- Topic 6 Tutorial Equilibrium Pulley Systems Note That It May Be Useful To Review The Notes On Pages 7 12 To 7 15 Reg 1 (107.87 KiB) Viewed 39 times
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Topic 6 Tutorial - Equilibrium & Pulley Systems Note that it may be useful to review the notes on pages 7-12 to 7-15 regarding the solving of equilibrium equations before attempting these problems. 50° A 250 kg 1. Determine the values of M1, M2, T and 0 required for equilibrium of the shown system. (NOTE: All ropes other than those attached to the 250 kg mass may be assumed vertical. Also, the shape of the 250 kg mass is not drawn to scale and is representative only). 50 kg M M Ans: Mi = 554.7 kg, M2 = 100 kg, T = 1814 N, 0 = 69.70 B 0 8 kg 2. Determine the vertical reaction at A, VA, and the vertical reaction at B, VB. 4 kg Ans: VA= 35.94 N; VB = 165.2 N 9.5 m 14 kg 12 m
А 3. The pulley system shown has a cable secured at the centre of pulley P2 and a cantilever beam labelled member CD. Determine the force T which is necessary for static equilibrium. Also determine the horizontal reaction force at point D, Hp, the vertical reaction force at point D, VD, and the moment reaction at D, MD. Neglect friction between sliding surfaces. B T=? P3 P2 P4 5m Ans: T = 6867 N, Hp = 0, VD=-13734 N, Mp=68670 Nm (CCW) M M=2100kg 3 kN M 1.5 m 4. The system shown in the figure for this question has a pin at location A and is in static equilibrium. For 2 kN/m the resultant reaction at A, determine the magnitude RA and orientation with respect to a conventional x axis. Also determine the magnitude of the moment 1.5 m 11m, M. Note that a triangular distributed load may be replaced by an equivalent point load having a magnitude of 72 x base x height and located at the centroid of the triangular shape, which is 2/3rds of the distance from tip of the triangle (see also pages 5-5 and 5-6 of these notes). 35 Ans: Ra = 2.56 kN, 0 = 196.4º, M = 2.47 kNm (CW)