Mass-Acceleration Diagram

Normal Forces

Horizontal Forces

The first issue that needs to be addressed is the unknown reactions at the floor. To determine these, begin with a free-body diagram as shown at the left.

Summing the forces in the x and y direction and summing the moments about the center, gives

     ΣF_x = F - f_A - f_B = ma_x

     ΣF_y = N_A - N_B - mg = 0

     ΣM_cg = 1.5N_B - 1.5N_A - 1.5F - 3f_A - 3f_B = 0

Noting that

     f_A = μ N_A             f_B = μ N_B   

The y-direction force equation and moment equation can be solved simultaneously for the normal forces N_A and N_B (the algebra has been omitted):

     N_A = mg (0.5 - μ) - 0.5F

     N_B = mg (0.5 + μ) + 0.5F

Noting that

     m = W/g = 200/32.2 = 6.211 slugs

the final reaction become

     N_A = 6.211 (32.2) (0.5 - 0.3) - 0.5 (75)
          = 2.5 slug-ft/s² = 2.5 lb

     N_B = 6.211 (32.2) (0.5 + 0.3) + 0.5 (75)
          = 197.5 slug-ft/s² = 197.5 lb

The acceleration of the wardrobe can now be determined using the x-direction force equation, giving

     a_x = 1/m (F - f_A - f_B)

          = 1/m (F - μ N_A - μ N_B)

          = 1/6.211 [75 - 0.3 (2.5) - 0.3 (197.5)]

         = 2.415 ft/s²

When the force F is just large enough to cause the wardrobe to tip, the normal force on the left leg is zero, and the right leg supports the entire weight of the wardrobe.

Substituting N_A = 0 and N_B = mg into moment equation, and rearranging gives,

     ΣM_cg = 1.5N_B - 1.5N_A - 1.5F - 3f_A - 3f_B = 0

    1.5 mg - 1.5 (0) - 3 μ N_A - 3 μ N_B = 1.5 F

    1.5 mg - 3 μ mg = 1.5 F

Solving for the minimum force required to tip the wardrobe gives

    F_min = mg (1 - 2μ)

           = 6.211 (32.2) [1 - 2(0.3)] = 80.0 lb