Friction - - Rigid Bodies in Translation (continued)

m = 4 kg g = 10 m/sec^2 P = 10 N μ = 1/4 θ = 30^o Block starts from rest.

Case 2: Check for sliding in the + x direction.

∑ F_x = m a_Cx , ˗ F ˗ mg sinθ + P cosθ = m a_Cx (1)

and ∑ F_y = 0 N ˗ P sinθ ˗ mg cosθ = 0 (2)

From (2) N = P sinθ + mg cosθ and for the data: N = 39.63 N

Also for sliding F = μ N = 9.91 N Insert into eq (1)

˗ P sinθ ˗ mg cosθ ˗ mg sinθ + P cosθ = m a_Cx

and for the data: a_Cx = ˗ 5.31 m/sec² Since a_Cx and F are in the same direction (˗ x)

(they do not oppose each other), sliding up the incline does not occur. Move on to Case 3.

Case 2: Check for sliding in the ˗ x direction. (Note: need a new free body diagram

showing friction acting in the + x direction as shown below).

∑ F_x = m a_Cx , F ˗ mg sinθ + P cosθ = m a_Cx (1)

and as before F = μ N Solve for a_Cx and use data: a_Cx = ˗ 0.3523 m/sec²

Here a_Cx < 0 and F > 0 i.e. they are in opposite directions .

So the block will slide down the sloped surface with an acceleration a_Cx = ˗ 0.3523 i m/sec²

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