Figure Q1(a) shows a frame structure supporting a mass of 180 kg from one end of cable that goes around a 1.2 m radius frictionless pulley of mass 350 kg. Member AC is rigidly attached to the wall at A and carries a distributed load with a linearly intensity varying from w = 800 N/m at D to 0 at E. A moment of 1500 Nm acts on the member BC at point G. Neglect the weights of the bars, and assume gravitational acceleration to be g = 10 m/s^2. (i) Determine the support reactions at A and B. (ii) Find the horizontal and vertical components of the force at joint C. (13 marks)
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The force due to the mass of 180 kg is F1 = 180 kg * 10 m/s^2 = 1800 N. This force is acting downwards at the end of the cable. The force due to the mass of the pulley is F2 = 350 kg * 10 m/s^2 = 3500 N. This force is acting downwards at the center of the Show more…
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