Step 1

We are given a beam $AB$ having a negligible mass and thickness, which is supportinguniform block. The beam $AB$ is pinned at point $A$ and rests on thetop of a post $BC$, having a negligible thickness. The following data is given:

The mass of the uniform block is $m = 200{\rm{ kg}}$.

The mass of the post $BC$ is ${m_1} = 20{\rm{ kg}}$.

The coefficients of static friction at $B$ and $C$ are ${\mu _B} = 0.4$ and ${\mu _C} = 0.2$ respectively.

We are asked to determine the minimum force $P$ needed to move the post.

 
Step 2

The free-body diagram of the member $AB$ is given by,

Here, ${A_x}$ is the horizontal component of the support reaction at point $A$, ${A_y}$ is the vertical component of the support reaction at point $A$, ${F_B}$ is the friction force acting at point $B$, and ${N_B}$is the normal reaction force at point $B$.

On taking moment about point A, we get:

\[\begin{array}{c} {N_B} \times \left( {1.5{\rm{ m}} + 1.5{\rm{ m}}} \right) - mg \times \left( {1.5{\rm{ m}}} \right) = 0\\ {N_B} \times \left( {3{\rm{ m}}} \right) = 200{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times \left( {1.5{\rm{ m}}} \right)\\ {N_B} = 981{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2} \times \left( {\frac{{1{\rm{ N}}}}{{{\rm{1 kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ {N_B} = 981{\rm{ N}} \end{array}\]
 
Step 3

The free-body diagram of the member $BC$ is given by,

Here, $P$ is the force applied to the post, ${F_C}$ is the friction force acting at point $C$, and ${N_C}$ is the normal reaction force at point $C$.

 
Step 4

At equilibrium condition, balancing all the forces acting on member $BC$ in the vertical direction, we have,

\[\begin{array}{c} {N_C} + P\sin \theta - {N_B} - {m_1}g = 0\\ {N_C} + P \times \left( {\frac{3}{5}} \right) - 981{\rm{ N}} - 20{\rm{ kg}} \times \left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right) \times \left( {\frac{{1{\rm{ N}}}}{{{\rm{1 kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right) = 0\\ {N_C} + 0.6P - 981{\rm{ N}} - 196.2{\rm{ N}} = 0\\ {N_C} + 0.6P - 1177.2{\rm{ N}} = 0 \end{array}\] …. (1)

Taking moments about point $C$ of the member $BC$, we get,

\[\begin{array}{c} {F_B} \times \left( {1.75{\rm{ m}}} \right) - P\cos \theta \times \left( {0.75} \right) = 0\\ {F_B} \times \left( {1.75{\rm{ m}}} \right) - P \times \left( {\frac{4}{5}} \right) \times \left( {0.75{\rm{ m}}} \right) = 0\\ 1.75{F_B} - 0.6P = 0\\ {F_B} = 0.343P \end{array}\]

Taking moments about point $B$ of the member $BC$, we get,

\[\begin{array}{c} P\cos \theta \times \left( {1{\rm{ m}}} \right) - {F_C} \times \left( {1.75{\rm{ m}}} \right) = 0\\ P \times \left( {\frac{4}{5}} \right) \times \left( {{\rm{1 m}}} \right) - {F_C} \times \left( {1.75{\rm{ m}}} \right) = 0\\ 0.8P - 1.75{F_C} = 0 \end{array}\] … (2)
 
Step 5

Assuming that the point $C$ slips if the force is applied, so according to the friction law, the friction force acting at point $C$ is given by,

\[\begin{array}{l} {F_C} = {\mu _C}{N_C}\\ {F_C} = 0.2{N_C} \end{array}\]

On substituting the value of ${F_C}$ in equation (2), we get,

\[\begin{array}{c} 0.8P - 1.75 \times 0.2{N_C} = 0\\ 0.8P - 0.35{N_C} = 0\\ {N_C} = 2.286P \end{array}\]

From equation (1) we have,

\[\begin{array}{c} 2.286P + 0.6P - 1177.2{\rm{ N}} = 0\\ 2.886P - 1177.2{\rm{ N}} = 0\\ P = 408{\rm{ N}} \end{array}\]
 
Step 6

Then the reaction force at point $C$ of the member $BC$ is given by,

\[\begin{array}{l} {N_C} = 2.286 \times 408{\rm{ N}}\\ {N_C} = 932.688{\rm{ N}} \end{array}\]

Then, the friction force acting at point $C$ of the member $BC$ is given by,

\[\begin{array}{l} {F_C} = 0.2 \times 932.688{\rm{ N}}\\ {F_C} = 186.538{\rm{ N}} \end{array}\]

Similarly, the friction force acting at point $B$ of the member $AB$ is given by,

\[\begin{array}{c} {F_B} = 0.343 \times 408{\rm{ N}}\\ {F_B} = 139.94{\rm{ N}} \end{array}\]
 
Step 7

Now according to the friction law, the maximum force required for the slipping to occur at point $B$, is given by,

\[\begin{array}{c} {\left( {{F_B}} \right)_{\max }} = {\mu _B}{N_B}\\ {\left( {{F_B}} \right)_{\max }} = 0.4 \times 981{\rm{ N}}\\ {\left( {{F_B}} \right)_{\max }} = 392.4{\rm{ N}} \end{array}\]

Comparing the friction force values at point $B$, we have,

\[{F_B} < {\left( {{F_B}} \right)_{\max }}\]

So, at point $B$ slipping will not occur which means that our assumption was correct that the slipping only occurs at point $C$.