A 1100-N crate is being pushed across a level floor at a constant speed by a for

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When the force is applied below the horizontal, the normal force, n is increased. When applied above the horizontal, it is decreased. In a) the forces acting horizontally are : SF = 0 = 350Ncos20.0° - µn µ = 329N / n---->eqn (1) The normal force is found from the net vertical forces : SF = 0 = n - 350Nsin20.0° - 1100N n = 1220N------>eqn (2) Plugging (2) into (1) : µ = 329N / 1220 = 0.270 b) The horizontal forces are : SF = ma = 350Ncos20.0° - 0.27n a = (329N - 0.270n) / (1100N / 9.80m/s²) = (329N - 0.270n) / 112kg------>eqn (1) The net forces vertically are : SF = 0 = n - 1100N + 350sin20.0° n = 980N------->eqn (2) Again, plugging (2) into (1) : a = [329N - 0.270(980N)] / 112kg = 0.575m/s²

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