Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 80P from Chapter 6 from Hibbeler's Engineering Mechanics.
Problem 80P
Step-by-Step Solution
We are given the force applied to the toggle clamp is $F$.
We are asked the vertical clamping force acting at E.
Step 2
The free body diagram of the toggle clamp is:
We have the distance between points C and D is $CD = \frac{a}{2}$.
We have the distance the applied force from point B is ${d_1} = 2a$.
We have the angle $\theta$ is $\theta = 30^\circ $.
Step 3
The moment of all the forces about point B is,
\[\begin{array}{c} \Sigma {M_B} = 0\\ \left[ \begin{array}{c} {F_{CD}}\cos \theta \left( {CD} \right) - {F_{CD}}\sin \theta \left( {CD} \right)\\ - F\left( {{d_1}} \right) \end{array} \right] = 0 \end{array}\]Step 4
Substitute the values in the above expression.
\[\begin{array}{c} \left[ \begin{array}{c} {F_{CD}}\cos 30^\circ \left( {\frac{a}{2}} \right) - {F_{CD}}\sin 30^\circ \left( {\frac{a}{2}} \right)\\ - F\left( {2a} \right) \end{array} \right] = 0\\ {F_{CD}} \times a = 10.93F \times a\\ {F_{CD}} = 10.93F \end{array}\]Step 5
The horizontal reaction at B by equilibrium condition is,
\[{B_x} - {F_{CD}}\sin \theta = 0\]Step 6
Substitute the values in the above expression.
\[\begin{array}{c} {B_x} - \left( {10.93F} \right)\sin 30^\circ = 0\\ {B_x} = 5.46F \end{array}\]Step 7
The free body diagram of component ABE is:
We have the length of AB is $AB = a$.
We have the length of AE is $AE = 1.5a$.
Step 8
The moment of all the forces at point A is,
\[{B_x}\left( {AB} \right) - {F_E}\left( {AE} \right) = 0\]Step 7
Substitute the values in the above expression.
\[\begin{array}{c} \left( {5.46F} \right)\left( a \right) - {F_E}\left( {1.5a} \right) = 0\\ {F_E} = 3.64F \end{array}\]