Applied Physics 9th Edition By Dale Ewen – Test Bank

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Applied Physics 9th Edition By Dale Ewen – Test Bank

44
Chapter 6
6.1
1. 80.0 kg m/s 2. 450 kg m/s 3. 765 slug ft/s 4. 3690 kg m/s 5. 9.5× 108kg m
s
6. 6.1× 105kg m
s
7. m =
Fw
a
=
(1.50 × 105N) 1kg m
s
2
1N








9.80m/ s
= 1.53× 104 kg
p = mv = (1.53× 104 kg)(4.50× 104m/ s)= 6.89 × 108 kgm
s
8. m =
Fw
a
=
(3200lb)
1slug ft
s
2
1lb



⎜⎜



⎟⎟
32.2ft / s
= 99slugs
p = mv = (99slugs) 60
mi
h

⎜⎝

⎟⎠
1h
3600s

⎜⎝

⎟⎠
5280 ft
1mi

⎜⎝

⎟⎠ = 8700slug ft
s
9. (a) p = mv = (180 slugs) 70.0
ft
s

⎜⎝

⎟⎠ = 12,600slug ft / s
(b) v =
p
m
=
12600slug ft
s
80.0slugs
= 158 ft / s
(c) Fw = mg = (180slugs)(32.2 ft / s2 )= 5800 lb; Fw = mg(80.0slugs)(32.2 ft / s2)= 2580lb
10. (a) p = mv = (1.0× 10−3 slug)(700 ft / s)= 0.700slug ft
s (b) v =
p
m
=
0.700slug ft
s
5.00× 10−4 slug
= 140 0 ft / s
11. (a) p = mv = (2300kg)(21.0m/ s) = 55,200kgm
s
(b) v =
p
m
=
55, 200kg m
s
1170kg
×
1km
1000m
×
3600s
1h
= 170 km /h
12. (a) p = mv = (0.50kg)(6.0m / s) = 3.0kgm
s (b) p = 0 because v =0 (c) 3.0 km m/s
13. v2 =
m1v1
m2
=
(0.060kg)(575m/ s)
4.50kg
= 7.67m / s 14. v2 =
m1v1
m2
=
(1.75kg)(30 0m / s)
4.500 kg
= 0.117m/ s
15. (a) v f = vi
2 + 2aavgs = 02 + 2(9.80m/ s2 )(10.0m) = 14.0m/ s
(b) p = mv = (125kg)(14.0m / s) = 1750kg m
s
45
16. (a) p = mv = (0.500kg)(75.0km / h) 1000m
1km

⎜⎝

⎟⎠
1h
3600s

⎜⎝

⎟⎠ = 10.4kg m
s
(b) p = mv = (0.500kg)(85.0km /h + 75.0km /h)×
1000m
1km

⎜⎝

⎟⎠
1h
3600s

⎜⎝

⎟⎠ = 22.2kgm
s
(c) p = mv = (0.500kg)(85.0km /h − 75.0km / h)×
1000m
1km

⎜⎝

⎟⎠
1h
3600s

⎜⎝

⎟⎠ = 1.39kgm
s
17. (a) t =
s
vavg
=
0.660m
230 m / s
= 0.00287s
(b) F =
mvf − mvi
t
=
(0.0750kg)(460 m/ s)− 0
0.00287s
= 12,0 00N
(c) Ft = (12,0 00N)(0.00287s) = 34.4kgm s
(d) p = mv = (0.0750kg)(460m / s)= 34.5kg m
s Note: Answer (d) differs slightly from (c) because
of rounding.
18. (a) t = s
vavg
= 0.550m
263m / s
= 0.00209s
(b) F =
mv f −mv i
t
= (0.0600kg)(525m/ s) − 0
0.00209s
= 15,100N
(c) Ft = (15,100N)(0.00209s) = 31.6kg m s
(d) p = mv = (0.0600kg)(525m / s) = 31.5kg m
s
19. (a) 95.0
km
h
×
1h
3600s
×
1000m
1km
= 26.4m / s
F =
mvf − mvi
t
= (1250kg)(0) − (1250kg)(26.4m/ s)
4.00s
= −8250N
(b) s =
1
2
v f + vi ( )t =
1
2
(0 + 26.4m/ s)(4.00s) = 52.8m
20. (a) 90.0kg
km
h
×
1h
3600s
×
1000m
1km
= 25.0m / s
25.0kg
km
h
×
1h
3600s
×
1000m
1km
= 6.94m/ s
F =
mvf − mvi
t
= (1350kg)(6.94m/ s) − (1350kg)(25.0m/ s)
4.00s
= 610 0N
(b) s =
1
2
v f + vi ( )t =
1
2
(6.9m / s + 25.0m / s)(4.00s) = 63.9m
(c) a =
v f − vi
t
= 69.4m / s − 25.0m/ s
4.00s
= −4.52m / s 2
t =
vf − vi
a
=
0m / s − 25.0m / s
−4.52m / s2 = 5.53s
46
21. 35.0
km
h
×
1h
3600s
×
1000m
1km
= 9.72m / s
F =
mvf − mvi
t
=
(3000kg)(0m / s) − (3000kg)(9.72m / s)
5.00s
= −5830N
22. 10.0
km
h
×
1h
3600s
×
1000m
1km
= 2.77m / s
F =
mvf − mvi
t
=
(5000kg)(0m / s) − (5000kg)(2.77m / s)
6.00s
= −2310N
6.2
1. v2′ =
m1v1 + m2v2 − m1v1′
m2
=
(0.500kg)(6.00m / s)+ (0.200kg)(0m/ s) + (0.500kg)(2.57m / s)
0.200kg
= 8.58m/ s, right
2. (a) v2′ =
m1v1 + m2v2 − m1v1′
m2
=
(625g)(4.00m / s) + (625g)(0m / s) − (625g)(0m/ s)
625g
= 4.00m / s , right
(b) same
3.
v2′ =
m1v1 + m2v2 − m1v1′
m2
=
(0.600kg)(4.00m/ s)+ (1.00kg)(−5.00m / s) − (0.600kg)(−7.25m/ s)
1.00kg
= 1.75m/ s
right
4. v2′ =
m1v1 + m2v2 − m1v1′
m2
=
(90.0g)(3.00m/ s) + (75.0g)(−8.00m/ s) − (90.0g)(−7.00m / s)
75.0g
= 4.00m/ s, left
5. v2′ =
m1v1 + m2v2 − m1v1′
m2
=
(98.0kg)(1.20m / s) + (125.0kg)(−0.750m / s) − (98.0kg)(−0.986m / s)
125kg
= 0.964m / s , right
6. v2′ =
m1v1 + m2v2 − m1v1′
m2
=
(85.0kg)(−1.30m / s) + (75.0kg)(0.965m / s) − (85.0kg)(0.823m / s)
75.0kg
= −1.44m / s , left
7. v’ =
m1v1 + m2v2
m1 + m2
=
(2.00 × 104 kg)(6.00m / s)+ (1.50 × 104 kg)(−4.00m/ s)
2.00× 104 kg + 1.50× 104 kg
=1.71 m/s, north
47
8. v’ =
m1v1 + m2v2
m1 + m2
=
(2.00 × 104 kg)(6.00m / s)+ (1.50 × 104 kg)(0m/ s)
2.00× 104 kg + 1.50× 104 kg
= 3.43m/ s, north
9. v’ =
m1v1 + m2v2
m1 + m2
=
(12.0kg)(6.00m / s)+ (4.00kg)(−3.00m/ s)
12.0kg + 4.00kg
= 3.75m/ s, right
10. v2 =
m1 + v2 ( )v’ −m1v1
m2
=
(15.0kg + 3.00m / s)(1.50m/ s)− (15.0kg)(5.00m / s)
3.00kg
= −16.0m / s (to the left)
11. v2 =
m1 + v2 ( )v’ −m1v1
m2
= (1650kg + 2450kg)(3.00m/ s)− (1650kg)(−12.0m / s)
2450kg
= 13.1m / s
12. v1 =
m1 + v2 ( )v’ −m2v2
m1
=
(16.0g + 4550g)(1.20m / s) − (4550g)(0m / s)
16.0g
= 342m/ s
13. (a) v’ =
m1v1 + m2v2
m1 + m2
=
(2450kg)(12.0m/ s) + (1650kg)(−8.00m/ s)
2450kg + 1650kg
= 3.95m/ s
(b) v’ =
m1v1 + m2v2
m1 + m2
=
(2450kg)(12.0m/ s) + (1650kg)(8.00m/ s)
2450kg + 1650kg
= 10.4m/ s
14. (a)
cos 40.0o =
p
A

p
A
p
A
‘ = (0.500kg)(0.800m / s) cos 40.0o = 0.306kgm / s
(b)
sin 40.0o =
p
B

p
A
p
B
‘ = (0.500kg)(0.800m / s)sin 30.0o = 0.257kgm / s
(c) PA ‘ = mAvA

v
A
‘ =
0.306kgm / s
0.500kg
= 0.612m / s
(d) PB ‘ = mBvB

v
B
‘ =
0.257kgm / s
0.500kg
= 0.514m / s
48
15. (a)
cos 35.0o =
p
A

p
B
p
A
‘ = (1000kg)(30.0m / s)cos 35.0o = 2.46×104 kgm / s
(b)
sin 35.0o =
p
B

p
A
p
B
‘ = (1000kg)(30.0m / s)sin 35.0o = 1.72×104 kgm / s
(c) PA ‘ = mAvA

v
A
‘ =
2.46×104 kgm / s
1000kg
= 24.6m / s
(d) PB ‘ = mBvB

v
B
‘ =
1.72×104 kgm / s
1000kg
= 17.2m / s
16. p = (1.20×105 kgm / s)2 + (8.50×104 kgm / s)2 = 1.47×105 kgm / s
17. (a)
cos 40.0o =
p
A

p
B
p
A
‘ = (950kg)(12.0m / s) cos 40.0o = 8.73x103kgm / s
(b)
sin 40.0o =
p
B

p
A
p
B
‘ = (950kg)(12.0m / s)sin 40.0o = 7.33×103 kgm / s
(c) PA ‘ = mAvA

v
A
‘ =
8.73×103 kgm / s
950kg
= 9.19m / s
(d) PB ‘ = mBvB

vB ‘ =
7.33x103kgm / s
950kg
= 9.19m / s
49
Chapter 6 Review Questions
1. b
2. d
3. The slow moving has a large mass and a small
velocity while the rifle bullet has a small mass and
a large velocity; the product of either is large.
4. They are the same.
5. The longer the bat (applied force) is on the ball,
the greater the impulse.
6. Total momentum in a system remains constant.
7. Momentum of the escaping gas molecules is
equal to the momentum of the rocket.
8. elastic
9. inelastic
10. They are equal.
Chapter 6 Review Problems
1. p = mv = (1475slugs)(57.0mi /h)×
5280 ft
1mi

⎜⎝

⎟⎠
1h
3600s

⎜⎝

⎟⎠ = 1.23× 105 slug ft
s
2. v =
p
m
=
5.50kgm s
27.0kg
= 0.204m / s 3. Ft = (125N)(2.00 min) 60s
1min

⎜⎝

⎟⎠ = 15, 0 00Ns
4. p = mv = (0.034kg)(250m / s) = 8.5kgm s 5. v1 =
m2v2
m1
=
(0.00400kg)(625m / s)
4.50kg
= 0.556m/ s
6. (a) v f = vi
2 + 2aavgs = 02 + 2(9.80m/ s2 )(7.5m) = 12m/ s
(b) p = mv = (150kg)(12m / s) = 1800kg m s
7. (a) t = s
vavg
= 0.750m
1625m / s
= 4.62 × 10−4 s
(b) F =
mvf − mvi
t
= (0.0150kg)(3250m / s)− 0
4.62× 10−4 s
= 106, 000N
(c) Ft = (106,000N)(4.62× 10−4 s)= 49.0Ns = 49.0kg m s
(d) p = mv = (0.0150kg)(3250m / s) = 48.8kg m s
50
8. F =
m vf − vi ( )
Δt
=
1250kg(8.33m / s − 31.9m/ s)
3.50s
= −8420N
(a) s =
1
2
v f + vi ( )t =
1
2
(8.33m/ s + 31.9m / s)3.50s = 70.4m
(b) Δt =
m vf − vi ( )
F
=
1250kg(0m/ s − 31.9m/ s)
−8420N
= 4.74s
9. v2′ =
m1v1 + m2v2 − m1v1′
m2
=
(575g)(3.50m / s) + (425g)(0m / s) − (425g)(4.03m/ s)
575g
= 0.521m / s,right
10. v’ =
m1v1 + m2v2
m1 + m2
=
(2.25× 104 kg)(5.50m/ s) + (3.00× 104 kg)(−1.50m/ s)
2.25× 104 kg + 3.00 × 104 kg
= 1.50m / s,east
11. v2′ =
m1v1 + m2v2 − m1v1′
m2
=
(0.195kg × 4.50m / s)+ (0.125kg × −12.0m/ s) − (0.195kg × −8.40m/ s)
0.125kg
= 8.12m/ s, right
12. p’ = (9.50×104 kgm / s)2 + (1.05×105 kgm / s)2 = 1.42×105 kgm / s
13. (a)
cos 37.0o = pA

pA
pA ‘ = (0.35kg)(0.75kg) cos 37.0o = 0.210kgm / s
(b)
sin 37.0o = pB

pA
pB ‘ = (0.35kg)(0.75kg)sin 37.0o = 0.158kgm / s
(c) pA ‘ = mAvA ‘
VA ‘ =
0.210kgm / s
0.35kg
= 0.600m / s
51
(d) pB ‘ = mBvB ‘
VB ‘ =
0.158kgm / s
0.35kg
= 0.451m / s
Chapter 6 Applied Concepts
1. (a) Impulse= m vf ( − vi)= 0.123slugs(57.3 ft / s + 11.5ft / s) = 8.46slugs ft s
(b) the outgoing velocity is less, thereby reducing the change in momentum and the impulse.
2. (a) Δpadult = m vf − vi ( )= 68.4kg(0m / s − 24.6m / s) = −1680Ns
Δpchild = m vf − vi ( )= 34.2kg(0m/ s − 24.6m/ s) = −841Ns
(b) Fadult =
mΔv
Δt
=
−1680Ns
0.564s
= −2980N
Fchild =
mΔv
Δt
=
−841Ns
0.260s
= −3230N
3. (a) F =
m vf − vi ( )
Δt
=
70.8kg(0 + 18.5m / s)
0.355s
= 3690N
(b) F =
m vf − vi ( )
Δt
=
70.8kg(9.75m / s + 18.5m/ s)
1.98s
= 1010N
4. (a) vboat =
mSallyvSally
mboat
=
125lb × 3.50 ft / s
65.5lb
= 6.68 ft / s
(b) It is easier to step out of a heavier canoe. The canoe has a greater inertia and does not move
backwards with as large a velocity.
5. (a) v1 =
m1 + m2 ( )v’ −m2v2
m1
=
(3230kg × −4.31m/ s)− (1510kg × 21.0m / s)
1720kg
= −26.5m/ s = −95.4km / h
(The negative sign represents west.)
(b) Yes, the jeep was speeding
52
Chapter 7
7.1
1. 30 N (right) 2. 735 n (left) 3. (a) 400 N (b) 50 N 4. (a) 525 N (b) 75 N
5. x-comp y-comp
F1 0 N 1350 N
F2 925 N 0N
FR 925 N 1350 N
tanα =
1350N
925N
α = 55.6° = θ
FR = (925N)2 + (1350N)2 = 1650N
6. x-comp y-comp
F1 0 lb -1150 lb
F2 805 lb 0 lb
FR 805 lb -1150 lb
tanα =
1150lb
805lb
α = 55.0°
θ = 360° − 55.0° = 305.0°
FR = (805lb)2 + (−115lb)2 = 1400lb
7. x-comp y-comp
F1 0 N 1000 N
F2 1500 N 0 N
FR 1500 N 1000 N
tanα =
1000N
1500N
α = 33.7°
FR = (1500N)2
+ (1000N)2
= 1800N
53
8. x-comp y-comp
F1 -100 N 0 N
F2 -50.0N 0 N
F3 0 N 0 N
FR -150 N 175 N
tanα =
175N
−150N
α = 49.4°
FR = (−150N)2
+ (175N)2 = 230N
9. First find the x − and y − components of F2
x-comp y-comp
F1 -1570 lb 0 lb
F2 -523 lb 1740 lb
FR −210 0lb 1740 lb
tanα =
1740lb
2100lb
α = 39.6°
θ = 180° − 39.6° = 140.4°
FR = (−2100lb)2 + (1740lb)2 = 2730lb
10. x-comp y-comp
F1 -1950N 0 N
F2 -1920 N -3330 N
FR -3870 N -3330 N
tanα =
3330N
3870N
α = 40.7°
θ = 180° + 40.7° = 220.7°
FR = (−3870N)2 + (−3330N)2 = 5110N
54
11. x-comp y-comp
F1 2550 N 0 N
F2 -1580 N 2730 N
F3 −350 0N 1140 N
FR -2530 N 3870 N
tanα =
3870N
2530N
α = 56.8°
θ = 180° − 56.8° = 123.2°
FR = (−2530N)2 + (−3870N)2 = 4620N
12. x-comp y-comp
F1 2660 lb 0 lb
F2 -1920 lb 1450 lb
F3 -831 lb -2280 lb
FR -91 lb -830 lb
tanα =
830lb
91lb
α = 83.7°
θ = 180° + 83.7° = 263.7°
FR = (−91lb)2 + (−830lb)2 = 835lb
13. x-comp y-comp
F1 1150 N 0 N
F2 0 N 875 N
F3 -1260 N -725 N
FR -110 N 150 N
tanα =
150N
110N
α = 53.7°
θ = 180° − 53.7° = 126.3° fromF1
FR = (−110N)2 + (150N)2 = 190N
14. x-comp y-comp
F1 2750 lb 0 lb
55
F2 2380 lb 1375 lb
F3 1375 lb 2380 lb
F4 0 lb 2750 lb
FR 6505 lb 6505 lb
tanα =
6505lb
6505lb
α = 45.0°
FR = (6505lb)2 + (6505lb)2 = 9200lb
7.2
1. 100 N 2. 100 lb 3. 260 N 4. 500 N 5. 690 0N 6. 720 lb
7. 570 N 8. No 9. Yes 10. 4.5 tons 11. 322N 12. 698N
Note: the sum of the x-components equation is written first; the sum of the y-components is written second.
13. −F1 + (100 N)(cos 45.0°) = 0
70.7N = F1
F2 +[−(100 N)(sin 45.0°)]= 0
F2 = 70.7N
14. −F1 + (950 N)(cos 30.0°) = 0
823N = F1
−F2 + [−(950 N)(sin 30.0°)]= 0
475N = F2
15. −F1(cos 30.0°)+ (500l b)= 0
577lb = F1
F1(sin 30.0°) + −F2 ( )= 0
289lb = F2
16. F1 + [−(1000 lb)(cos10.0°)]= 0
F1 = 985lb
F2 + [−(1000 lb)(sin10.0°)]= 0
F2 = 174lb
17. F2(cos 60.0°) + (−250l b)= 0
F2 = 500 lb
F1 + −F2 ( )(sin 60.0°) = 0
F1 = 433lb
56
18. F1(cos 20.0°)+ −F2 ( )= 0
F1(sin 20.0°) + (−400 N)= 0
F1 = 1170N
Then F2 = 110 0N
19. −T1(cos 20.0°) + T2(cos 20.0°) = 0
T2(cos 20.0°) = T1(cos 20.0°)
T2 = T1
T1(sin 20.0°)+ T2 (sin 20.0°) + (−500l b)= 0
2T1(sin 20.0°) = 500 lb
T1 = 731lb = T2
20. −T1(cos10.0°) + T2(cos10.0°) = 0
T2(cos10.0°) = T1(cos10.0°)
T2 = T1
T1(sin10.0°)+ T2 (sin10.0°) + (−500l b)= 0
2T1(sin10.0°) = 500l b
T1 = 1440lb = T2
21. −T1(cos 20.0°) + T2(cos 30.0°) = 0
T2 =
T1(cos 20.0°)
cos 30.0°
T1(sin 20.0°)+ T2 (sin 30.0°) + (−500l b)= 0
T1(sin 20.0°)+
T1(cos 20.0°)
cos 30.0°
(sin 30.0°) = 500l b
T1 = 565lb
T2 = 613lb
22. E(cos 25.0°)+ (−T ) = 0
E(cos 25.0°) = T
E(sin 25.0°) + (−8900 N)= 0
E = 21,100N
T = (21,100N)(cos 25.0°) = 19,100N
57
23. E(cos 30.0°)+ (−T ) = 0
E(cos 30.0°) = T
E(sin 30.0°) + (−1500 N)= 0
E =C = 300 0lb
T = (300 0lb)(cos 30.0°) = 260 0lb
24. 20lb + [−M(cos 45.0°)]= 0
28lb = M
25. B + (−16,200N)(cos 70.0°) = 0
B = 5540N
26. E + (−T )(cos 40.0°) = 0
T (sin 40.0°) + (−750 N)= 0
T = 1170N
E = (1170N)(cos 40.0°) = 896N
27. sum of x-components= 0
E(cos 58.0°)+ (−T )(cos 33.0°) = 0
sum of y-components= 0
E(sin 58.0°) + (−T )(sin 33.0°) + (−1850lb) = 0
0.530E − 0.839T = 0
0.848E − 0.545T = 1850
Note: Solve the above equations simultaneously.
E = 3690lb = C
T = 2330lb
58
28. sum of x-components= 0
E(cos 34.7°)+ (−T )(cos19.6°) = 0
sum of y-components= 0
E(sin 34.7°) + (−T )(sin19.6°) + (−11,500N) = 0
0.822E − 0.942T = 0
0.569E − 0.335T = 11,500
Note: Solve the above questions simultaneously.
E= 41,600 N = C
T= 36,300 N
7.3
1. T = Fst = (16.0lb)(6.00 ft) = 96.0lb ft 2. T = Fst
= (100 N)(0.420 ft) = 42.0N m
3. st =
T
F
=
60.0N m
30.0N
= 2.00m 4. F =
T
st
=
35.7lb ft
0.0240ft
= 1490lb
5. F =
T
st
=
65.4N m
35.0cm
= 187N 6. T = Fst = (630 N)(0.740m) = 466N m
7. F =
T
st
=
38.0N m
0.0237m
= 1.60 × 103N 8. T = Fst = (56.2lb)(1.50 ft) = 84.3lb ft
9. st =
T
F
=
25.0Nm
70.0N
= 0.357m 10. T = Fst = (112N)(0.350m) = 3.92N m
11. F =
T
st
=
175lb ft
110 ft
= 159lb 12. F =
T
st
=
14.5N m
25.0cm
100cm
1m

⎜⎝

⎟⎠ = 58.0N
13. F =
T
st
=
12.0N m
0.0300m
= 400 N 14. F =
T
st
=
30.0N m
0.29m
= 103N
15. F =
T
st
=
65.0N m
0.30m
= 217N 16. F =
T
st
=
27.0N m
0.300m
= 90.0N
17. (a) F =
T
st
=
25lb ft
1.0 ft
= 25lb (b) It is halved. 18. st =
T
F
=
13Nm
28N
= 0.46m
19. F =
T
st
=
40.0Nm
0.300m
= 133N 20. It is halved. The applied force and the length of the
torque arm is inversely proportional.
21. F =
T
st
=
60.0N m
0.325m
= 171N 22. F =
T
st
=
55.0N m
0.325m
= 169N
59
7.4
1. F = 100 lb 2. F = 200 N 3. F + 200 N = 700 N;F = 500 N 4. F = 200 N + 150 N = 350 N
5. 900 N = 450 N + F;F = 450 N 6. 650 lb = 100 lb + 250 lb + F;F = 300 lb
7. F + 210 0N = 750 N + 1500 lb + 250 N = 400 N
8. 50.0N + 35.0N + 15.0N = 10.0N + F + 75.0N = 15.0N
9.
Fw ( )d1 ( )= F2 ( )d2 ( )
(90.0kg)(9.80m / s2)(3.00m) = F2(8.00m)
F2 = 331N
W = mg = (90.0kg)(9.80m / s2)= 882N
F1 = 882N − 331N = 551N
10.
F2(50.0 ft)= (500 0lb)(20.0ft) + (400 0lb)(40.0 ft)
F2 = 520 0lb
520 0lb + F1 = 400 0lb + 500 0lb
F1 = 380 0lb
11. F2(27.0m) = (240 0kg)(9.80m/ s2 )
(6.00m)+ (150 0kg)(9.80m/ s2 )(10.0m)
F2(27.0m) = 2.88× 105kg m2 / s2
F2 = 1.07× 104N
Fup = Fdown
Ft + Fc = F1 + F2 or F1 = Ft + Fc −F2
F1 = 2.88 × 105N −1.07× 104N
F1 = 2.77 × 105N
12. F2(2.50m) = (165kg)(9.80m/ s2 )(1.00m)
F2 = 647N
F1 + 647N = 1620N
F1 = 970N
60
13.
(20.0kg)(9.80m / s 2)+ (40.0kg)(9.80m / s2)= Fw
Fw = 588N
(588N)x = (196N)(8.00m)
x = 2.67m
14.
F1 + F2 = (75.0kg)(9.80m / s 2)+ (75.0kg)(9.80m / s2)+ (21.0kg)(9.80m/ s2 )= 1680N
so, F1 = F2 = 840N
15.
F1 + F2 = (75.0kg)(9.80m / s 2)+ (21.0kg)(9.80m / s2)+ (90.0kg)(9.80m/ s2 )= 1820N
Select F1 for point of rotation.
F2(12.0m) = (75.0kg)(9.80m/ s2 )(3.00m)+ (21.0kg)(9.80m/ s2 )(6.00m)+ (90.0kg)(9.80m/ s2 )(9.00m)
F2 = 948N
F1 = 1820N − 948N = 872N
16.
F1 + F2 = (65.0kg)(9.80m / s 2)+ (18.0kg)(9.80m / s2)+ (95.0kg)(9.80m/ s2 )= 1740N
Select F1 for point of rotation.
F2(10.0m) = (65.0kg)(9.80m/ s2 )(2.00m)+ (18.0kg)(9.80m/ s2 )(3.50m)+ (95.0kg)(9.80m/ s2 )(6.00m)
F2 = 748N
F1 = 1740N − 748N = 992N
61
17. (1.0m)(76.0kg)(9.80m / s2 ) = F2 (2.22m)
F2 = 335N = ma
m =
335N
9.80m / s2 = 34.2kg
(76.0kg)(9.80m / s2 ) − 335N = 410N
F1 = 410N = ma
m =
410N
9.80m / s2 = 41.8kg
18. (0.75m)(12.60kg)(9.80m / s2 ) = F2 (2.00m)
F2 = 46.3N = ma
m =
46.3N
9.80m / s2 = 4.72kg
(12.60kg)(9.80m / s2 ) − 46.3N = 77.18N
F1 = 77.18N = ma
m =
77.18N
9.80m / s2 = 7.88kg
7.5
1. F1 = 22.6 2. Fw = 42.0
3. Select F1 for point of rotation.
F2(15.0ft) = (165lb)(7.00ft) + (22.0lb)(7.50 ft)
F2 = 88.0lb
F1 + F2 = Fp + FB
88.0lb + F1 = 165lb + 22.0lb
F1 = 99lb
4 Select F1 for point of rotation.
F1 + F2 = 720N
F2(2.0m) = (720N)(0.50m)
F2 = 180N
F1 = 720N −180N = 540N
62
5. Select F1 for point of rotation.
F2 (3.30m) = 1.50 × 103 ( N)(1.30m)
F2 = 591N
591N + F1 = 1.50 × 103N
F1 = 909N
6. F2(10.0ft) = (650l b)(4.00ft) + (75.0lb)(5.00 ft)
F2 = 298lb
298lb + F1 = 650lb + 75.0lb
F1 = 427lb
7. F1 + F2 = 187, 2000N
F2(9.00m) = (98, 00 0N)(4.00m) + (889, 200N)(4.50m)
F2 = 88, 200N
Therefore F1 = 99, 0 00N
8.
(200 lb)(12.0ft) = (155lb)(9.00ft) + F0 ( )(4.00ft) + (75.0lb)(6.00 ft)
F0 = 139lb
9.
F2(4.40 ft)= (14.0lb)(1.90ft) + (29.0lb)(2.20 ft)+ (125lb)(3.40 ft)
F2 = 117lb
117lb + F1 = 14.0lb + 29.0lb + 125lb
F1 = 51lb
63
10.
F2(5.00m) = (735N)(2.00m) + (294N)(2.50m)
F2 = 441N
F1 + 441N = 735N + 294N
F1 = 588N
11.
F1(21.0m) = (1.57 × 105 )(10.5m) + (3.43× 104 )(7.00m)
F1 = 8.99× 104N
8.99× 104N + F2 = 1.57× 105N + 3.43× 104N
F2 = 1.01× 105N
12.
F1 (5.00m) = (360N)(2.50m)
F1 = 180N
13. F1(4.00m) = (315N)(1.50m)
F1 = 118N
F2 + 118N = 315N
F2 = 197N
14. Fx = (3500 kg)(9.80m/ s2)
FT = (2.60× 104 kg)(9.80m/ s2)
F1(32.0m) = (2.55× 105N)(16.0m)+ (3.43× 104 N)(15.0m)
F1 = 1.44 × 105N
F2 + 1.44 × 105N = 2.55× 105N + 3.43× 104N
Therefore F2 = 1.45× 105N
64
15.
Fx = (295kg)(9.80m/ s2)
Fx = (295kg)(9.80m/ s2)
F1(4.00m) = (441N)(2.00m)+ (2890N)(1.00m)
F1 = 943N
F2 + 943N = 441N + 2890N
F2 = 2390N
16.
F1 + F2 = (325kg)(9.80m / s 2)+ (125kg)(9.80m / s2)= 4410N
Select F1 for point of rotation.
F2(4.00m) = (325kg)(9.80m/ s2 )(1.00m)+ (125kg)(9.80m/ s2 )(2.00m)
F2 = 1410N
F1 = 4410N −1410N = 300 0N
Note: Since the length of the beam is not given, we need 1.00m, 1.00m, 2.00m as the lengths. Any length
ratios of 1:1:2 will work. Try any other set.
17.
(2.00m)F2 = (1550N)(1.00m)+ (245N)(2.00m)
F2 = 1020N
F1 + 1020N = 1550N + 245N
F1 = 775N
18.
(a) F4 = 255N + 975N + 375N = 1605N
(b) (975N)(2.50m) + (375N)(3.50m) = (1605N)(x)
x = 2.34 m
65
19.
F6 = 1525N (up)
x = distance from A of F6
(375N)(1.00m) + (1175N)(3.00m)+ (1850 N)(4.00m) = 625N(7.00m)+ (1525N)(x)
4.54 m =x
20.
F6 = 1375N (down)
x= distance from A’ of F6
(500 N)(1.00m) + (750N )(2.20m)+ (2375N)(3.95m)+ (1375N)(x) = (3750 N)(4.45m)
x = 3.75 m from A’ and 4.75 m from A
Chapter 7 Review Questions
1. b
2. b
3. a
4. c
5. c
6. a
7. b
8. b
9. No; e.g., bridge
10. They are equal
11. Equilibrium is the condition of a body where
the net force acting on it is zero.
66
12. Toward the center of the earth.
13. They are in equilibrium.
14. It is a diagram showing how forces act on a
body.
15. No, only when the pedals are parallel to the
ground.
16. Even if the vector sum of the opposing forces
is zero, they must also be positioned so there is no
rotation in the system.
17. Choosing a point through which a force acts to
eliminate a variable.
18. (a) stacking bridges
(b) riding a bicycle
(c) lifting any object
(d) hitting a baseball
(e) leaning into the wind
19. No; only if the object is of uniform
composition and shape.
20. The support closer to the bricks.
Chapter 7 Review Problems
1. 569 N (left) 2. (a) 500 lb (b) 50 lb
3. x- comp y- comp
F1 0 N 1500 N
F2 -3400 N 0 N
FR -3400 N 1500 N
tan A =
1500N
3400N
A = 24°
θ = 180° − 24° = 156°
c = (−3400N)2 + (1500N)2 = 3700N
4. x- comp y- comp
F1 5080 lb 0 lb
F2 2550 lb -3640 lb
FR 7630 lb -3640 lb
tan A =
3640lb
7630lb
A = 25.5°
θ = 360° − 25.5° = 334.5°
c = (7630lb)2 + (−3640lb)2 = 8450lb
67
5. x- comp y- comp
F1 54,600 N 0 N
F2 0 N 54,600 N
F3 -38,600 N 38,600 N
FR 16, 0 00N 93,600 N
tan A = 93,200N
16,000N
A = 80.3° =θ
c = (16, 00 0N)2 + (93, 200N)2 = 94, 600N
6. x- comp y- comp
F1 1250 N 0 N
F2 -313 N 541 N
F3 -1600 N 925 N
FR -660 N 1466 N
tan A =
1466N
660N
A = 65.8°
θ = 180° − 65.8° = 114.2°
c = (−660N)2 + (1466N)2 = 1610N
7. 220 N 250 N 8. 6.0 tons
340 N 160 N
180 N 420 N
560 N 830 N
130 0N
-830 N
470 N
9. Fx + 375+ 150 = 0
Fx = −525
10. tan A = 600 N
900 N
11. F1 + [−(1650N)(cos 67.0°)]= 0
A = 33.7° =θ F1 = 645N
c = (900N )2 + (600 N)2 = 1080N F2 + [−(1650N)(sin 67.0°)]= 0
F2 = 1520N
68
12. F1(cos 65.0°)+ −F2 ( )= 0
F1(sin 65.0°) + (−4750lb) = 0
F1 = 5240lb
F2 = 2210lb
13.
E + (−T )(cos 40.0°) = 0
T (sin 40.0°) + (−1150N) = 0
T = 1790N
E = 1370N
14.
−T1 ( )(cos 30.0°)+ T2(cos 45.0°) = 0
T1(sin 30.0°)+ T2 (sin 45.0°) + (−475lb) = 0
−0.866T1 + 0.707T2 = 0
−0.500T1 + 0.707T2 = 475
T1 = 348lb
T2 = 426lb
T3 = 475lb
15.
−T1 ( )(cos 50.0°)+ T2 = 0
T1(sin 50.0°)+ (−2200N) = 0
T1 = 2900N
T2 = 1900N
T3 = 2200N
16. E(cos 30.0°)+ −T1 ( )(cos 30.0°) = 0
E = T1
E(sin 30.0°) + T1 ( )(sin 30.0°) + (−6250N) = 0
2E(sin 30.0°) = 6250N
E = 6250N = T1 = T2
69
17. T = Fst = (53.0N)(0.500m) = 26.5N m 18. F =
T
st
=
81.0lb ft
3.00 ft
= 27.0lb
19. (200 kg)(9.80m / s2)= 1080N + T 20. 450 lb – 290 lb =160 lb
880 N = T
21. 100.0kg 22. 68.0 cm x 3/4 = 51.0 cm 23.
800 0kg + 320 0kg
2
= 560 0kg
24. F1 + F2 = 7.84 × 104N + 3.14 × 104N = 11.0 × 104N
F1(26.0m) = (7.84 × 104N)(13.0m) + (3.14 × 104N)(7.00m)
F1 = 47, 700N
F2 = 62,300N
25.
(392N)(0.700m)+ (21.6N)(1.35m)+ (539N)(2.20m) = (127N)(1.45m)+ F2(2.70m)
F2 = 483N
F1 + F2 + 127N = 392N + 21.6N + 539N
F1 + F2 = 826N
F1 = 343N
26. (3475N)(2.00m) + (1125N)(3.00m) = F(4.00m)
F = 2580N
Chapter 7 Applied Concepts
1. (a) τ = Fxd = 894N × 2.57m = 230 0N m
(b) τ = Fxd = (894N × sin 30°)2.57m = 1150N m
(c) Continually push perpendicular to the door.
70
2. (a) tan−1 5.50ft
33.5ft

⎜⎝

⎟⎠ = 9.32° ;
9.32° × 2 = 18.6°
(b) F =
Fg
cosθ
= 57.5lb
cos 9.32°
= 58.3lb
(c) tan−1 5.50ft
5.75ft

⎜⎝

⎟⎠ = 43.7°
F =
F2
cosθ
=
57.5ft
cos 43.7°
= 79.5lb
(d) The angle between the ropes is increased so that Sean and Greg not only pull up but also pull
horizontally.
3. (a) Δτ =τ new −τ old = (25.5N × 0.127m)− (25.5N × 0.0374m) = 2.28N m
(b) F =
τ
d
=
(25.5N × 0.0374m)
0.127m
= 7.51N
ΔF = Foriginal − Fnew = 25.5N − 7.51N = 18.0N less force
4. (a) F = 85.8N cos 45.0° = 60.7N
τ = F × d = 60.7N × 0.750m = 45.5N m which will not support the torque.
(b) Reduce the angle between the wall and the bracket.
5. (a) F1 + F2 = 350 N + 877N = 1230N
F1 = 1230N −F2
F2(1.53m) = (350 N × 2.75m)+ (877N × 5.50kg)
F2 = 3780N
F1 = 1230N − 3780N = −2550N (negative sign means that the bracket pulls down)
(b) Between the bracket and the fulcrum.