Step 1

We are given that the coefficient of static friction between B and D is ${\mu _s} = 0.2$, the coefficient of static friction between A and B is $\mu _s^{'} = 0.1$ and the angle made by wedge with the horizontal is $\theta = 15^\circ $. The weight of load is $W = 3000\;{\rm{lb}}$.

We are asked to calculate the reversed force required to pull out wedge.

 
Step 2

The free body diagram of the wedge B is shown as:

Here, $W$ is the weight of load, ${F_B}$ is the frictional force acting on bottom surface, ${N_B}$ is the normal force on bottom surface, ${F_D}$ is the frictional force acting on side D and ${N_D}$ is the normal force acting on side D.

 
Step 3

According to free body diagram, the net force acting on wedge B along x-axis is given as:

\[\begin{array}{c} \sum {F_x} = 0\\ - {F_B}\cos \theta + {N_B}\sin \theta - {N_D} = 0 \end{array}\]…… (1)
 
Step 4

To calculate the frictional force acting on bottom surface we use the formula:

\[{F_B} = \mu _s^{'}{N_B}\]
 
Step 5

Substitute the known value in the equation (1):

\[\begin{array}{c} - \mu _s^{'}{N_B}\cos \theta + {N_B}\sin \theta - {N_D} = 0\\ {N_D} = {N_B}\left( { - \mu _s^{'}\cos \theta + \sin \theta } \right) \end{array}\]…… (2)
 
Step 6

According to free body diagram, the net force acting on wedge B along y-axis is given as:

\[\begin{array}{c} \sum {F_y} = 0\\ {N_B}\cos \theta + {F_B}\sin \theta + {F_D} - W = 0 \end{array}\]…… (3)
 
Step 7

To calculate the frictional force acting on side D we use the formula:

\[{F_D} = {\mu _s}{N_D}\]
 
Step 8

Substitute the known value in equation (3):

\[{N_B}\cos \theta + \mu _s^{'}{N_B}\sin \theta + {\mu _s}{N_D} - W = 0\]
 
Step 9

Substitute the known values in the equation:

\[\begin{array}{c} {N_B}\cos \theta + \mu _s^{'}{N_B}\sin \theta + {\mu _s}{N_B}\left( { - \mu _s^{'}\cos \theta + \sin \theta } \right) - W = 0\\ {N_B}\left( {\cos \theta + \mu _s^{'}\sin \theta - {\mu _s}\mu _s^{'}\cos \theta + {\mu _s}\sin \theta } \right) = W\\ {N_B} = \frac{W}{{\left( {\cos \theta + \mu _s^{'}\sin \theta - {\mu _s}\mu _s^{'}\cos \theta + {\mu _s}\sin \theta } \right)}} \end{array}\]
 
Step 10

Substitute the known values in the equation:

\[\begin{array}{c} {N_B} = \frac{{\left( {3000\;{\rm{lb}}} \right)}}{{\left( {\cos 15^\circ + \left( {0.1} \right)\sin 15^\circ - \left( {0.1} \right)\left( {0.2} \right)\cos 15^\circ + \left( {0.2} \right)\sin 15^\circ } \right)}}\\ = \frac{{\left( {3000\;{\rm{lb}}} \right)}}{{\left( {0.966 + \left( {0.1} \right)\left( {0.259} \right) - \left( {0.02} \right)\left( {0.966} \right) + \left( {0.2} \right)\left( {0.259} \right)} \right)}}\\ = \frac{{\left( {3000\;{\rm{lb}}} \right)}}{{\left( {1.024} \right)}}\\ = 2929\;{\rm{lb}} \end{array}\]
 
Step 11

Substitute the known value in equation (2):

\[\begin{array}{c} {N_D} = \left( {{\rm{2929}}\;{\rm{lb}}} \right)\left( { - 0.1\cos 15^\circ + \sin 15^\circ } \right)\\ = \left( {{\rm{2929}}\;{\rm{lb}}} \right)\left( { - \left( {0.1} \right)\left( {0.966} \right) + 0.259} \right)\\ = \left( {{\rm{2929}}\;{\rm{lb}}} \right)\left( {0.1624} \right)\\ = 475\;{\rm{lb}} \end{array}\]
 
Step 12

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

Here, ${\bf{P}}$ is the required force, ${F_C}$ is the frictional force acting on bottom surface and ${N_C}$ is the normal force acting on bottom surface.

 
Step 13

According to free body diagram, the net force acting on wedge along y-axis is given as:

\[\begin{array}{c} \sum {F_y} = 0\\ {N_C} - {F_B}\sin \theta - {N_B}\cos \theta = 0 \end{array}\]
 
Step 14

Substitute the known value in equation (2):

\[\begin{array}{c} {N_C} - \left( {\mu _s^{'}{N_B}} \right)\sin \theta - {N_B}\cos \theta = 0\\ {N_C} = {N_B}\left( {\cos \theta + \mu _s^{'}\sin \theta } \right) \end{array}\]
 
Step 15

Substitute the known values in the equation:

\[\begin{array}{c} {N_C} = \left( {{\rm{2929}}\;{\rm{lb}}} \right)\left( {\cos 15^\circ + \left( {0.1} \right)\sin 15^\circ } \right)\\ = \left( {{\rm{2929}}\;{\rm{lb}}} \right)\left( {0.966 + \left( {0.1} \right)\left( {0.259} \right)} \right)\\ = \left( {{\rm{2929}}\;{\rm{lb}}} \right)\left( {0.992} \right)\\ = 2905\;{\rm{lb}} \end{array}\]
 
Step 16

According to free body diagram, the net force acting on wedge along x-axis is given as:

\[\begin{array}{c} \sum {F_x} = 0\\ - {\bf{P}} + {F_C} + {F_B}\cos \theta - {N_B}\sin \theta = 0 \end{array}\]…… (3)
 
Step 17

To calculate the spring force acting on bottom surface we use the formula:

\[{F_C} = {\mu _s}{N_C}\]
 
Step 18

Substitute the known value in equation (3):

\[\begin{array}{c} - {\bf{P}} + {\mu _s}{N_C} + \mu _s^{'}{N_B}\cos \theta - {N_B}\sin \theta = 0\\ {\bf{P}} = {\mu _s}{N_C} + {N_B}\left( {\mu _s^{'}\cos \theta - \sin \theta } \right) \end{array}\]
 
Step 19

Substitute the known values in the equation:

\[\begin{array}{c} {\bf{P}} = \left( {0.2} \right)\left( {{\rm{2905}}\;{\rm{lb}}} \right) + \left( {{\rm{2929}}\;{\rm{lb}}} \right)\left( {\left( {0.1} \right)\cos 15^\circ - \sin 15^\circ } \right)\\ = 581\;{\rm{lb}} + \left( {{\rm{2929}}\;{\rm{lb}}} \right)\left( {\left( {0.1} \right)\left( {0.966} \right) - 0.259} \right)\\ = 581\;{\rm{lb}} - 475\;{\rm{lb}}\\ = 106\;{\rm{lb}} \end{array}\]