Step 1

We are given that the weight of block A is ${W_A} = 100\;{\rm{N}}$, the weight of block B is ${W_B} = 100\;{\rm{N}}$, the weight of block C is ${W_C} = 200\;{\rm{N}}$ and the coefficient of static friction is ${\mu _s} = 0.2$.

We are asked to calculate the force required to move the block B.

 
Step 2

The free body diagram of the block A is shown as:

Here, ${F_A}$ is the frictional force acting on block A, ${N_A}$ is the reaction force acting on block A, ${W_A}$ is the weight on block A and $T$ is the tension in rope.

 
Step 3

According to the free body diagram of the block A, the net forces acting on the block A along the y-axis is given as:

\[\begin{array}{c} \sum {F_y} = 0\\ {N_A} - {W_A} = 0\\ {N_A} = {W_A} \end{array}\]
 
Step 4

Substitute the known value in the equation:

\[{N_A} = 100\;{\rm{N}}\]
 
Step 5

To calculate the frictional force acting on block A we use the formula:

\[{F_A} = {\mu _s}{N_A}\]
 
Step 6

Substitute the known values in the formula:

\[\begin{array}{c} {F_A} = \left( {0.2} \right)\left( {100\;{\rm{N}}} \right)\\ = 20\;{\rm{N}} \end{array}\]
 
Step 7

The free body diagram of the block A and B is shown as:

Here, ${F_B}$ is the frictional force acting on block B, ${N_B}$ is the reaction force acting on block B, ${W_B}$ is the weight on block B and $P$ is the force applied to block B.

 
Step 8

According to the free body diagram of the block B, the net forces acting on the block B along the y-axis is given as:

\[\begin{array}{c} \sum {F_y} = 0\\ {N_B} - {W_A} - {W_B} = 0\\ {N_B} = {W_A} + {W_B} \end{array}\]
 
Step 9

Substitute the known value in the equation:

\[\begin{array}{c} {N_B} = 100\;{\rm{N}} + 100\;{\rm{N}}\\ = 200\;{\rm{N}} \end{array}\]
 
Step 10

To calculate the frictional force acting on block B we use the formula:

\[{F_B} = {\mu _s}{N_B}\]
 
Step 11

Substitute the known values in the formula:

\[\begin{array}{c} {F_B} = \left( {0.2} \right)\left( {200\;{\rm{N}}} \right)\\ = 40\;{\rm{N}} \end{array}\]
 
Step 12

According to the free body diagram of the block B, the net forces acting on the block B along the x-axis is given as:

\[\begin{array}{c} \sum {F_x} = 0\\ P - {F_A} - {F_B} = 0\\ P = {F_A} + {F_B} \end{array}\]
 
Step 13

Substitute the known value in the equation:

\[\begin{array}{c} P = 20\;{\rm{N}} + 40\;{\rm{N}}\\ = 60\;{\rm{N}} \end{array}\]
 
Step 14

The free body diagram of the block A, B and C together is shown as:

Here, ${F_C}$ is the frictional force acting on block C, ${N_C}$ is the reaction force acting on block C and ${W_C}$ is the weight on block C.

 
Step 15

According to the free body diagram of the block C, the net forces acting on the block C along the y-axis is given as:

\[\begin{array}{c} \sum {F_y} = 0\\ {N_C} - {W_A} - {W_B} - {W_C} = 0\\ {N_C} = {W_A} + {W_B} + {W_C} \end{array}\]
 
Step 16

Substitute the known value in the equation:

\[\begin{array}{c} {N_C} = 100\;{\rm{N}} + 100\;{\rm{N}} + 200\;{\rm{N}}\\ = 400\;{\rm{N}} \end{array}\]
 
Step 17

To calculate the frictional force acting on block C we use the formula:

\[{F_C} = {\mu _s}{N_C}\]
 
Step 18

The coefficient of static friction for block C is ${\mu _s} = 0.1$.

Substitute the known values in the formula:

\[\begin{array}{c} {F_C} = \left( {0.1} \right)\left( {400\;{\rm{N}}} \right)\\ = 40\;{\rm{N}} \end{array}\]
 
Step 19

According to the free body diagram of the system, the net forces acting on the system along the x-axis is given as:

\[\begin{array}{c} \sum {F_x} = 0\\ P - {F_A} - {F_B} = 0\\ P = {F_A} + {F_B} \end{array}\]
 
Step 20

Substitute the known value in the equation:

\[\begin{array}{c} P = 20\;{\rm{N}} + 40\;{\rm{N}}\\ = 60\;{\rm{N}} \end{array}\]