Physics: Principles with Applications (7th Edition)

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

Chapter 9 - Static Equilibrium; Elasticity and Fracture - Problems: 8

Answer

Part A) The tension in the rope is 230N Part B) The tension in the rope is 2100N

Work Step by Step

Part A) We will sum the forces vertically to find the tension in the rope, T. First we will find the angle of depression of the rope. $tan(\theta)=\frac{1.5m}{3.3m}\approx24.4^{\circ}$ $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $-19kg*9.8m/s^2+T*sin\theta+Tsin\theta=0$ $T=\frac{19kg*9.8m/s^2}{2*sin\theta}$ $T=\frac{19kg*9.8m/s^2}{2*sin24.4^{\circ}}$ $T\approx230N$ Part B) Same as part A but use .15m instead of 1.5m $tan(\theta)=\frac{.15m}{3.3m}\approx2.60^{\circ}$ $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $-19kg*9.8m/s^2+T*sin\theta+Tsin\theta=0$ $T=\frac{19kg*9.8m/s^2}{2*sin\theta}$ $T=\frac{19kg*9.8m/s^2}{2*sin2.60^{\circ}}$ $T\approx2100N$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.