The slender rod AB shown has a mass of m770 kgand is being s

The slender rod AB shown has a mass of m=77.0 kgand 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=2.55 kNm and has an effective moment arm of r=0.130 m. The dimensions shown in the figure are l=2.70 m and h=1.40 m. Assume the pulley is frictionless and massless.

Part A - Angular Acceleration of the Rod

Determine the angular acceleration of the rod the instant the rope and pulley system have pulled the rod through an angle of =2.10.

Express your answer to three significant figures and include the appropriate units.

Part B - Normal Component of the Reaction at A

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.

Part C - Tangential Component of the Reaction at A

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.

Solution

a) Using the geometry of the figure and dimensions, the angles are 62.6 degre at the pulley C and 27.4 at the point B

Torque 2.55KNm, r=0.13 m means a force of 19615.38 N=F, on the cable pulling at B

When the rod turns thru an angle of 2.1 degrees, assuming the tension is same, the component of tension normal to the rod at B is F cos (60.5)deg = 9659 N

This is a moment of 9659* 2.7 Nm about A

Use the equation for angular acceleration given the MI of the rod about one end is 1/12Ma^2 + Ma^2/4 = 1/3Ma^2

I = 1/3 *77*2.7^2=187.1 Nm^2

187.1*(ang accln) =torque=F.2.7=9659*2.7

ang accln=139.4 r/s^2

b)resolving the external forces in the direction of N

Tension*cos29.5+Wsin2.1 = Normal reaction=9659*cos29.5 + 77*9.81*sin2.1 newton=899 N approx

c) Tangential comp = W cos2.1 + T cos(60.5)=754.9+ 9659*.492.4=5511 N