Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - Problems - Page 103: 30


(a) The required horizontal force is 9.70 N (b) The tension in the wire is 265 N

Work Step by Step

We can find the angle $\theta$ that the wire makes with the vertical. $sin(\theta) = \frac{0.15 ~m}{4.0 ~m}$ $\theta = sin^{-1}( \frac{0.15 ~m}{4.0 ~m}) = 2.1^{\circ}$ (a) The vertical component of tension $T_y$ is equal to the weight $mg$. $\frac{T_x}{T_y} = tan(\theta)$ $T_x = T_y~tan(\theta)$ $T_x = mg~tan(\theta)$ $T_x = (27~kg)(9.80~m/s^2)~tan(2.1^{\circ})$ $T_x = 9.70 ~N$ The required horizontal force will be equal to the horizontal component of the tension which is 9.70 N (b) $\frac{T_x}{T} = sin(\theta)$ $T = \frac{T_x}{sin(\theta)}$ $T = \frac{9.70~N}{sin(2.1^{\circ})}$ $T = 265~N$ The tension in the wire is 265 N
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