Question

(2) A horizontal rope is attached to a uniform beam as shown. The beam is hinged at the bottom. From the end of the beam, a mass is hanging. The entire system is at rest. A) What is the tension in the horizontal rope if the beam has a mass of 10 kg? B) what are the horizontal and C) vertical components of the force that the hinge applies to the beam? Answer B with "into wall" or "out from wall' and C with "upward" or "downward" included in your answers. Answer each part to 3 significant figures. rope 40 o 15 kg

Answer 1

Consider Moments about Point 'A'. \r\n Therefore T(AB) = 10 times 9.81(EF) + (15 times 9.81)(OD) equation (i) T = Tension in rope. Now: AB = 2 cos 40 degrees (in m) EF = 1.5 sin 40 degrees (in m) OD = 3 sin 40 degrees (in m) equation (ii) From (i), (ii) T(2 cos 40 degrees) = 10 times 9.81 times (1.5 sin 40 degrees) + 15 times 9.81 times (3 sin 40 degrees) Therefore T(2 cos 40 degrees) = 9.81 sin 40 degrees[10 times 1.5 + 15 time s3] 2T(cos 40 degrees) = 588.6(sin 40 degrees) Therefore T = 588.6/2 (sin 40 degrees/cos 40 degrees) Therefore T = 294.3(tan 40 degrees) Therefore T = 246.947 N tension in rope equation (A) sigma F_H = 0 ... (sum of horizontal forces) Therefore R_H = T implies R_H = 246.947 N out From the wall equation (B) (Horizontal component at hinge) \r\n Also sigma F_v = 0 Therefore