A block which weights 100 lbs resets on an inclined plane, and is attached to a

Question

A block which weights 100 lbs resets on an inclined plane, and is attached to a cable which passes over a fixed pipe. The free end of the cable is subjected to a force P of 200 lbs. The inclined surface the pipe both have coefficients of static friction equal to 0. 3, and coefficients of dynamic friction equal to Perform an analysis to determine whether the system is in equilibrium. Determine the friction force which acts on the block. Determine the tension in the cable at the attachment point on the block.

Explanation / Answer

The picture got cut on the sides and hence dynamic friction coefficient value is missing. I am taking it as . You may substitute its value.

Angle of wrap over pulley = 90 + 30 = 120 deg = 2/3 radians

Downward force on block along the plane due to gravity = mgSin30

Normal force on block perpendicular to plane = mg Cos30

Friction force on block along the plane = 0.3*mgCos30

a) For static equilbrium, mgSin30 - 0.3*mgCos 30 = T2 (T2 is tension in rope in the portion conncted with block)

or, T2 = 100*32.2*sin30 - 0.3*100*32.2*Cos30 = 773.4 lbf = 773.4/32.2 = 24 lb

Also, T1/T2 = e^(0.3*) where T1 is the pulling force on free end

T1/24 = e^(0.3*2/3)

T1 = 45 lb

Since 200 lb > 45 lb, the block will move up along the plane. and is not in static equilbrium.

b) Fric force on block = *mgCos30 = *100*Cos30 = *86.6 lb

c) 200/T2 = e^(*2/3)

so, T2 = 200*e^(-*2/3)

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