Recent content by panda02

centre of mass of board at the centre of the board = at distance 6m from left end At no rotation condition and equillibsium IM= net moment of. force = 0 If, = net horizontal force =- ity = net vestical force =0 a. To prevent the board from rotating, the total torque on one side of the pivot...

Torque_left = 100 kg * 1 m = 100 kg·m Torque_right = 200 kg * (13 - x) m = 200(13 - x) kg·m 100 kg·m = 200(13 - x) kg·m 100 = 200(13 - x) x = 12.5 meters pivot = 12.5 m