College Physics 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.

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