Step 1

We are given the following data:

The mass of drum is $m = 25\;{\rm{kg}}$.

The radius of the drum is $r = 125\;{\rm{mm}}$.

The value of force $P$ is $P = 85\;{\rm{N}}$

The value of moment $M$ is $M = 50\;{\rm{N}} \cdot {\rm{m}}$.

The coefficient of static friction between the drum and brake bar is ${\mu _s} = 0.4$.

The coefficient of kinetic friction between the drum and brake bar is ${\mu _k} = 0.3$.

We are asked to determine the horizontal and vertical components of reaction at the pin $O$.

 
Step 2

We will draw a free body diagram of the brake drum.

Here, ${O_x}$ is the horizontal reaction force at point $O$, ${O_y}$ is the vertical reaction force at point $O$, ${F_B}$ is the friction force at contact surface $B$, ${N_B}$ is the normal reaction force at point $B$ and $W$ is the weight of the brake drum.

 
Step 3

We will take the sum of the moment about point $O$ and equate to zero.

\[\begin{array}{c} \sum {{M_O}} = 0\\ \left[ {M - \left( {{F_B}r} \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ {\left( {50\;{\rm{N}} \cdot {\rm{m}}} \right) - \left( {{F_B}} \right)\left( {125\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)} \right] = 0\\ \left[ {\left( {50\;{\rm{N}} \cdot {\rm{m}}} \right) - \left( {{F_B}} \right)\left( {0.125\;{\rm{m}}} \right)} \right] = 0\\ {F_B} = 400\;{\rm{N}} \end{array}\]
 
Step 4

We will draw a free body diagram of the brake bar.

Here, ${A_x}$ is the horizontal reaction force at point $A$, ${A_y}$ is the vertical reaction force at point $A$, ${N_B}$ is the normal reaction force at point $B$ and ${F_B}$ is the friction force at contact surface $B$.

We will take the sum of the moment about point $A$ and equate to zero.

\[\begin{array}{c} \sum {{M_A}} = 0\\ \left[ {\left( P \right)\left( {{x_1} + {x_2}} \right) + \left( {{F_B}} \right)\left( {{y_1}} \right) - \left( {{N_B}} \right)\left( {{x_2}} \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {85\,{\rm{N}}} \right)\left[ {\left( {300{\rm{ + 700}}} \right)\;{\rm{mm}}\; \times \left( {\frac{{{{10}^{ - 3}}\,{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)} \right]\\ + \left( {400\;{\rm{N}}} \right)\left( {500\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\\ - \left( {{N_B}} \right)\left( {700\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right) \end{array} \right] = 0\\ \left[ {85 + 200 - 0.7{N_B}} \right]\;{\rm{N}}\;{\rm{ = }}\,{\rm{0}}\\ {N_B} \approx 407.14\;{\rm{N}} \end{array}\]
 
Step 5

The formula to calculate the maximum friction force at $B$ is given by,

\[{\left( {{F_B}} \right)_{\max }} = {\mu _s}{N_B}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {\left( {{F_B}} \right)_{\max }} = \left( {0.4} \right)\left( {407.14\;{\rm{N}}} \right)\\ = 162.86\;{\rm{N}} \end{array}\]
 
Step 6

Since, ${F_B} > {\left( {{F_B}} \right)_{\max }}$, the drum slips at point $B$ and rotates. Therefore, the coefficient of kinetic friction should be used.

The formula to calculate the friction force at contacting point $B$ is given by,

\[{F_B} = {\mu _k}{N_B}\]

Substitute all the known values in the above formula.

\[{F_B} = \left( {0.3} \right)\left( {{N_B}} \right)\]
 
Step 7

We will take the sum of the moment about point $A$ and equate to zero.

\[\begin{array}{c} \sum {{M_A}} = 0\\ \left[ {\left( P \right)\left( {{x_1} + {x_2}} \right) + \left( {{F_B}} \right)\left( {{y_1}} \right) - \left( {{N_B}} \right)\left( {{x_2}} \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {85\,{\rm{N}}} \right)\left[ {\left( {300{\rm{ + 700}}} \right)\;{\rm{mm}}\; \times \left( {\frac{{{{10}^{ - 3}}\,{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)} \right]\\ + \left( {0.3{N_B}} \right)\left( {500\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\\ - \left( {{N_B}} \right)\left( {700\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right) \end{array} \right] = 0\\ \left[ {85 + \left( {0.15{N_B}} \right) - 0.7{N_B}} \right]\;{\rm{ = }}\,{\rm{0}}\\ {N_B} \approx 154.54\;{\rm{N}} \end{array}\]
 
Step 8

We will take the sum of the force along the vertical direction and equate to zero.

\[\begin{array}{c} \sum {{F_y}} = 0\\ \left[ {{O_y} - \left( {mg} \right) - {N_B}} \right] = 0 \end{array}\]

Here, $g$ is the gravitational acceleration $\left( {g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)$.

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ {{O_y} - \left( {25\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) - \left( {154.54} \right)} \right]\;{\rm{N}}\;{\rm{ = }}\;0\\ \left[ {{O_y} - \left( {245.25} \right) - \left( {154.54} \right)} \right]\;{\rm{N}}\;{\rm{ = }}\;0\\ {O_y} = 400\;{\rm{N}} \end{array}\]
 
Step 9

We will take the sum of the force along the horizontal direction and equate to zero.

\[\begin{array}{c} \sum {{F_x}} = 0\\ \left[ {{F_B} - {O_x}} \right] = 0\\ \left[ {\left( {0.3{N_B}} \right) - {O_x}} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ {\left( {0.3} \right)\left( {154.54\;{\rm{N}}} \right) - {O_x}} \right] = 0\\ {O_x} = \left( {46.4} \right)\;{\rm{N}} \end{array}\]