Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 91P from Chapter 6 from Hibbeler's Engineering Mechanics.
Problem 91P
Step-by-Step Solution
We are given the mass of shovel $m = 1.25\;{\rm{Mg}}$.
We are asked to determine the force created in the hydraulic cylinder $EF$ and $AD$.
Step 2
The free body diagram of the member is shown below:
The length from center of gravity to $H$ is ${d_1} = 0.5\;{\rm{m}}$.
The horizontal length from point $E$ to $H$ is ${d_2} = \left( {1.5\;{\rm{m}}} \right)\sin 30^\circ $.
To find the force we will find the moment about $H$ by using relation,
\[\begin{array}{c} \Sigma {M_H} = 0\\ - W \times {d_1} + {F_{EF}} \times {d_2} = 0\\ - mg \times {d_1} + {F_{EF}} \times {d_2} = 0 \end{array}\]On plugging the values in the above relation, we get,
\[\begin{array}{c} - \left( {1.25\;{\rm{Mg}} \times \frac{{{{10}^3}\;{\rm{kg}}}}{{1\;{\rm{Mg}}}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {0.5\;{\rm{m}}} \right) + {F_{EF}} \times \left( {1.5\;{\rm{m}}} \right)\sin 30^\circ = 0\\ {F_{EF}} = 8175\;{\rm{N}} \end{array}\]Step 3
The free body diagram of the member is shown below:
The length from point $C$ to $D$ is ${d_3} = 0.25\;{\rm{m}}$.
The horizontal length from point $C$ to $H$ is ${d_4} = \left( {2\;{\rm{m}}} \right)\cos 10^\circ $.
To find the required force we will find the moment about $C$ by using relation,
\[\begin{array}{c} \Sigma {M_C} = 0\\ {F_{AD}}\cos 40^\circ \times {d_3} - W \times \left( {{d_1} + {d_4}} \right) = 0 \end{array}\]On plugging the values in the above relation, we get,
\[\begin{array}{c} {F_{AD}}\cos 40^\circ \times \left( {0.25\,{\rm{m}}} \right) - \left( {1.25\;{\rm{Mg}} \times \frac{{{{10}^3}\;{\rm{kg}}}}{{1\;{\rm{Mg}}}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {\left( {2\;{\rm{m}}} \right)\cos 10^\circ + 0.5\,{\rm{m}}} \right) = 0\\ {F_{AD}} = 158130.04\;{\rm{N}} \end{array}\]