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Holt Physics
Problem 4A NET EXTERNAL FORCE PROBLEM
Two soccer players kick a ball at the same instant. One player kicks with a force of 65 N to the north, while the other player kicks with a force of 88 N to the east. In what direction does the ball travel?
SOLUTION 1. DEFINE
Given:
F1 = 65 N north F2 = 88 N east
Unknown: Diagram:
q =?
F1 = 65 Ν
Ν F2 = 88 Ν
2. PLAN
3. CALCULATE
Select a coordinate system and apply it to the free-body diagram. Choose the positive x-axis to align with east and the positive y-axis to align with north. Find the x and y components of all vectors. F1,x = 0 N
F1,y = 65 N
F2,x = 88 N
F2,y = 0 N
Find the net external force in both the x and y directions. Fx,net = ΣFx = F1,x + F2,x = 0 N + 88 N = 88 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fy,net = ΣFy = F1,y + F2,y = 65 N + 0 N = 65 N Find the direction of the net external force. Use the tangent function to find the angle q of Fnet. Fy,net 65 N q = tan−1 = tan−1 = 36° Fx, net 88 N
q = 36° north of east 4. EVALUATE
The direction is about three-fourths of the way to the midpoint (45°) between north and east. This corresponds closely to the ratio of 65 N to 88 N (0.74).
ADDITIONAL PRACTICE 1.
Two tugboats pull a barge across the harbor. One boat exerts a force of 7.5 × 104 N north, while the second boat exerts a force of 9.5 × 104 N at 15.0° north of west. Precisely, in what direction does the barge move?
2. Three workers move a car by pulling on three ropes. The first worker exerts a force of 6.00 × 102 N to the north, the second a force of 7.50 × 102 N to the east, and the third 6.75 × 102 N at 30.0° south of east. In what precise direction does the car move? Problem 4A
Ch. 4–1
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3. Four forces are acting on a hot-air balloon: F1 = 2280.0 N up, F2 = 2250.0 N down, F3 = 85.0 N west, and F4 = 12.0 N east. What is the precise direction of the net external force on the balloon? 4. What is the magnitude of the largest net force that can be produced by combining a force of 6.0 N and a force of 8.0 N? What is the magnitude of the smallest such force? 5. Two friends grab different sides of a videotape cartridge and pull with forces of 3.0 N to the east and 4.0 N to the south, respectively. What force would a third friend need to exert on the cartridge in order to balance the other two forces? What would be that force’s precise direction? 6. A four-way tug-of-war has four ropes attached to a metal ring. The forces on the ring are as follows: F1 = 4.00 × 103 N east, F2 = 5.00 × 103 N north, F3 = 7.00 × 103 N west, and F4 = 9.00 × 103 N south. What is the net force on the ring? What would be that force’s precise direction?
8.
A shopper pushes a grocery cart by exerting a force on the handle. If the force equals 76 N at an angle of 40.0° below the horizontal, how much force is pushing the cart in the forward direction? What is the component of force pushing the cart against the floor?
9.
Two paramedics are carrying a person on a stretcher. One paramedic exerts a force of 350 N at 58° above the horizontal and the other paramedic exerts a force of 410 N at 43° above the horizontal. What is the magnitude of the net upward force exerted by the paramedics?
10. A traffic signal is supported by two cables, each of which makes an angle of 40.0° with the vertical. If each cable can exert a maximum force of 7.50 × 102 N, what is the largest weight they can support?
Ch. 4–2
Holt Physics Problem Bank
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. A child pulls a toy by exerting a force of 15.0 N on a string that makes an angle of 55.0° with respect to the floor. What are the vertical and horizontal components of the force?
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Forces and the Laws of Motion
Chapter
4
Additional Practice 4A
Givens 1. F1 = 7.5 × 104 N north
Solutions Fx,net = Fx = F1(cos q1) + F2(cos q2)
F2 = 9.5 × 104 N at 15.0° north of west
Fx,net = (7.5 × 104 N)(cos 90.0°) + (9.5 × 104 N)(cos 165.0°)
q1 = 90.0°
Fy,net = Fy = F1(sin q1) + F2(sin q2)
q2 = 180.0° − 15.0° = 165.0°
Fy,net = (7.5 × 104 N)(sin 90.0°) + (9.5 × 104 N)(sin 165.0°)
Fx,net = −9.2 × 104 N
Fy,net = 7.5 × 104 N + 2.5 × 104 N = 10.0 × 104 N
Fy,net q = tan−1 Fx,net q= 2. F1 = 6.00 × 102 N north
q1 = 90.0°
Fx,net = Fx = F1(cos q1) + F2(cos q2) + F3(cos q3) = (6.00 × 102 N)(cos 90.0°) + (7.50 × 102 N)(cos 0.00°) + (6.75 × 102 N)[cos(−30.0°)] Fx,net = 7.50 × 102 N + 5.85 × 102 N = 13.35 × 102 N Fy,net = Fy = F1(sin q1) + F2(sin q2) + F3(sin q3) = (6.00 × 102 N)(sin 90.0°) + (7.50 × 102 N)(sin 0.00°) + (6.75 × 102 N)[sin (−30.0°)]
q2 = 0.00°
Fy,net = 6.00 × 102 N + (−3.38 × 102 N) = 2.62 × 102 N
q3 = −30.0°
Fy,net 2.62 × 102 N q = tan−1 = tan−1 13.35 × 102 N Fx,net
q= Copyright © by Holt, Rinehart and Winston. All rights reserved.
= −47°
47° north of west
F2 = 7.50 × 102 N east F3 = 6.75 × 102 N at 30.0° south of east
10.0 × 104 N = tan−1 −9.2 × 104
3. F1 = 2280.0 N upward F2 = 2250.0 N downward F3 = 85.0 N west F4 = 12.0 N east
F2 = 8.0 N
11.1° north of east
Fy,net = Fy = F1 + F2 = 2280.0 N + (−2250.0 N) = 30.0 N Fx,net = Fx = F3 + F4 = −85.0 N + 12.0 N = −73.0 N
30.0 N Fy,net q = tan−1 = tan−1 = −22.3° −73.0 N Fx,net q=
4. F1 = 6.0 N
22.3° up from west
Fmax = F1 + F2 = 6.0 N + 8.0 N Fmax = 14.0 N Fmin = F2 − F1 = 8.0 N − 6.0 N Fmin =
2.0 N
V
Section Five—Solution Manual
V Ch. 4–1
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Givens
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5. F1 = 3.0 N east F2 = 4.0 N south
Fx,net = F1 + F3(cos q) = 0 F3(cos q) = −F1 = −3.0 N Fy,net = F2 + F3(sin q) = 0 F3(sin q) = −F2 = −(−4.0 N) = 4.0 N
2 2 2 F3 = [F co s q )] si n q )]2 = (−3 .0 N )2 +(4 .0 N )2 = 9. 0 N 6 N = 25 N2 +[F +1 3( 3(
F3 =
5.0 N
F3(sin q) 4.0 N q = tan−1 = tan−1 = −53° −3.0 N F3(cos q) q= 6. F1 = 4.00 × 103 N east 3
F2 = 5.00 × 10 N north F3 = 7.00 × 103 N west F4 = 9.00 × 103 N south
53° north of west
Fx,net = F1 + F3 = 4.00 × 103 N + (−7.00 × 103 N) = −3.00 × 103 N Fy,net = F2 + F4 = 5.00 × 103 N + (−9.00 × 103 N) = −4.00 × 103 N
2 N)2 +(− N)2 Fnet = (F x,ne (Fy,n et)2 = (− 3.0 0×103 4. 00 ×103 t )+
Fnet = 9 06 N2+ 06 N2 = 25 06 N2 .00×1 16. 0×1 .0 ×1 Fnet =
5.00 × 103 N
Fy,net −4.00 × 103 N q = tan−1 = tan−1 −3.00 × 103 N Fx,net
q = 53.1° south of west
7. F1 = 15.0 N q = 55.0°
Fy = F(sin q) = (15.0 N)(sin 55.0°) Fy = 12.3 N Fx = F(cos q) = (15.0 N)(cos 55.0°) Fx = 8.60 N
8. F = 76 N
Fx = 58 N Fy = F(sin q) = (76 N)(sin 40.0°) Fy = 49 N
9. F1 = 350 N
Fy,net = F1(sin q1) + F2(sin q2) = (350 N)(sin 58°) + (410 N)(sin 43)
q1 = 58.0°
Fy,net = 3.0 × 102 N + 2.8 × 102 N
F2 = 410 N
Fy,net = 580 N
q2 = 43° 10. F1 = 7.50 × 102 N
Fy,net = Fg = F1(cos q1) + F2(cos q2)
q1 = 40.0°
Fy,net = (7.50 × 102 N)(cos 40.0°) + (7.50 × 102 N) [cos(−40.0°)]
F2 = 7.50 × 102 N
Fg = 575 N + 575 N = 1.150 × 103 N
q2 = −40.0°
V
V Ch. 4–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fx = F(cos q) = (76 N)(cos 40.0°)
q = 40.0°
