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## Applied Physics 9th Edition By Dale Ewen – Test Bank A+

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Applied Physics 9th Edition By Dale Ewen – Test Bank A+

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

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.

Only 0 units of this product remain