A 41.7-kg sign is suspended by two wires, as the drawing shows. Wire 1 makes and angle of 42.2deg with the horizontal and wire 2 makes an angle of 58.8deg. Find the tension in wire 1 and wire 2.
To solve this, you have to find 2 equations, because there are 2 unknowns. Assume, Tension in wire 1 = T1 Vertical component of T1 = T1 sin42.2 Horizontal component of T1 = T1 cos42.2 Tension in wire 2 = T2 Vertical component of T2 = T2 sin58.8 Horizontal component of T2 = T2 cos58.8 Since the sign board is in equilibrium, the forces must balance each other. Horizontally, the horizontal components of both forces are balancing each other. Therefore, T2 cos58.8 = T1 cos42.2 T2 = (T1 cos42.2) / (cos58.8) -------------------(1) For the vertical components of the wire, both are balancing the weight of the sign. T1 sin42.2 + T2 sin58.8 = (41.7)(9.8) T1 sin42.2 + T2 sin58.8 = 408.66N ----------------(2) Combine (1) into (2), T1 sin42.2 + (T1 cos42.2)(sin58.8) / (cos58.8) = 408.66 T1 ( sin 42.2 + cos 42.2 tan 58.8) = 408.66 T1 = 215.66N Subst T1 = 215.66N into (1), T2 = (T1 cos42.2) / (cos58.8) = 308.4N Hope it is correct ^^
