Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 116: 62

Answer

Magnitude of $\vec{T}_R=435.8N$ and it is directed at angle of $5.64^o$ to the horizontal.

Work Step by Step

Take rightward and upward to be our $+i$ and $+j$ directions, respectively. We rewrite all the forces in unit-vector notation: 1) Weight: $m\vec{g}=(-151N)j$ 2) Left tension: $\vec{T}_L=-(447\cos14N)i+(447\sin14N)j$ $$\vec{T}_L=(-433.72N)i+(108.14N)j$$ 3) Right tension: $\vec{T}_R=ai+bj$ As there are no more forces, we have $$\sum \vec{F}=m\vec{g}+\vec{T}_L+\vec{T}_R$$ $$\sum \vec{F}=(a-433.72N)i+(b-42.86N)j$$ Since the limb is static, there is no acceleration whatsoever, which means $\sum \vec{F}=0$ Therefore, $a-433.72N=0$ and $a=433.72N$ $b-42.86N=0$ and $b=42.86N$ - Magnitude of $\vec{T}_R=\sqrt{433.72^2+42.86^2}=435.8N$ - Direction: let's call $\theta$ the angle relative to the horizontal, we have $\tan\theta=\frac{42.86}{433.72}=0.0988$ so $\theta=5.64^o$
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