Solution a)

Free Body Diagram for Wedge A

The solution requires finding the smallest magnitude of P required to move the column. First, consider the forces acting on wedge A. Force equilibrium provides two equations,

     ΣF_x = 0
     N₁ - f₂ cos15 - N₂ sin15 = 0

     ΣF_y = 0
     N₂ cos15 - f₁ - W_col - W_A - f₂ sin15 = 0
     N₂ cos15 - f₁ - 1,000 - 100 - f₂ sin15 = 0

Since motion is impending, the friction forces are

     f₁ = μ N₁ = 0.3 N₁
     f₂ = μ N₂ = 0.3 N₂

If these two friction equations are used, the four equations can be reduced to two equations and two unknowns, N₁ and N₂.

     N₁ - 0.3 N₂ cos15 - N₂ sin15 = 0
     N₂ cos15 - 0.3 N₁ - 0.3 N₂ sin15 = 1,100

     N₁ = 0.5486 N₂
     0.8883 N₂ - 0.3 N₁ = 1,100

Solving gives

     N₂ = 1,520 lb     and     N₁ = 833.3 lb

and the friction forces are

     f₂ = 456.0 lb     and     f₁ = 250.0 lb

Now that the forces on the bottom surface of Wedge A are known, wedge B can be analyzed. First, sum the forces in the vertical direction, to give,

     ΣF_y = 0
     f₂ sin15 + N₃ - W_B - N₂ cos15 = 0
     (456) sin15 + N₃ - 100 - (1,520) cos15 = 0

     N₃ = 1,450 lb   and     f₃ = 435 lb

Finally, P can be determined by summing forces on wedge B in the horizontal,

     ΣF_x = 0
     f₃ + f₂ cos15 + N₂ sin15 - P = 0
     (435) + (456) cos15 + (1,520) sin15 - P = 0

Solving for P gives,

     P = 1,269 lb

Solution b)

Wedge A and B Interaction

The applied force and lateral constraints have been removed. As a result, the direction of potential motion has changed, and the frictional forces will reverse direction.

The equations of equilibrium on wedge A will produce

     ΣF_x = 0
     f₂ cos15 - N₂ sin15 = 0
     f₂ = 0.2679 N₂

     ΣF_y = 0
     N₂ cos15 + f₂ sin15 - W_col - W_A = 0
     N₂ cos15 + (0.2679 N₂) sin15 - 1,000 - 100 = 0

Since the status of the wedge is being analyzed, it cannot be assumed that f₂ = μN₂. The resulting forces are

     N₂ = 1,063 lb    and     f₂ = 285 lb

The friction due to impending motion is determined as

     f_{2 max} = μ N₂ = 318.9 lb

Since this value is greater than the friction due to equilibrium, wedge A must be in equilibrium.

The minimum coefficient of friction value required to maintain equilibrium is 0.268.

Since the top wedge is not moving, it is impossible for the bottom wedge to move. This can be verified with a free-body diagram of wedge B.

Also, if the column force were to increase due to the weight from the ceiling, the wedge would still not move. Feel free to experiment with this and other factors in the simulation.