Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 2FP from Chapter 17 from Hibbeler's Engineering Mechanics.
Problem 2FP
Step-by-Step Solution
We are given the mass $m = 80\;{\rm{kg}}$ and height of center of mass is $h = 1.5\,{\rm{m}}$.
We are asked to determine the acceleration of the cabinet and the normal reactions on the pair of rollers at A and B.
Step 2
The free body diagram of the picture is as follows:
Here, g is the gravitational acceleration, ${N_A}$ is the normal reaction at point A and ${N_B}$ is normal reaction at point B.
Step 3
To find the force in x-direction, we will use the relation,
\[\begin{array}{c} \Sigma {F_x} = ma\\ - mg\sin 15^\circ = ma\\ a = - \left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right)\sin 15^\circ \\ a = 2.539\,{\rm{m/}}{{\rm{s}}^2} \end{array}\]To find the normal forces we will take the moment at point G,
\[\begin{array}{c} \Sigma M = 0\\ {N_A}{d_1} - {N_B}{d_2} = 0 \end{array}\]Here, ${d_1}$ is the distance from point A to G whose value is ${d_1} = 0.5\,{\rm{m}}$ and ${d_2}$ is the distance from point B to G whose value is ${d_2} = 0.5\,{\rm{m}}$.
Step 4
On plugging the values in the above relation, we get,
\[\begin{array}{c} {N_A}\left( {0.5\;{\rm{m}}} \right) - {N_B}\left( {0.5\;{\rm{m}}} \right) = 0\\ {N_A} = {N_B} \end{array}\]To find the force in y-direction, we will use the relation,
\[\begin{array}{c} \Sigma {F_y} = 0\\ {N_A} + {N_B} - W\cos 15^\circ = 0\\ {N_A} + {N_A} - mg\cos 15^\circ = 0\\ 2{N_A} = mg\cos 15^\circ \end{array}\]On plugging the values in the above relation, we get,
\[\begin{array}{c} 2{N_A} = \left( {80\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\cos 15^\circ \\ {N_A} = 379.02\;{\rm{N}} \end{array}\]