Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 1RP from Chapter 4 from Hibbeler's Engineering Mechanics.
Problem 1RP
Chapter:
Problem:
The boom has a length of 30 ft, a weight of 800 lb, and mass center...
Step-by-Step Solution
Step 1
Step 2
Step 3
We are given the following data.
The moment at the point $A$ is ${M_{\rm{A}}} = 20 \times \left( {{{10}^3}} \right)\;{\rm{lb}} \cdot {\rm{ft}}$.
The length of the boom is $L = 30\;{\rm{ft}}$.
We are asked to estimate the maximum load $W$.
Step 2
The free-body diagram of the boom is given below.
Step 3
By taking the moment about the point $A$, the value of the maximum load $W$ can be obtained.
\[\begin{array}{c} {M_{\rm{A}}} = \left( {800\;{\rm{lb}}} \right) \times \left( {16\;{\rm{ft}} \times \cos 30^\circ } \right) + \left( {W \times 32\;{\rm{ft}} \times \cos 30^\circ } \right)\\ \left( {20 \times {{10}^3}\;{\rm{lb}} \cdot {\rm{ft}}} \right)\;{\rm{ = }}\left( {\;11085.13\;{\rm{lb}} \cdot {\rm{ft}}} \right) + \left( {27.71W\;{\rm{ft}}} \right)\\ \left( {27.71W\;{\rm{ft}}} \right)\;{\rm{ = }}\;\left( {{\rm{8914}}{\rm{.87}}\;{\rm{lb}} \cdot {\rm{ft}}} \right)\\ W = \frac{{\left( {{\rm{8914}}{\rm{.87}}\;{\rm{lb}} \cdot {\rm{ft}}} \right)}}{{\left( {27.71\;{\rm{ft}}} \right)}}\\ \approx \left( {321.7\;{\rm{lb}}} \right) \end{array}\]