Step 1

We are given the following data:

The diameter of the pin is $d = 0.5\;{\rm{in}}{\rm{.}}$.

The value of angle is $\theta = 45^\circ $.

The magnitude of the applied load is $F = 50\;{\rm{lb}}$.

The magnitude of $P$ is $P = 41\;{\rm{lb}}$.

We are asked to determine the coefficient of static friction between the pin and the bell crank.

 
Step 2

The free-body diagram of the bell crank is given below.

Here, $M$ represent the moment, $R$ is the resultant force.

The crank is on the verge of rotating counterclockwise so the moment $M$ must rotate clockwise.

From the free-body diagram,

The distances ${x_1}$ is ${x_1} = 10\;{\rm{in}}{\rm{.}}$.

The distances ${y_1}$ is ${y_1} = 12\;{\rm{in}}{\rm{.}}$.

 
Step 3

We will take the sum of the forces along the $x$ direction and equate to zero to obtain the resultant force in the $x$ direction ${R_{\rm{x}}}$.

\[\begin{array}{c} \sum {{F_{\rm{x}}}} = 0\\ \left[ { - {R_{\rm{x}}} + \left( {P\cos \theta } \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ { - {R_{\rm{x}}} + \left( {41\;{\rm{lb}}} \right)\left( {\cos 45^\circ } \right)} \right] = 0\\ {R_{\rm{x}}} \approx 28.99\;{\rm{lb}} \end{array}\]
 
Step 4

We will take the sum of the forces along the $y$ direction and equate to zero to obtain the resultant force in the $y$ direction ${R_y}$.

\[\begin{array}{c} \sum {{F_{\rm{y}}}} = 0\\ \left[ {{R_{\rm{y}}} - F - \left( {P\sin \theta } \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} {R_{\rm{y}}} - \left( {50\;{\rm{lb}}} \right)\\ - \left( {41\;{\rm{lb}}} \right)\left( {\sin 45^\circ } \right) \end{array} \right] = 0\\ \left[ {{R_{\rm{y}}} - \left( {78.99\;{\rm{lb}}} \right)} \right] = 0\\ {R_{\rm{y}}} = 78.99\;{\rm{lb}} \end{array}\]
 
Step 5

The formula to calculate the reaction force is given by,

\[R = \sqrt {{{\left( {{R_{\rm{x}}}} \right)}^2} + {{\left( {{R_{\rm{y}}}} \right)}^2}} \]

Here, $R$ represent the resultant reaction force.

Substitute all the known values in the above formula.

\[\begin{array}{c} R = \sqrt {{{\left( {28.99\;{\rm{lb}}} \right)}^2} + {{\left( {78.99\;{\rm{lb}}} \right)}^2}} \\ = \sqrt {7079.84\;{\rm{l}}{{\rm{b}}^{\rm{2}}}} \\ \approx 84.14\;{\rm{lb}} \end{array}\]
 
Step 6

The formula to calculate the radius of the pin is given by,

\[r = \frac{d}{2}\]

Here, $r$ represent the radius of the pin.

Substitute all the known values in the above formula.

\[\begin{array}{c} r = \frac{{\left( {{\rm{0}}{\rm{.5}}\;{\rm{in}}{\rm{.}}} \right)}}{2}\\ = 0.25\;{\rm{in}}{\rm{.}} \end{array}\]
 
Step 7

We will take the sum of the moment about hinge point and equate to zero to obtain the angle between the pin and the bell crank ${\phi _s}$.

\[\begin{array}{c} \sum M = 0\\ \left[ {\left( F \right)\left( {{x_1}} \right) - \left( R \right)\left( {r\sin {\phi _{\rm{s}}}} \right) - P\left( {{y_1}} \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {50\;{\rm{lb}}} \right)\left( {10\;{\rm{in}}{\rm{.}}} \right)\\ - \left( {84.14\;{\rm{lb}}} \right)\left( {0.25\;{\rm{in}}{\rm{.}}} \right)\left( {\sin {\phi _{\rm{s}}}} \right)\\ - \left( {41\;{\rm{lb}}} \right)\left( {12\;{\rm{in}}{\rm{.}}} \right) \end{array} \right] = 0\\ \left[ \begin{array}{l} \left( {8\;{\rm{lb}} \cdot {\rm{in}}{\rm{.}}} \right)\\ - \left( {21.035\;{\rm{lb}} \cdot {\rm{in}}{\rm{.}}} \right)\left( {\sin {\phi _{\rm{s}}}} \right) \end{array} \right] = 0\\ \sin {\phi _{\rm{s}}} = 0.3803\\ {\phi _{\rm{s}}} = 22.35^\circ \end{array}\]
 
Step 8

The formula to calculate the coefficient of static friction is given by,

\[{\mu _{\rm{s}}} = \tan {\phi _{\rm{s}}}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {\mu _{\rm{s}}} = \tan {\phi _{\rm{s}}}\\ = \tan \left( {22.35^\circ } \right)\\ = 0.41 \end{array}\]