Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 8P from Chapter 3 from Hibbeler's Engineering Mechanics.
Problem 8P
Step-by-Step Solution
We are given the following data:
The cord $ABC$ and $BD$ each supports a magnitude of a maximum load of ${T_{{\rm{ABC}}}} = {T_{{\rm{BD}}}} = T = 100\;{\rm{lb}}$.
We are asked to estimate the maximum weight of the crate and the angle $\theta$ for equilibrium.
Step 2
The free-body diagram of the system is given below.
Here, $W$ represent the maximum weight of the crate.
According to the force equilibrium along the $x$-axis and $y$-axis,
\[\begin{array}{c} \sum {{F_{\rm{x}}}} = 0\\ W \times \left( {\frac{5}{{13}}} \right) - {T_{{\rm{BD}}}}\cos \theta = 0\\ \frac{{5W}}{{13}} - \left( {100\;{\rm{lb}}} \right) \times \cos \theta = 0\\ \left( {100\;{\rm{lb}}} \right)\cos \theta = \frac{{5W}}{{13}} \end{array}\]And,
\[\begin{array}{c} \sum {{F_{\rm{y}}}} = 0\\ {T_{{\rm{BD}}}}\sin \theta - W - W \times \left( {\frac{{12}}{{13}}} \right) = 0\\ \left( {100\;{\rm{lb}}} \right) \times \sin \theta - \frac{{25W}}{{13}} = 0\\ \left( {100\;{\rm{lb}}} \right) \times \sin \theta = \frac{{25W}}{{13}} \end{array}\]Step 3
On dividing the above two equations with each other, the value of the angle $\theta$ can be obtained.
\[\begin{array}{c} \frac{{\left( {100\;{\rm{lb}}} \right) \times \sin \theta }}{{\left( {100\;{\rm{lb}}} \right) \times \cos \theta }} = \frac{{\frac{{25W}}{{13}}}}{{\frac{{5W}}{{13}}}}\\ \tan \theta = \frac{{25}}{5}\\ \tan \theta = 5\\ \theta \approx 78.7^\circ \end{array}\]Step 4
The value of the maximum weight of the crate can be obtained by using the known values as,
\[\begin{array}{c} \left( {100\;{\rm{lb}}} \right) \times \sin \theta = \frac{{25W}}{{13}}\\ \left( {100\;{\rm{lb}}} \right) \times \sin \left( {78.7^\circ } \right) = \frac{{25W}}{{13}}\\ \left( {98.06\;{\rm{lb}}} \right){\rm{ = }}\frac{{25W}}{{13}}\\ W \approx 51\;{\rm{lb}} \end{array}\]