Step 1

We are given the coefficient of static friction between the drum and brake bar is ${\mu _s} = 0.4$, the moment is $M = 35\;{\rm{N}} \cdot {\rm{m}}$, and the mass of the drum is $m = 25\;{\rm{kg}}$.

We are asked to determine the smallest force P that needs to be applied to the brake bar in order to prevent the drum from rotating and the corresponding horizontal and vertical components of reaction at pin O.

 
Step 2

The free body diagram of the brake bar is shown as:

We have the horizontal distance between point A and applied force is $x = 300\;{\rm{mm}} + 700\;{\rm{mm}} = 1000\;{\rm{mm}}$.

We have the vertical distance between point A and applied force is $y = 500\;{\rm{mm}}$.

We have the distance between points A and B is $AB = 700\;{\rm{mm}}$.

 
Step 3

The free body diagram of the brake drum is shown as:

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

 
Step 4

The formula to calculate the friction force at point B is,

\[{F_B} = {\mu _s}{N_B}\] ... (1)
 
Step 5

The formula to calculate the moment about point A is,

\[\begin{array}{c} Px - {N_B}\left( {AB} \right) + {F_B}\left( y \right) = 0\\ Px - {N_B}\left( {AB} \right) + \left( {{\mu _s}{N_B}} \right)\left( y \right) = 0 \end{array}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} P\left( {1000\;{\rm{mm}}} \right) - {N_B}\left( {700\;{\rm{mm}}} \right)\\ + \left( {0.4{N_B}} \right)\left( {500\;{\rm{mm}}} \right) \end{array} \right] = 0\\ P = 0.5{N_B} \end{array}\] ... (2)
 
Step 7

The formula to calculate the moment about point O is,

\[M - {F_B}\left( r \right) = 0\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} \left( {35\;{\rm{N}} \cdot {\rm{m}}} \right) - {F_B}\left( {125\;{\rm{mm}} \times \frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right) = 0\\ {F_B} = 280\;{\rm{N}} \end{array}\]
 
Step 9

Substitute the values in the equation (1).

\[\begin{array}{c} 280\;{\rm{N}} = \left( {0.4} \right){N_B}\\ {N_B} = 700\;{\rm{N}} \end{array}\]
 
Step 10

Substitute the value of ${N_B}$ in the equation (2).

\[\begin{array}{l} P = 0.5\left( {700\;{\rm{N}}} \right)\\ P = 350\;{\rm{N}} \end{array}\]
 
Step 11

The formula to calculate the forces acting in horizontal direction over the drum is,

\[{O_x} + {F_B} = 0\]
 
Step 12

Substitute the values in the above expression.

\[\begin{array}{c} {O_x} + \left( {280\;{\rm{N}}} \right) = 0\\ {O_x} = - 280\;{\rm{N}} \end{array}\]
 
Step 13

The formula to calculate the forces acting in vertical direction over the drum is,

\[{O_y} - mg - {N_B} = 0\]

Here, g is the acceleration due to gravity having a standard value of $9.81\;{\rm{m/}}{{\rm{s}}^2}$.

 
Step 14

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{l} {O_y} - \left( {25\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {\frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ - \left( {700\;{\rm{N}}} \right) \end{array} \right] = 0\\ {O_y} = 945.2\;{\rm{N}} \end{array}\]