## College Physics (4th Edition)

Published by McGraw-Hill Education

# Review & Synthesis: Chapters 1-5 - Review Exercises - Page 192: 10

#### Answer

(a) The magnitude of the wind's force exerted on the tire is $0.21~W$ (b) The tension in the rope is $1.02~W$

#### Work Step by Step

(a) The vertical component of the tension in the rope is equal in magnitude to the weight. $W = T~cos~\theta$ The horizontal component of the tension is equal in magnitude to the force of the wind $F_w$: $F_w = T~sin~\theta$ We can divide the second equation by the first equation: $\frac{F_w}{W} = \frac{sin~\theta}{cos~\theta}$ $F_w = W~tan~\theta$ $F_w = W~tan~12^{\circ}$ $F_w = 0.21~W$ The magnitude of the wind's force exerted on the tire is $0.21~W$ (b) We can find the tension in the rope: $T~cos~\theta = W$ $T = \frac{W}{cos~\theta}$ $T = \frac{W}{cos~12^{\circ}}$ $T = 1.02~W$ The tension in the rope is $1.02~W$

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