Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 12 - Exercises and Problems - Page 224: 51

Answer

The proof is below.

Work Step by Step

We will call the bottom of the latter the axis of rotation. Since any force at the axis of rotation does not apply a torque, this allows us to ignore the normal force due to the ground as well as the force of friction. We call the length of the ladder l, and we set the positive torques (which we will call torques causing clockwise motion) equal to the negative torques. This gives: $lMgsin(\theta^{\circ})+\frac{l}{2}mgsin(\theta^{\circ})=l\mu (m+M)gsin(90+\theta)^{\circ}$ $Msin(\theta^{\circ})+\frac{1}{2}msin(\theta^{\circ})=\mu (m+M)sin(90+\theta)^{\circ}$ We now consider the two extremes. We act as if there is no mass, for we must consider the limit as the mass approaches 0: $\frac{1}{2}msin(\theta^{\circ})=\mu (m)sin(90+\theta)^{\circ}$ $\frac{1}{2}msin(\theta^{\circ})=\mu (m)cos(\theta)^{\circ}$ $\mu=\frac{1}{2}tan(\theta)$ Thus, if $\mu\lt \frac{1}{2}tan(\theta)$, no one can climb the ladder. We now consider if M is infinite: $Msin\theta=Msin(90+\theta)$ $Msin\theta=\mu Mcos(90+\theta)$ $\mu = tan\theta$ Thus, if $\mu\geq \frac{1}{2}tan(\theta)$, anyone can climb the ladder.
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.