A person pushes a cabinet at a height of 3 feet with a horizontal force (P) of 3

Question

A person pushes a cabinet at a height of 3 feet with a horizontal force (P) of 30 pounds up a 5 degree slope as shown. The center of gravity of the cabinet is at 2 feet in height and 1.5 feet from each leg. The coefficient of kinetic friction is 0.15. If the cabinet was on a horizontal surface the application of the force (P) would be directly above the center of leg A. Assume that the cabinet slides without tipping. (5) draw x, y axes and free body diagram. (5) Normal force at A (5) Normal force at B (10) Acceleration tangential to the slope

Explanation / Answer

Normal reaction = N'
friction, f = mu*N'
Applying force balance, Pcos(5) - mu*N' - mgsin(5) = ma
and Psin(5) + mgcos(5) = N' = 30*sin(5) + 60*cos(5) = 62.38635 N
a =[30*cos(5) - 0.15*62.38635 - 60*sin(5)]/60 = 0.25497 inch/s

Normal reaction at A = R
Normal reaction at B = N
using moment balance, Pcos(5)*1.5 - Psin(5)*1.5 + R*1.5 - N*1.5 + mu*R*2 + mu*N*2 = 0
40.9067 + 1.5(R - N) + mu(R + N)*2 =
now R + N = N' = 62.38635
39.7484 = N - R

so Reaction at B = 51.067375 N
so Reaction at A = 11.3189 N

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