Question

The uniform 15-m pole has a mass of 200 kg and and B. ESSAGE NY INSTRUCTOR FULL SCREEN supported by its smooth ends against the vertical walls and by the tension PRINTER VERSION BACK NE in the vertical cable Compute the magnitudes of the reactions at A T 10m 5 m Answers

Answer 1

Consider the free-body diagram of the pole Тя е -- -- Orsa yang - Ilm The weight of the pole acts at the center of mass of the pole, which is at 7.5m from point À along the rod. From the above diagram we have Q. 90° - 8 Shq, sh (90* - ) - cos @ = Ilm SinQ - Coso- Using SnO - N-cos? We get sind = 11 - 12 = 1104 15m 5 15

In equilibpium, the net yorce and the net torque on the pole must be zero. For net vestical force to be zero, we requise - To mgao → Tengo For the net horizontal force to be zero, we require The dorque due to a force F acting at a dirtance ğ from a reference point is Z = 8 x Š The magnitude are related as Z - 8 Fsino where e is the angle between of and E

The direction of torque perpendicular to both 5 and i For the pole, the direction of torque will either be out of the screen or indo the screen. We take torque pointing out of the page to be positive, and that pointing into the page to be negative. The too que above point A must be zero 57 sinQ – 7.5 mg since +15F SIDO Using O, and sin Q = , sino - Vich We get 5mga tis - 75 mg tits + 15 Fe Jean so 1 mg - mg + 1104 Fs to 1 mg + 104 Foto => Fe - Ilmainen

Substituting m=20069, 929.8ms? we get 15=352.4 N Using equation © F - Fe = 352.4 N