Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 82P from Chapter 6 from Hibbeler's Engineering Mechanics.
Problem 82P
Step-by-Step Solution
We are given the force applied to the handles of the vise grip is $F = 5\;{\rm{lb}}$.
We are asked the compressive force developed on the smooth bolt shank A at the jaws.
Step 2
The free body diagram of the force applied to the handles of the vise grip is:
We have the distance between points E and C is $EC = 1\;{\rm{in}}$.
We have the distance of force applied from point C is $d = 3\;{\rm{in}}$.
We have the distance between points B and E is $BE = 1.75\;{\rm{in}}$.
Step 3
The angle $\theta$ is,
\[\tan \theta = \frac{{BE}}{d}\]Step 4
Substitute the values in the above expression.
\[\begin{array}{c} \tan \theta = \frac{{1.75\;{\rm{in}}}}{{3\;{\rm{in}}}}\\ \theta = 30.25^\circ \end{array}\]Step 5
The moment of all the forces about point E is,
\[\begin{array}{c} \Sigma {M_E} = 0\\ Fd - {F_{CD}}\sin \theta \left( {EC} \right) = 0 \end{array}\]Step 6
Substitute the values in the above expression.
\[\begin{array}{c} \left( {5\;{\rm{lb}}} \right)\left( {4\;{\rm{in}}} \right) - {F_{CD}}\sin 30.25^\circ \left( {1\;{\rm{in}}} \right) = 0\\ {F_{CD}} = 39.70\;{\rm{lb}} \end{array}\]Step 7
The horizontal reaction at E by equilibrium condition is,
\[{E_x} - {F_{CD}}\cos \theta = 0\]Step 8
Substitute the values in the above expression.
\[\begin{array}{c} {E_x} - \left( {39.70\;{\rm{lb}}} \right)\cos 30.25^\circ = 0\\ {E_x} = 34.30\;{\rm{lb}} \end{array}\]Step 9
The free body diagram of the jaw of the vise grip is:
We have the vertical distance between points A and B is $AB' = 0.75\;{\rm{in}}$.
We have the horizontal distance between points A and B is $AB = 1.5\;{\rm{in}}$.
Step 10
The compressive force developed on the smooth bolt shank A at the jaws by moment about point B is,
\[\begin{array}{c} \Sigma {M_B} = 0\\ \left[ \begin{array}{l} {R_A}\sin 20^\circ \left( {AB'} \right) + {R_A}\cos 20^\circ \left( {AB} \right)\\ - {E_x}\left( {BE} \right) \end{array} \right] = 0 \end{array}\]Step 11
Substitute the values in the above expression.
\[\begin{array}{c} \left[ \begin{array}{l} {R_A}\sin 20^\circ \left( {0.75\;{\rm{in}}} \right) + {R_A}\cos 20^\circ \left( {1.5\;{\rm{in}}} \right)\\ - \left( {34.30\;{\rm{lb}}} \right)\left( {1.75\;{\rm{in}}} \right) \end{array} \right] = 0\\ {R_A} = 36\;{\rm{lb}} \end{array}\]