A stick of weight 2 lb is moving under the action of the known force P = 10 lb a

Question

A stick of weight 2 lb is moving under the action of the known force P = 10 lb as shown at right. At point A, the stick is sliding with friction along the horizontal surface. The kinetic coefficient of friction between the stick and stationary surface at A is mu_k = 0.05. At the instant, the angular velocity of the rod is 2 rad/s clockwise. Your objective will be to derive a set of equations that will determine the reaction forces on the stick, the acceleration of the center of gravity of the stick, and the acceleration of point A on the stick. You may assume that the stick does not lose contact with the horizontal surface.

Explanation / Answer

Force balance on the stick
P - f = ma
and N = mg
and N = mu*mg
so, 10 = (0.05*32.1740 + a)2
a = 0.8913 m/s/s

Torque balance about the point of contact
P*sqroot(2) - 2*sqroot(2)/2 = 2*4*alpha/3
alpha = 4.772 angular acc

f = 3.2147 N
N = 64.294 N

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