Wedge Used to Raise
and Lower A

The vertical position of a machine (block A) is adjusted by moving wedge B. The coefficient of static friction between all surfaces is 0.3. Determine the horizontal force, P, acting on wedge B, that is required to

a) raise the block A (acting on the right side), and
b) lower the block A (acting on the left side).

The first part of this problem asks for the the smallest value of the force, P, to raise the machine. This force will act on the right side of the wide of block A, as shown, in order to push block A upward. As with all static problems, a free-body diagram will help identify all forces acting on an object.

Since there is a known force of 1,500 lb acting on block A, this object will be analyzed first. All forces acting on the block 'A' are shown in the free-body diagram on the left. The frictional forces are,

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

Applying the equilibrium equations gives,

     ΣF_x = 0
     N₁ - f₂ cos 12 - N₂ sin 12 = 0     
     N₁ - 0.3 N₂ (0.9781) - 0.2079 N₂ = 0
     N₁ = 0.5013 N₂

     ΣF_y = 0
     N₂ cos 12 - f₁ - 1,500 - f ₂ sin 12 = 0 
     0.9781 N₂ - 0.3 N₁ - 1,500 - 0.3 N₂ (0.2079)= 0 
     0.9158 N₂ - 0.3 N₁ = 1,500

Solving above two equations gives,

     N₂ = 1,960 lb
     N₁ = 982.4 lb              

and the frictional forces are,

     f₂ = 587.9 lb
     f₁ = 294.7 lb

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

     ΣF_y = 0
     N₃ + f₂ sin 12 - N₂ cos 12 = 0
     N₃ + 587.9 (0.2079) - 1,960 (0.9781) = 0
     N₃ = 1,795 lb and f₃ = 538.5 lb

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

     ΣF_x = 0
     N₂ sin 12 + f₃ + f₂ cos 12 - P = 0
     1,960 (0.2079) + 538.5 + 587.9 (0.9781) = P

     P = 1,521 lb

The second part of this problem asks for the the smallest value of the force, P, to lower the machine. This force will act on the left side of the wide of block A, as shown, in order to lower block A downward.

Since force P is acting in the opposite direction than in the previous solution, the forces acting on block A will be in opposite directions. The new forces acting on the block A are shown in the free-body diagram on the left. Frictional forces are given by,

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

Applying the equilibrium equations gives,

     ΣF_x = 0
     f₂ cos 12 - N₂ sin 12 - N₁= 0     
     (0.3) (0.9781) N₂ - 0.2079 N₂ - N₁ = 0
     N₁ = 0.08553 N₂ 

     ΣF_y = 0
     N₂ cos 12 + f₁ - 1,500 + f₂ sin 12 = 0 
     0.9781N₂ + 0.3N₁ - 1,500 + 0.3N₂ (0.2079) = 0 
     1.0405N₂ + 0.3N₁ = 1,500 

Solving above two equations gives,

     N₁ = 120.1 lb
     N₂ = 1,407 lb

and the frictional forces are,

     f₁ = 36.03 lb
     f₂ = 422.1 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
     N₃ - f₂ sin 12 - N₂ cos 12 = 0
     N₃ - 422.1 (0.2079) - 1,407 (0.9781) = 0

     N₃ = 1,464 lb and f₃ = μ N₃ = 439.2 lb

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

     ΣF_x = 0
     N₂ sin 12 - f₃ - f₂ cos 12 + P = 0
     1,407 (0.2079) - 439.2 - 422.1 (0.9781) - P = 0

     P = -559.5 lb
        = 559.5 lb ←