N/m2 )" (0.0025 m)2= 1.4 #10$2 m

The rod stretches 1.4 cm. Assess: This result makes sense.

P8.31. Prepare: Equation 8.6 relates the quantities in question; solve it for !L, which is what we want to know.

Look up Young’s modulus for steel in Table 8.1: Y

steel= 20 !10

10N/m

2.

Assume a circular cross section: A = !r

2= ! (D/2)2

= ! (0.0125 m)2= 4.91"10#4 m2.

We are also given that L = 500 m. Compute

F = w = mg = (3000 kg)(9.8 m/s2 ) = 29,400 N.

Solve:

!L =LF

AY=

(500 m)(29,400 N)

(4.91"10#4 m2 )(20 "1010 N/m2 )= 0.15 m = 15 cm

Assess: 15 cm is quite a stretch, but 3000 kg (times g) is quite a bit of weight, and 500 m is quite a long cable, so the answer is reasonable. The design of the shaft would have to take this 15 cm stretch into account. Also check to see that the units work out.

P8.32. Prepare: Solve Equation 8.7 for !L . The force is 3 mg. Solve:

!L =F L

AY=

3(80 kg)(9.8 m/s2 )(0.52 m)

(5.2 "10#4 m2 )(1.6 "1010 N/m2 )= 0.15 mm

Assess: That’s a small compression, but bone is strong, and we wouldn’t want it to compress much more than that.

Equilibrium and Elasticity 8-17

P8.33. Prepare: Equation 8.6 relates the quantities in question; the fractional decrease in length will be

!L/L, so rearrange the equation so !L/L is isolated.

Look up Young’s modulus for Douglas fir in Table 8.1: Y

Douglas fir= 1!1010 N/m2.

The total cross section will be three times the area of one leg:

Atot

= 3(!r2 ) = 3 ! D

2( )2"

#$%&'

= 3! (0.010 m)2

= 9.42 !10

"4m2

Compute F = w = mg = (75 kg)(9.8 m/s2 ) = 735 N.

Solve:

!L

L=

F

AY=

735 N

(9.42 "10#4 m2 )(1"1010 N/m2 )= 7.8"10#5

This is a 0.0078% change in length. Assess: We were not given the original length of the stool legs, but regardless of the original length, they decrease in length by only a small percentage—0.0078%—because F isn’t large but A is.

P8.34. Prepare: The load supported by a concrete column creates compressive stress in the concrete column. The weight of the load produces tensile stress given by F/A, where A is the cross-sectional area of the concrete column and F equals the weight of the load. Solve: From the definition of Young’s modulus,

Y =F/A

!L/L" !L =

F

A

#

$%&

'(L

Y

#

$%&

'(=

200,000 kg ) 9.8 m/s2

* (0.25 m)2

#

$%&

'(3.0 m

3)1010 N/m2

#

$%&

'(= 1.0 mm

Assess: A compression of 1.0 mm of the concrete column by a load of approximately 200 tons is reasonable.

P8.35. Prepare: The beam is a rigid body of length 3.0 m and the student is a particle.

!F

1and

!F

2are the

normal forces on the beam due to the supports,

!w

beamis the weight of the beam acting at the center of gravity,

and

!w

studentis the student’s weight. The student is 1 m away from support 2.

Solve: To stay in place, the beam must be in both translational equilibrium (!F

net=

!0 N) and rotational equilibrium

(!net = 0 Nm). The first condition is

!F

y= "w

beam" w

student+ F

1+ F

2= 0 N # F1 + F2 = w

beam+ w

student= (100 kg + 80 kg) (9.80 m/s2 ) = 1764 N

Taking the torques about the left end of the beam, the second condition is

!wbeam (1.5 m) – wstudent (2.0 m) + F2 (3.0 m) = 0 N m

! (100 kg)(9.8 m/s2)(1.5 m) – (80 kg)(9.8 m/s2)(2.0 m) + F2 (3.0 m) = 0 Nm ! F2 = 1013 N

From F1 + F2 = 1764 N, we get F1 = 1764 N – 1013 N = 750 N. Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point on the body of interest.

8-18 Chapter 8

P8.36. Prepare: The structure is a rigid body.

!F

hand

!F

vare the horizontal and vertical components of the

force on the beam by the wall, !T is the tension in the cable,

!w

Wis the worker’s weight, and

!w

Bis the weight of the

beam acting at its center of gravity. We pick the left end of the beam as our pivot point. We don’t need to know the forces Fh and Fv because the pivot point passes through the line of application of Fh and Fv and therefore these forces do not exert a torque.

Solve: For the beam to stay in equilibrium, the net torque about this point is zero. We can write

!about left end = – wB (3.0 m) – wW (4.0 m) + (T sin 150°)(6.0 m) = 0 N m

Using wB = (1450 kg)(9.8 m/s2) and wW = (80 kg)(9.80 m/s2), the torque equation can be solved to yield T = 15,000 N. The tension in the cable is slightly more than the cable rating. The worker should be worried.

P8.37. Prepare: Neglect the weight of the arm. The arm is 45! below horizontal which introduces sin45! in

each term, but it will cancel out. Solve:

!" = T (14 cm)sin 45! #(10 kg)(9.8 m/s2 )(35 cm)sin 45! = 0 $ T = 860 N

Assess: This answer is in the general range of the results in the example.

P8.38. Prepare: Compute torques around the knees to find the normal force on the hands. Then use !F = 0 to find the normal force on the knees.

Solve: (a)

!" = (580 N)(54 cm) # n(76 cm) = 0 $ n = 412 N

Divide this by two to find the normal force on each hand: 412 N / 2 = 206 N ! 210 N (b) The sum of the upward normal force on the hands and the upward normal force on the knees must equal the weight. Therefore the normal force on both knees is 580 N ! 412 N = 168 N. Divide this by two to get the force on each knee: 168 N / 2 = 84 N. Assess: It makes sense that the force on the hands is greater than the force on the knees since the center of gravity is closer to the arms than the knees.

Equilibrium and Elasticity 8-19

P8.39. Prepare: Assume equilibrium and compute the torques around the point labeled “pivot.” The angle between the lever arm and the force is 60° for the weight of the torso and the head and arms. Solve: (a)

!" = T2

3L

#

$%&

'(sin 12° ) (320 N)

1

2L

#

$%&

'(sin 60° ) (160 N)(L)sin 60° = 0

T =(320 N)(L)sin 60°

2

3Lsin 12°

= (480 N)sin 60°

sin 12°= 2000 N

(b) Align the x -axis with the spine to find the force from the pelvic girdle.

!F

x= F

p.g ." (320 N)cos 60° " (160 N)cos 60° " (2000 N)cos 12° = 0

F

p.g .= 2196 N ! 2200 N

Assess: These are large forces, which is why they can cause back injuries. Squat when you lift.

P8.40. Prepare: Following the example, let !d be the new distance from the pivot to the woman's center of gravity, with

!F = 260N. We want to know

!d " d. We know from the example that d = 0.9167 m.

Solve:

!d =L !F "

1

2Lw

b

w=

(2.5 m)(260 N) "1

2(2.5 m)(60 N)

600 N= 0.9583 m

Now !d " d = 0.9167 m " 0.9583 m = 0.0416 m # 4.2 cm.

Assess: A few centimeters seems reasonable.

P8.41. Prepare: Assume that the flagpole has a uniform diameter so that its center of mass is at the center,

7.5 m/2 = 3.75 m from either end. Since the pole is motionless it is in equilibrium, so !

net= 0.

The magnitude of the counterclockwise torque due to the tension in the rope must equal the magnitude of the clockwise torque due to the weight of the flagpole. Use Equation 7.5 for the torques:

! = rF sin" .

Solve: The clockwise torque is due to the weight of the flagpole. Draw a quick right triangle to see that the angle between r and F (that is, w) is 60°.

!

clockwise= rF sin" = rwsin" = r(mg)sin" = (3.75 m)(28 kg)(9.8 m/s2 )sin60° = 891 N #m

In the counterclockwise direction we want to know the force (tension in the rope), so solve the torque equation for F and put in the previous result for the torque. This time r is the distance from the pivot at the bottom of the pole to where the rope is attached at the top of the pole, or 7.5 m.

Frope

=!

r sin"=

891 N #m

(7.5 m)(sin20°)= 350 N

Assess: The man must exert 350 N of force because the angle 20° is so small. One way to increase the angle (and the sine of the angle) is to use a longer rope and stand farther back. This will slightly decrease the needed force. By doing this r is not changed; it is still the length of the pole if the rope is attached to the top of the pole.

P8.42. Prepare: Use both conditions of equilibrium on just one leg of the ladder—say the left one. The right leg will exert a force F on the top of the left leg along the direction of the right leg. Due to the wheels, we will assume no friction from the floor. Because the chain is light we will assume it is massless and horizontal. Each leg of the ladder weighs

mg = 100N.

8-20 Chapter 8

Solve: Sum the horizontal forces on the left leg of the ladder. There is none from the floor since we assumed no friction because of the wheels.

!F = T " F cos 65° = 0 # F =T

cos 65°

Compute the torques on the left leg around its wheel on the floor.

!" = FL sin 130° # mgL

2

$

%&'

()sin 25° #T

L

2

$

%&'

()sin 115° = 0

Substitute for F and cancel L.

T

cos 65°sin 130° !T

1

2

"

#$%

&'sin 115° = mg

1

2

"

#$%

&'sin 25°

Solve for T .

T =

mg

2sin 25°

sin 130°

cos 65°!

sin 115°

2

"#$

%&'

= 15.54 N ( 16 N

Assess: This answer isn't much compared to the weight of the ladder, but that is why a light chain can suffice. The tension in the chain would be larger if the apex half-angle were larger.

P8.43. Prepare: Model the beam as a rigid body. For the beam not to fall over, it must be both in translational

equilibrium (!F

net=

!0 N) and rotational equilibrium

(!!

net= 0 N "m). The boy walks along the beam a distance x,

measured from the left end of the beam. There are four forces acting on the beam.

!F

1and

!F

2are from the two

supports,

!w

bis the weight of the beam, and

!w

Bis the weight of the boy.

Equilibrium and Elasticity 8-21

Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium is

–wb (2.5 m) + F2 (3.0 m) – wBx = 0 N !m

–(40 kg)(9.80 m/s2)(2.5 m) + F2 (3.0 m) – (20 kg)(9.80 m/s2)x = 0 N !m

The equation for translation equilibrium is

!F

y= 0 N = F

1+ F

2" w

b" w

B# F

1+ F

2= w

b+ w

B= (40 kg + 20 kg)(9.8 m/s2 ) = 588 N

Just when the boy is at the point where the beam tips, F1 = 0 N. Thus F2 = 588 N. With this value of F2, we can simplify the torque equation to

–(40 kg)(9.80 m/s2)(2.5 m) + (588 N)(3.0 m) – (20 kg)(9.80 m/s2)x = 0 N !m ! x = 4.0 m

Thus, the distance from the right end is 5.0 m – 4.0 m = 1.0 m.

P8.44. Prepare: Compute the torques around the point where the wheel touches the curb. Because the height of the curb is

R / 2, the angle a radius from the center to the curb makes with the horizontal is 30° (it therefore

makes a 60° angle with the vertical). When we just barely lift the wheel off the ground the ground does not exert a force (or a torque) on the wheel. Solve:

!" = mgR sin 60° # FR sin 30° = 0 $ F = mgsin 60°

sin 30°= 1.7mg

Assess: It makes sense that the force required would be proportional to the weight of the wheel, and more than it would take to just lift it straight up.

P8.45. Prepare: Assume that the spring is ideal and it obeys Hooke’s law. We also model the 5.0 kg mass as a particle. We will use the subscript s for the scale and sp for the spring. With the y-axis representing vertical positions, pictorial representations and free-body diagrams are shown for parts (a) through (c).

Solve: (a) The scale reads the upward force F

s on mthat it applies to the mass. Newton’s second law gives

!(F

on m)

y= F

s on m" w = 0 # F

s on m= w = mg = (5.0 kg)(9.8 m/s2 ) = 49 N

(b) In this case, the force is

!(Fon m

)y= F

s on m+ F

sp" w = 0 # 20 N + k$y " mg = 0 # k = (mg " 20 N)/$y

= (49 N " 20 N)/0.02 m = 1450 N/m

(c) In this case, the force is

!(F

on m)

y= F

sp" w = 0 # k$y " mg = 0 # $y = mg /k = (49 N)/(1450 N/m) = 0.0338 m = 3.4 cm

8-22 Chapter 8

P8.46. Prepare: Assume both the springs obey Hooke’s law and that they are arranged equal distances from the center of gravity of the box. Solve: Since both springs have the same spring constant, each stretches the same amount. The net force in the y-direction is zero, so

F + F ! mg = 0

or F = mg/2.

Using this to calculate the stretch of one of the spring gives

!y = "F

k= "

mg

2k= "

(2.00 kg)(9.80 m/s2 )

2(240 N/m)= "4.08#10"2 m

Each spring stretches downward 4.08 cm. Assess: Note that if the springs were not equidistant from the center of gravity of the box, the two forces would be different.

P8.47. Prepare: Both springs are compressed by the same amount: !x = 1.00 cm. Each spring obeys

Hooke’s law (we assume we are in the linear region of the springs) and so exerts a force back on the block with a

magnitude of F

sp= k!x. The net spring force will simply be the sum of the two individual spring forces.

Solve:

(Fsp

)1

= k1!x = (12.0 N/cm)(1.00 cm) = 12.0 N

(Fsp

)2

= k2!x = (5.4 N/cm)(1.00 cm) = 5.4 N

(Fsp

)tot

= (Fsp

)1+ (F

sp)

2= 12.0 N + 5.4 N = 17.4 N

Assess: We have purposefully omitted the negative sign in Hooke’s law since it only reflects the fact that the force and the stretch are in opposite directions—something we had kept in mind, but did not worry about since we only needed the magnitude of the forces. These two springs are said to be in parallel, and they are equivalent to one spring whose spring constant is the sum of the spring constants of the two parallel springs.

P8.48. Prepare: Assume each spring obeys Hooke’s law and that the block is prevented from rotating. Solve: The net force that the springs exert on the block is the sum of the forces exerted by each spring. Labeling the spring with the smaller equilibrium length spring 1,

F

1+ F

2= !k"x

1! k"x

2= !k("x

1+ "x

2) = !(130 N/m)((0.005 m) + (0.025)) = !3.9 N

Assess: Compare to Problem 8.46.

P8.49. Prepare: When the block is in its equilibrium position, !!F =

!0. So the force exerted to the right by the

left spring must be balanced by a force exerted to the left by the right spring. The force exerted by the left spring is

(F

sp)

1= (10 N/m)(0.020 m) = 0.20 N.

In part (b) the left spring will exert a force to the left (because it will be stretched to the right); the right force will also exert a force to the left since it will be quite compressed. Remember, the block is no longer in equilibrium,

and so we do not expect !!F =

!0.

Solve: (a) The force exerted by the right spring must be 0.20 N. This will be produced if

!x2

=F

sp

k2

=0.20 N

20 N/m= 0.010 m = 1.0 cm

(b) The spring on the left will be stretched 15 cm ! 2.0 cm = 13 cm beyond its unstretched length. The force will

be toward the left.

(F

sp)

1= k

1!x

1= (10 N/m)(0.13 m) = 1.3 N

The spring on the right will be compressed 15 cm +1.0 cm = 16 cm from its uncompressed length. This force will

also be toward the left.

Equilibrium and Elasticity 8-23

(F

sp)

2= k

2!x

2= (20 N/m)(0.16 m) = 3.2 N

Both forces are to the left so we simply add them up.

F

net= (F

sp)

1+ (F

sp)

2= 1.3 N + 3.2 N = 4.5 N

to the left. Assess: In part (b) the right spring exerted a greater force because its k is greater and its length is farther from its unstretched/uncompressed length.

P8.50. Prepare: Assume both springs obey Hooke’s law. Solve: Refer Figure P8.50. Consider spring 2. The net force on this spring must be zero, so the force the block exerts on spring 2 must have the same magnitude as the force spring 1 exerts on spring 2. Therefore the forces on both springs are the same and equal to the force of the block on spring 2.

F = !k

1"x

1= !k

2"x

2

Using !x

2= !x " !x

1and solving for

!x

1in terms of

!x,

! x1

=k

2!x

k1+ k

2

So

F = !k1"x

1=

k1k

2

k1+ k

2

"x =(10 N/m)(20 N/m)

10 N/m + 20 N/m(0.15 m) = 1.0 N

Assess: This result makes sense. The spring with the smaller spring constant stretches more.

P8.51. Prepare: We will model the student (S) as a particle and the spring as obeying Hooke’s law. The only two forces acting on the student are his weight and the force due to the spring.

Solve: According to Newton’s second law the force on the student is

!(F

on S)

y= F

spring on S" w = ma

y# F

spring on S= w + ma

y= mg + ma

y= (60 kg)(9.8 m/s2

+ 3.0 m/s2 ) = 768 N

since F

spring on S= F

S on spring= k!y, k!y = 768 N. This means !y = (768 N)/(2500 N/m) = 0.307 m = 0.31 m.

P8.52. Prepare: Assume the spring obeys Hooke’s law. Solve: See the following figure.

8-24 Chapter 8

At the bottom of the bounce, the child has two forces on him, the force of gravity downwards and the force of the spring upward. There is a net force in the y-direction that causes his upward acceleration. Newton’s second law applied to the child gives

F

net= F

sp! mg = k"y ! mg = ma

y

Solving for the compression of the spring,

!y =ma

y+ mg

k=

(25 kg)(9.8 m/s2 ) + (25 kg)(9.80 m/s2 )

2.0 "104 N/m= 2.5"10#2 m

The spring compresses by 2.5 cm. Assess: This result makes sense. The force on the spring must be enough to overcome gravity and cause an acceleration upward equal to the acceleration of gravity.

P8.53. Prepare: There seem to be a lot of pieces to this puzzle; where does one start? Since the only horizontal

force on the right block is due to the spring, let’s start there: F

sp= k!x = (1000 N/m)(0.015 m) = 15 N.

We apply Newton’s second law to find the acceleration of the right block. F

net= F

sp= ma:

a = F

sp/m = 15 N/3.0 kg = 5.0 m/s2

For part (b) we note that since the spring’s compression is constant over the time interval the two blocks must move together; i.e., they must have the same acceleration, so the left block’s acceleration is also 5.0 m/s2. Also note that while the spring pushes to the right on the right block, it pushes to the left on the left block. Solve: (a) We now apply Equation 2.11 (since the acceleration is constant) to find the final velocity of the right block at

t = 1.0 s.

(v

x)

f= (v

x)

i+ a

x!t = 3.2 m/s + (5.0 m/s2 )(1.0s) = 8.2 m/s

(b) Think about a simple free-body diagram for the left block and apply Newton’s second law, noting that F

spis

subtracted because it is to the left.

Fnet

= F ! Fsp

= ma

F = Fsp

+ ma = 15 N + (3.0 kg)(5.0 m/s2 ) = 30 N

Assess: This is an interesting situation to analyze from a slightly different perspective. Think about both blocks making up a system (the parts of which are connected by the spring). It will take a certain amount of force F to accelerate that system at 5.0 m/s2. Now isolate the right block and call it a new system. It has half the mass of the old system and so will need only half the force to accelerate it at the same rate. That is, F must accelerate twice as much mass as Fsp does on the right block.

P8.54. Prepare: Equation 8.6 applies. Solve: Equation 8.6 has the form of Hooke’s law with the effective spring constant related to the length, cross-sectional area, and Young’s modulus of the material.

F =YA

L

!

"#$

%&'L

The spring constant is

Equilibrium and Elasticity 8-25

k =YA

L=

Y!r 2

L=

(11"1010 N/m2 )! (2.5"10#3 m)2

5.0 m= 4.3"105 N/m

Assess: This is a relatively large spring constant compared to a normal spring.

P8.55. Prepare: Call the distance from the pivot to the machinery at maximum distance x. Compute the

torques around the pivot at the left end of the plank. The plank's center of gravity is at

3.5 m

2. The cross section

area of the rope is

A = !7.0 mm

2

"

#$%

&'

2

= 3.848(10)5

m2. The maximum tension the rope can support is

T = (6.0 !107 N/m2 )(3.848!10"5 m2 ) = 2309 N.

Solve:

!" = T (3.5 m) # (100 kg)(9.8 m/s2 )3.5 m

2

$

%&'

()# (800 kg)(9.8 m/s2 )x = 0

x =

(2309 N)(3.5 m) ! (100 kg)(9.8 m/s2 )3.5 m

2

"#$

%&'

(800 kg)(9.8 m/s2 )= 81 cm

Assess: 81 cm isn't very far along the plank, but the machinery is heavy. We could have moved the machinery farther out if the rope had been thicker.

P8.56. Prepare: Model the tendon as a rod, for which Equation 8.5 gives

k =YA

L

We are given L = 15 cm = 0.15 m and

A = 110 mm

2= 1.1!10

"4

m2.

Look up Young’s modulus for tendons in Table 8.4: Y = 0.15!10

10N/m2.

Solve:

k =YA

L=

(0.15!1010 N/m2 )(1.1!10" 4 m2 )

0.15 m= 1.1!106 N/m

Assess: This is a large spring constant, which means tendons don’t stretch much under a given force. The units cancel properly.

P8.57. Prepare: Model the disk as a short wide rod. We are asked for the strain—the fractional change in length of the disk. We can solve for !L/L from Equation 8.7.

Assume that the disk is circular so that the area is A = !R2= ! (D/2)2

= ! (0.020 m)2= 0.00126 m2.

The force is half the weight of the person: F =12 mg =

12 (65 kg)(9.8 m/s2 ) = 319 N.

Young’s modulus for cartilage is not given in the chapter, but is in the problem: Y = 1.0 !106 N/m2.

Solve: Solve Equation 8.6 for !L/L.

!L

L=

F

YA=

319 N

(1.0 "1010 N/m2 )(0.00126 m2)= 0.000025 = 0.0025%

Assess: This means the disk compresses by only a tiny amount. This seems reasonable. Notice that the actual thickness of the disk, given as 0.50 cm, is not needed in the calculation of the fractional compression.

P8.58. Prepare: If you stretch a material until it fails, we call the measured ultimate stress the tensile strength. The tensile strength of tendon is given as 100 ! 106 N/m2.

The stress is defined as F/A (see Equation 8.6). So the ultimate stress is F/A = 100 !10

6N/m2.

8-26 Chapter 8

We are given that A = 130 mm

2= 1.3!10

"4m2.

Solve: Solve for F.

F = A(100 !106 N/m2 ) = (1.3!10" 4 m2 )(100 !106 N/m2 ) = 13,000 N

To answer the question we want to know what fraction of 13,000 N is 3900 N.

3900 N

13,000 N= 0.30 = 30%

Assess: We are glad to get an answer in this range. Had we computed a maximum tension less than 3900 N, we would know we did something wrong: 30% of the maximum tension the tendon can handle seems about right.

P8.59. Prepare: We’ll use the data from Example 8.10: m

original= 70 kg and

A

original= 4.8!10"4 m2.

The femur is not solid cortical bone material; we model it as a tube with an inner diameter and an outer diameter. Look up Young’s modulus for cortical bone in Table 8.1. Solve: (a) Both the inner and outer diameters are increased by a factor of 10; however, the cross-sectional area of the bone material does not increase by a factor of 10. Instead, because A = !R2, the outer cross-sectional area and the inner cross-sectional area (the “hollow” of the tube) both increase by a factor of 100. But this means that the cross-sectional area of the bone material (the difference of the outer and inner areas) also increases by a

factor of 100. So the new area is A

new= 100(4.8!10"4 m2) = 4.8!10"2 m2.

(b) Since volume is a three-dimensional concept, if we increase each linear dimension by a factor of 10 then the

volume increases by a factor of 103

= 1000.We assume the density of the man is the same as before, so his mass

increases by the same factor as the volume: m

new= 1000(70 kg) = 70 000 kg.

(c) We follow the strategy of Example 8.10. The force compressing the femur is the man’s weight, F = mg =

(70,000 kg)(9.8 m/s2 ) = 690,000 N. The resulting stress on the femur is

F

A=

690,000 N

4.8!10"2

m2

= 1.4 !107

N/m2

A stress of 1.4 !10

7N/m2 is 14% of the tensile strength of cortical bone given in Table 8.4.

Assess: This scaling problem illustrates clearly why animals of different sizes have different proportions. Because the volume scales with the cube of the linear dimensions and the area scales with the square of the linear dimensions then the force in F/A grows more quickly than the cross sectional area does.

P8.60. Prepare: The amount of stretch is given by Equation 8.6:

!L =FL

YA

where F is the weight of the spider: w = mg = (0.50 g)(9.8 m/s2 ) = 0.0049 N.

We are given L = 0.12 m and

A = !R

2= ! (D/2)2

= ! (0.15 mm/2)2= 1.767 "10#8 m2.

Look up Young’s modulus for spider silk in Table 8.4: Y = 0.2 !10

10N/m.

If you stretch a material until it fails, we call the measured ultimate stress the tensile strength. The tensile

strength of spider silk is given as 1000 !10

6N/m2.

Solve: (a)

!L =FL

YA=

(0.0049 N)(0.12 m)

(0.2 "1010 N/m)(1.767 "10#8 m2 )= 1.7 "10#5 m = 0.017 mm

(b) The stress is defined as F/A (see Equation 8.6). So the ultimate stress is F/A = 1000 !10

6N/m2. Solve for F.

F = A(1000 !106 N/m2 ) = (1.767 !10"8 m2 )(1000 !106 N/m2 ) = 18 N

Assess: The thread of silk can support many times the weight of the spider! As the chapter says, “spider silk has a greater ultimate stress than bone!”

Equilibrium and Elasticity 8-27

P8.61. Prepare: For equilibrium the sum of the forces must be zero. Solve: For one foot:

!F

y= n "

mg

2= 0 # n =

mg

2=

(60kg)(9.8m/s2 )

2= 294N

So the correct choice is B. Assess: This is half the weight of a typical person.

P8.62. Prepare: Compute the torques around the ankle pivot. Solve:

!" = T (5cm) # (294N)(15cm) = 0 $ T = 882N

The correct choice is D. Assess: That's quite a bit of force, but tendons are strong.

P8.63. Prepare: Use Equation 8.7 and solve for !L .

Solve:

!L =FL

AY=

(882N)(0.15m)

(110mm2 )(0.15"1010 N/m2 )= 8.0 "10#4 m = 0.8mm

The correct choice is B. Assess: We expect the tendon to stretch a tiny bit, but not much.