Kinetics ˗ Body Translating in a Plane

Solution: (Continued)

For FBD 1: ↑ ∑ F_y = 0 , N ˗ Mg = 0 (1)

→ ∑ F_x = M a_B , F = M a_B (2)

For FBD 2: → ∑ F_x = m a_C , P ˗ F = m a_C So P = F + m a_C (3)

For sliding to impend F = μ N = μ Mg So by eq. (2) μMg = M a_B and a_B = μ g

But if sliding motion impends a_B = a_C and from eq. (3) P = μ (M + m) g (result)

For tipping to impend the normal force acts at the corner A of block B. Then summing moments

taking counterclockwise as positive: Note: α = 0 k since block B is translating.

M_B = 0 k F(h/2) ˗ Mg(h/6) = 0 and F = (1/3) Mg (4)

Then eqs (2) and (4) give a_B = (1/3) g (5)

Again if tipping impends a_C = a_B and from eq. (3 )

P = (1/3)Mg + m (1/3)g or P = (1/3)(M+m)g (result if tipping impends)

Now if friction is such that μ = 1/4 sliding impends when P = (1/4) (M + m)g.

So for μ = 1/4, sliding will impend before tipping. (result)

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