The slender rod AB shown has a mass of m=65 kg and is being supported by a rope

Question

The slender rod AB shown has a mass of m=65 kg and is being supported by a rope and pulley system stationed at C. Starting from rest (in the position shown), the rope and pulley system tug on the rod causing it to rotate about A. The torque applied to the pulley is T=2.75kN*m and has an effective moment arm of r = .160m. The dimensions shown in the figure are l=2.6 m and h=1.6 m. Assume the pulley is frictionless and massless.

A) Determine the angular acceleration of the rod the instant the rope and pulley system have pulled the rod through an angle of theta=4 degrees

B) Determine the normal component of the reaction the rod exerts on the pin at A. Use the coordinate system set up by the free-body diagram below.

C) Determine the tangential component of the reaction the rod exerts on the pin at A. Use the coordinate system set up by the free-body diagram below.

Explanation / Answer

I = moment of inertia about A =m*l^2/12+m*(0.5l)^2 =m*l^2/3 =146.47... F= Torque/arm =17.2kN... angle beta =32.67.....therefore balancing momentum about A, we get angular acceleration =(F*l*sin(beta)-W*0.5*l*cos(theta))/I =159 rad/sec^2

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