Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 372: 100

Answer

a) $T=229N$ b) $F_x=216N$ c) $F_y=78.5N$

Work Step by Step

(a) We know that $T_y=\frac{mg}{2}$ We plug in the known values to obtain: $T_y=\frac{(16Kg)(9.81m/s^2)}{2}=78.5N$ and $T_x=\frac{T_y}{tan 20^{\circ}}$ $\implies T_x=\frac{78.5N}{tan 20^{\circ}}$ $T_x=216N$ Now $T=\sqrt{T_x^2+T_y^2}$ $\implies T=\sqrt{(216N)^2+(78.5N)^2}$ $T=229N$ (b) and (c) We know from part (a) that the horizontal component of the force is $F_x=229\times cos(20^{\circ})=216N$ and the vertical component of the force is $F_y=229\times sin(20^{\circ})=78.5 N$.
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.