Asked by ylopez
A 20 kg slab is to be pulled up a plane inclined 20 degrees with the horizontal at a constant speed by a force that is directed 15 degrees above the surface of the inclined plane. Determine the magnitude of the force necessary to perform the task. The coefficient of friction between the block and the plane is 0.2.
Answers
Answered by
Henry
Ws = M*g = 20 * 9.8 = 196 N.
Fp = 196*sin20 = 67 N.
Fn = 196*Cos20 - F*sin15 = 184-0.26F.
Fs = 0.2(184-0.26F) = 36.8-0.052F. = Static friction.
F*Cos15-Fp-Fs = M*a.
0.97F-67-(36.8-0.052F) = M*0 = 0.
0.97F-67-36.8+0.052F = 0,
1.022F = 36.8, F = 36 N.
Fp = 196*sin20 = 67 N.
Fn = 196*Cos20 - F*sin15 = 184-0.26F.
Fs = 0.2(184-0.26F) = 36.8-0.052F. = Static friction.
F*Cos15-Fp-Fs = M*a.
0.97F-67-(36.8-0.052F) = M*0 = 0.
0.97F-67-36.8+0.052F = 0,
1.022F = 36.8, F = 36 N.
Answered by
Anonymous
In the 0.97F-67-(36.8-0.052F) = M*0 = 0, where the 0.97 came from?
Answered by
Henry
Cos15 = 0.966.
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