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In a first order system of pulleys there are three movable pulleys. What is the effort required to raise a load of 6000 N ? Assume efficiency of the system to be 80%. If the same load is to be raised using 520 N, find the number of movable pulleys that are necessary. Assume a reduction of efficiency of 5% for each additional pulley used in the system. |
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Answer» VR = 2n , where n is the number of movable pulleys. VR = 23 = 8 Now, MA = η × VR = 0.8 × 8 = 6.4 \(\frac WP\) = 6.4 P = \(\frac W{6.4}\) = \(\frac {6000}{6.4}\) P = 937.5 N In the second case, Effort = 520 N Efficiency η = 0.80 – n1 × 0.05 where n1 = number of additional pulleys required and equal to (n – 3). MA = η × VR \(\frac WP\) = η × VR W = P × η × 2n = P(0.8 – n1 × 0.05) × 2n = P[0.8 – (n – 3) × 0.05] 2n By going for a trial and error solution, starting with one additional pulley i.e., totally with four pulleys, W = 520 [0.8 – (4 – 3) × 0.05] 24 = 6240 N i.e., if four pulleys are used, a load of 6240 N can be raised with the help of 520 N effort. ∴ Number of movable pulleys required = 4 |
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