Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 43P from Chapter 11 from Hibbeler's Engineering Mechanics.
Problem 43P
Step-by-Step Solution
We are given the mass of a truck as $m = 20\;{\rm{Mg}}$ and the mass center at point G.
We are asked to determine the steepest grade $\theta $ along which the truck can park without overturning and also to investigate the stability in the position.
Step 2
The free-body diagram of the truck can be drawn as:
Here, y is the distance between center of gravity and the center of a truck wheel.
From the above figure, the distance y can be given as:
\[y = \left( {1.5\;{\rm{m}}} \right)\sin \theta + \left( {3.5\;{\rm{m}}} \right)\cos \theta \]To calculate the potential energy, we have:
\[V = Wy\]Substitute the known value in the above expression, we get:
\[V = W\left[ {\left( {1.5\;{\rm{m}}} \right)\sin \theta + \left( {3.5\;{\rm{m}}} \right)\cos \theta } \right]\]On differentiating the above expression with respect to $\theta $, we get:
\[\frac{{dV}}{{d\theta }} = W\left[ {1.5\cos \theta - 3.5\sin \theta } \right]\] … (1)Step 3
For equilibrium, we have:
\[\frac{{dV}}{{d\theta }} = 0\]Hence,
\[\begin{array}{c} W\left[ {1.5\cos \theta - 3.5\sin \theta } \right] = 0\\ 1.5\cos \theta = 3.5\sin \theta \\ \tan \theta = 0.4286\\ \theta = 23.2^\circ \end{array}\]On again differentiating the equation (1) with respect to $\theta $, we get:
\[\begin{array}{c} \frac{{{d^2}V}}{{d{\theta ^2}}} = W\left[ { - 1.5\sin \theta - 3.5\cos \theta } \right]\\ W\left( { - 1.5\sin 23.2^\circ - 3.5\cos 23.2^\circ } \right)\\ - 3.808W < 0 \end{array}\]Therefore, the truck will be unstable.