A shop sign is suspended from a uniform pole fixed to a wall and supported by a cable. The mass of the sign is 66 kg, and it's attached to the pole 2.1 m from the wall. The pole's mass is 18 kg and its length is 2.5 m. When the cable is anchored to a windowsill so that the angle between the pole and the cable is ? = 22°, the tension in the cable is 1700 N, well below its breaking tension of 2500 N. The shop owner is worried that the windowsill rigging is unsafe, though, and would like to anchor the cable to a bolt drilled firmly into the wall below the windowsill. However, lowering the point of attachment will increase the tension necessary to support the pole and the sign. If the bolt is inserted at the highest possible point below the windowsill, the cable will make an angle of 15° with the pole. Question Will a stronger cable be required, or will this cable be able to hold the pole and sign if anchored at the lower position? Hints The net torque on the pole must be zero in order for it to be in equilibrium. Once you've completed your FBD for the pole, choose an axis/pivot point for calculating torques. Think carefully about where each force is applied to the pole, direction of each torque (CCW is positive, CW is negative). Beware of treating torques as if they were forces, and vice versa – they're related, but not at all equal to one another. You can't use N2L to find the tension by summing forces, since there is a completely unknown force exerted on the pole by the wall; though of course once you know the tension required to keep the pole in equilibrium, you can use N2L to determine the unknown force from the wall.