College Physics for AP® Courses 2e

# Chapter 9

## Problems & Exercises

1.

a) $46.8 N·m46.8 N·m$

b) It does not matter at what height you push. The torque depends on only the magnitude of the force applied and the perpendicular distance of the force’s application from the hinges. (Children don’t have a tougher time opening a door because they push lower than adults, they have a tougher time because they don’t push far enough from the hinges.)

3.

23.3 N

5.

Given:

$m1 = 26.0 kg, m2= 32.0 kg, ms= 12.0 kg, r1 = 1.60 m, rs= 0.160 m, find (a) r2, (b) Fp m1 = 26.0 kg, m2= 32.0 kg, ms= 12.0 kg, r1 = 1.60 m, rs= 0.160 m, find (a) r2, (b) Fp$

a) Since children are balancing:

$net τ cw = – net τ ccw ⇒ w 1 r 1 + m s gr s = w 2 r 2 net τ cw = – net τ ccw ⇒ w 1 r 1 + m s gr s = w 2 r 2$

So, solving for $r2r2$ gives:

$r2 = w1r1+msgrs w2 = m1 gr1 + ms grs m2 g = m1r1+msrs m2 = (26.0 kg) (1.60 m) + (12.0 kg) (0.160 m) 32.0 kg = 1.36 m r2 = w1r1+msgrs w2 = m1 gr1 + ms grs m2 g = m1r1+msrs m2 = (26.0 kg) (1.60 m) + (12.0 kg) (0.160 m) 32.0 kg = 1.36 m$

b) Since the children are not moving:

$net F = 0 = F p – w 1 – w 2 – w s ⇒ F p = w 1 + w 2 + w s net F = 0 = F p – w 1 – w 2 – w s ⇒ F p = w 1 + w 2 + w s$

So that

$Fp = ( 26.0 kg+ 32.0 kg+ 12.0 kg ) ( 9.80 m/ s2 ) = 686 N Fp = ( 26.0 kg+ 32.0 kg+ 12.0 kg ) ( 9.80 m/ s2 ) = 686 N$
6.

$F wall = 1.43 × 10 3 N F wall = 1.43 × 10 3 N$

8.

a) $2.55×103 N, 16.3º to the left of vertical (i.e., toward the wall)2.55×103 N, 16.3º to the left of vertical (i.e., toward the wall)$

b) 0.292

10.

$F B = 2.12 × 10 4 N F B = 2.12 × 10 4 N$

12.

a) 0.167, or about one-sixth of the weight is supported by the opposite shore.

b) $F=2.0×104NF=2.0×104N$, straight up.

14.

a) 21.6 N

b) 21.6 N

16.

350 N directly upwards

19.

25

50 N

21.

a) $MA=18.5MA=18.5$

b) $Fi=29.1 NFi=29.1 N$

c) 510 N downward

23.

$1 . 3 × 10 3 N 1 . 3 × 10 3 N$

25.

a) $T=299 NT=299 N$

b) 897 N upward

26.

$F B = 470 N; r 1 = 4.00 cm; w a = 2.50 kg; r 2 = 16.0 cm; w b = 4.00 kg; r 3 = 38.0 cm F E = w a r 2 r 1 − 1 + w b r 3 r 1 − 1 = 2.50 kg 9.80 m / s 2 16.0 cm 4.0 cm – 1 + 4.00 kg 9.80 m / s 2 38.0 cm 4.00 cm – 1 = 407 N F B = 470 N; r 1 = 4.00 cm; w a = 2.50 kg; r 2 = 16.0 cm; w b = 4.00 kg; r 3 = 38.0 cm F E = w a r 2 r 1 − 1 + w b r 3 r 1 − 1 = 2.50 kg 9.80 m / s 2 16.0 cm 4.0 cm – 1 + 4.00 kg 9.80 m / s 2 38.0 cm 4.00 cm – 1 = 407 N$

28.

$1.1 × 10 3 N θ = 190 º ccw from positive x axis 1.1 × 10 3 N θ = 190 º ccw from positive x axis$

30.

$F V = 97 N, θ = 59º F V = 97 N, θ = 59º$

32.

(a) 25 N downward

(b) 75 N upward

33.

(a) $FA=2.21×103NFA=2.21×103N$ upward

(b) $FB=2.94×103NFB=2.94×103N$ downward

35.

(a) $Fteeth on bullet=1.2×102NFteeth on bullet=1.2×102N$ upward

(b) $FJ=84 NFJ=84 N$ downward

37.

(a) 147 N downward

(b) 1680 N, 3.4 times her weight

(c) 118 J

(d) 49.0 W

39.

a) $x-2=2.33 mx-2=2.33 m$

b) The seesaw is 3.0 m long, and hence, there is only 1.50 m of board on the other side of the pivot. The second child is off the board.

c) The position of the first child must be shortened, i.e. brought closer to the pivot.

## Test Prep for AP® Courses

1.

(a)

3.

Both objects are in equilibrium. However, they will respond differently if a force is applied to their sides. If the cone placed on its base is displaced to the side, its center of gravity will remain over its base and it will return to its original position. When the traffic cone placed on its tip is displaced to the side, its center of gravity will drift from its base, causing a torque that will accelerate it to the ground.

5.

(d)

7.
1. FL = 7350 N, FR = 2450 N
2. As the car moves to the right side of the bridge, FL will decrease and FR will increase. (At exactly halfway across the bridge, FL and FR will both be 4900 N.)
9.

The student should mention that the guiding principle behind simple machines is the second condition of equilibrium. Though the torque leaving a machine must be equivalent to torque entering a machine, the same requirement does not exist for forces. As a result, by decreasing the lever arm to the existing force, the size of the existing force will be increased. The mechanical advantage will be equivalent to the ratio of the forces exiting and entering the machine.

11.
1. The force placed on your bicep muscle will be greater than the force placed on the dumbbell. The bicep muscle is closer to your elbow than the downward force placed on your hand from the dumbbell. Because the elbow is the pivot point of the system, this results in a decreased lever arm for the bicep. As a result, the force on the bicep must be greater than that placed on the dumbbell. (How much greater? The ratio between the bicep and dumbbell forces is equal to the inverted ratio of their distances from the elbow. If the dumbbell is ten times further from the elbow than the bicep, the force on the bicep will be 200 pounds!)
2. The force placed on your bicep muscle will decrease. As the forearm lifts the dumbbell, it will get closer to the elbow. As a result, the torque placed on the arm from the weight will decrease and the countering torque created by the bicep muscle will do so as well.
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