Step 1

We are given the weight of stone block $A$ as ${W_A} = 60{\rm{0}}\;{\rm{lb}}$, the weight of stone block $B$ as ${W_B} = 150\;{\rm{lb}}$,the weight of stone block $C$ as ${W_C} = 50{\rm{0}}\;{\rm{lb}}$, the coefficient of static friction between the blocks as ${\mu _s} = 0.3$, and the coefficient of static friction between the floor and blocks as ${\mu _s}' = 0.5$.

We are asked to determine the smallesthorizontal force applied to the block.

 
Step 2

We will draw a free body diagram of the three stone blocks.

Here, $N$ is the reaction and $f = {\mu _s}'N$ is the friction force.

 
Step 3

We will resolve the forces in the vertical direction.

\[\begin{array}{c} \sum {{F_y}} = 0\\ N - {W_A} - {W_B} - {W_C} = 0\\ N = {W_A} + {W_B} + {W_C} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} N = \left( {60{\rm{0}}\;{\rm{lb}}} \right) + \left( {150\;{\rm{lb}}} \right) + \left( {50{\rm{0}}\;{\rm{lb}}} \right)\\ N = 1250\;{\rm{lb}} \end{array}\]
 
Step 4

We will resolve the forces in the horizontal direction.

\[\begin{array}{c} \sum {{F_x}} = 0\\ f - P = 0\\ {\mu _s}' \times N - P = 0\\ P = {\mu _s}' \times N \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} P = 0.5\left( {1250\;{\rm{lb}}} \right)\\ P = 625\;{\rm{lb}} \end{array}\]
 
Step 5

We will draw a free body diagram of the stone block $A$.

Here, ${N_A}$ is the reaction force, ${N_1}$ is the reaction force, and ${f_A} = {\mu '_s}{N_A}$ is the horizontal friction force, and ${f_1} = {\mu _s}{N_1}$ is the vertical friction force.

 
Step 6

We will resolve the forces in vertical direction.

\[\begin{array}{c} \sum {{F_y}} = 0\\ {N_A} + {f_1} - {W_A} = 0\\ {N_A} + {\mu _s}{N_1} - {W_A} = 0\\ {N_A} = {W_A} - {\mu _s}{N_1} \end{array}\]

Substitute the given value in the above equation.

\[{N_A} = 600\;{\rm{lb}} - 0.3{N_1}\] … (1)
 
Step 7

We will resolve the forces in horizontal direction.

\[\begin{array}{c} \sum {{F_x}} = 0\\ {f_A} - {N_1} = 0\\ \left( {{\mu _s}'} \right){N_A} - {N_1} = 0\\ {N_1} = \left( {{\mu _s}'} \right){N_A} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {N_1} = 0.5\left( {600\;{\rm{lb}} - 0.3{N_1}} \right)\\ {N_1} = 300\;{\rm{lb}} - 0.15{N_1}\\ {N_1} = 261\;{\rm{lb}} \end{array}\]
 
Step 8

We will draw a free body diagram of the stone block $B$.

Here, ${N_2}$ is the reaction force, ${N_1}$ is the reaction force, and ${f_2} = {\mu _s}{N_2}$ is the friction force, and ${f_1} = {\mu _s}{N_1}$ is the friction force.

 
Step 9

We will resolve the forces in vertical direction.

\[\begin{array}{c} \sum {{F_y}} = 0\\ {N_2}\sin 45^\circ - {f_1} - {f_2}\cos 45^\circ - {W_B} = 0\\ {N_2}\sin 45^\circ - {\mu _s}{N_1} - {\mu _s}{N_1}\cos 45^\circ - {W_B} = 0\\ {N_2} = \frac{{{W_B} + {\mu _s}{N_1}\left( {1 - \cos 45^\circ } \right)}}{{\sin 45^\circ }} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {N_2} = \frac{{150\;{\rm{lb}} + 0.3\left( {261\;{\rm{lb}}} \right)\left( {1 - \cos 45^\circ } \right)}}{{\sin 45^\circ }}\\ {N_2} = 244.6\;{\rm{lb}} \end{array}\]
 
Step 10

We will draw a free body diagram of the stone block $C$.

Here, ${f_C} = \left( {{\mu _s}'} \right){N_C}$ is the friction force.

 
Step 11

We will resolve the forces in vertical direction.

\[\begin{array}{c} \sum {{F_y}} = 0\\ {N_C} - {W_C} + {f_2}\cos 45^\circ - {N_C}\sin 45^\circ = 0\\ {N_C} - {W_C} + {\mu _s}{N_2}\cos 45^\circ - {N_C}\sin 45^\circ = 0\\ {N_C} = \frac{{{W_C} - {\mu _s}{N_2}\cos 45^\circ }}{{\left( {1 - \sin 45^\circ } \right)}} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {N_C} = \frac{{{\rm{500}}\;{\rm{lb}} - 0.3\left( {244.6\;{\rm{lb}}} \right)\cos 45^\circ }}{{\left( {1 - \sin 45^\circ } \right)}}\\ {N_C} = 1530\;{\rm{lb}} \end{array}\]
 
Step 12

We will resolve the forces in horizontal direction.

\[\begin{array}{c} \sum {{F_x}} = 0\\ {N_2}\cos 45^\circ + {f_2}\sin 45^\circ - P + {f_C} = 0\\ {N_2}\cos 45^\circ + {\mu _s}{N_2}\sin 45^\circ - P + \left( {{\mu _s}'} \right){N_C} = 0\\ P = {N_2}\cos 45^\circ + {\mu _s}{N_2}\sin 45^\circ + \left( {{\mu _s}'} \right){N_C} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} P = \left( {244.6\;{\rm{lb}}} \right)\cos 45^\circ + 0.3\left( {244.6\;{\rm{lb}}} \right)\sin 45^\circ + \left( {0.5} \right)\left( {1530\;{\rm{lb}}} \right)\\ P = 989.8\;{\rm{lb}} \end{array}\]

So, the smallest force is $P = 625\;{\rm{lb}}$.