The uniform slender bar AB has a mass of 7 kg and swings in a vertical plane about the pivot at A. If angular velocity of the bar is 3 rad/s when θ = 35o , compute the force supported by the pin at A at that instant

49 N

Explanation:

The diagram of the bar is obtained online and attached to this solution.

The free body diagram is also attached.

Since the weight of the bar acts at the middle of the bar, the torque due to the weight of the bar is given by

τ = mgx

where x = (L/2) cos 35° = 0.45 × cos 35° = 0.3686 m

τ = (7)(9.8)(0.3686) = 25.29 Nm

The force acting on pin A = torque ÷ (length × sin 35°) = 25.29 ÷ (0.9 × sin 35°)

= 25.29 ÷ 0.5162 = 48.99 N = 49 N

Hope this Helps!!!

real, smaller and upside-down.   so the answer should be b.