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 223: 35

Answer

79 kilograms

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: $lMgsin(15^{\circ})+\frac{l}{2}mgsin(15^{\circ})=l\mu (m+M)gsin105^{\circ}$ $Msin(15^{\circ})+\frac{1}{2}msin(15^{\circ})=\mu (m+M)sin105^{\circ}$ $Msin(15^{\circ})+\frac{1}{2}(5)sin(15^{\circ})=(.26) (5+M)sin105^{\circ}$ $M\approx 76 \ kg$
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