1 answer

Learning Goal: To apply the equations of motion to a system that involves rotation about a...

Question:

Learning Goal: To apply the equations of motion to a system that involves rotation about a fixed axis and to use this informa

Part A - Angular Acceleration of the Rod Determine the angular acceleration of the rod the instant the rope and pulley system

Part C - Tangential Component of the Reaction at A Determine the tangential component of the reaction the rod exerts on the p

Learning Goal: To apply the equations of motion to a system that involves rotation about a fixed axis and to use this information to determine key characteristics. The slender rod AB shown has a mass of m = 51.0 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 = 2.85 kN.m and has an effective moment arm of r= 0.150 m. The dimensions shown in the figure are l = 2.40 m and h = 1.90 m. Assume the pulley is frictionless and massless. (Figure 1) Figure 1 of 1 h B
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 0 = 2.50°. Express your answer to three significant figures and include the appropriate units. ► View Available Hint(s) uA ? undo a = Value 'Units Submit 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. BE B Tw
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. h BE 3) В A, Express your answer to three significant figures and include the appropriate units. ► View Available Hint(s) UA ? At = Value Units Submit

Answers

SOLUTION 100 ma =510kg T = 2.85kn.m h 8-0.150m h = 1.90m l=2.40m 90+2.50 -2 sime sule By applysing 90-2.50 12.5691 h W Sin(e)= 28412:45 – 600 IAK = 28312.95 IA = me? me 2 51X2.4 3 IA 297.92 ugom? 97.92*= 28812.95 = 289114 sade 17=0 Ay 7 wsincoj +Fcos

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