Step 1

We are given the resultant couple moment on the crank is $M = 0$.

We are asked the magnitude of couple forces $F$.

 
Step 2

The following is the free body diagram of the system:

We have the distance between the points O and A is A is ${d_1} = 10\;{\rm{in}}$.

We have the distance between the points O and B is ${d_2} = 5\;{\rm{in}}$.

We have the distance between the points O and C is ${d_3} = 4\;{\rm{in}}$.

We have the distance between the points O and D is ${d_4} = 8\;{\rm{in}}$.

We have the angle $\theta $ is $\theta = 45^\circ $.

 
Step 3

The magnitude of moment at point O due to forces at point A is,

\[{M_1} = \left( {150\cos 30^\circ \;{\rm{lb}} \times {d_1}\sin \theta } \right) + \left( {150\sin 30^\circ \;{\rm{lb}} \times {d_1}\cos \theta } \right)\]
 
Step 4

Substitute the values in the above expression.

\begin{array}{l} {M_1} = \left( {150\cos 30^\circ \;{\rm{lb}} \times 10\sin 45^\circ \;{\rm{in}}} \right) + \left( {150\sin 30^\circ \;{\rm{lb}} \times 10\cos 45^\circ \;{\rm{in}}} \right)\\ {M_1} = 1448.9\;{\rm{lb}} \cdot {\rm{in}} \end{array}
 
Step 5

The magnitude of moment at point O due to forces at point B is,

\[{M_2} = - \left( {F\cos 30^\circ \times {d_2}\cos \theta } \right) - \left( {F\sin 30^\circ \times {d_2}\sin \theta } \right)\]
 
Step 6

Substitute the values in the above expression.

\begin{array}{l} {M_2} = - \left( {F\cos 30^\circ \times 5\cos 45^\circ \;{\rm{in}}} \right) - \left( {F\sin 30^\circ \times 5\sin 45^\circ \;{\rm{in}}} \right)\\ {M_2} = - 4.82F\;{\rm{in}} \end{array}
 
Step 7

The magnitude of moment at point O due to forces at point C is,

\[{M_3} = - \left( {F\cos 30^\circ \times {d_3}\cos \theta } \right) + \left( {F\sin 30^\circ \times {d_3}\sin \theta } \right)\]
 
Step 8

Substitute the values in the above expression.

\begin{array}{l} {M_3} = - \left( {F\cos 30^\circ \times 4\cos 45^\circ \;{\rm{in}}} \right) + \left( {F\sin 30^\circ \times 4\sin 45^\circ \;{\rm{in}}} \right)\\ {M_3} = - 1.03F\;{\rm{in}} \end{array}
 
Step 9

The magnitude of moment at point O due to forces at point A is,

\[{M_4} = - \left( {150\cos 30^\circ \;{\rm{lb}} \times {d_4}\sin \theta } \right) + \left( {150\sin 30^\circ \;{\rm{lb}} \times {d_4}\cos \theta } \right)\]
 
Step 10

Substitute the values in the above expression.

\begin{array}{l} {M_4} = - \left( {150\cos 30^\circ \;{\rm{lb}} \times 8\sin 45^\circ \;{\rm{in}}} \right) + \left( {150\sin 30^\circ \;{\rm{lb}} \times 8\cos 45^\circ \;{\rm{in}}} \right)\\ {M_4} = - 310.6\;{\rm{lb}} \cdot {\rm{in}} \end{array}
 
Step 11

The magnitude of net moment at point O due to forces is,

\[M = {M_1} + {M_2} + {M_3} + {M_4}\]
 
Step 12

Substitute the values in the above expression.

\begin{array}{c} 0 = 1448.9\;{\rm{lb}} \cdot {\rm{in}} - 4.82F\;{\rm{in}} - 1.03F\;{\rm{in}} - 310.6\;{\rm{lb}} \cdot {\rm{in}}\\ 5.85F\;{\rm{in}} = 1138.3\;{\rm{lb}} \cdot {\rm{in}}\\ F = 194.6\;{\rm{lb}} \end{array}