posted by Anonymous
A 20 kg sign is pulled by a horizontal force such that the single rope holding the sign makes an angle of 21 degrees with the vertical. Assuming the sign is motionless, find magnitude of the tension in the rope.
Well we know that our whole system should have a net torque of 0. So the Torque caused by tension needs to cancel with the torque of both the beam center of mass and the sign's torques. so TL-1/2Lmgsin(21)-Lmgsin(21)=0 the lengths cancel leaving T - 1/2mgsin(21)-mgsin(21)= 0