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The pulley arrangements of figures (a) and (b) are identical. The mass of the rope is negligible. In figure (a), the mass m is lifted up by attaching a mass 2m to the other end of the rope. In figure(b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. Calculate the accelerations in the two cases. |
Answer» Solution :In figure (a), for the motion of MASS m, T-mg=ma....... (1) For motion of MASS2 m figure (B) 2mg -T= (2m)a ........ (2) Adding equation (1) and (2), we GET 2mg - mg = 2ma + ma, mg = 3ma, a = g/3 (b) In figure `T^(1) - mg=ma^(1)` But `T^(1) = 2mg , :. 2mg - mg = a^(1)`, Therefore `a^(1)=g` |
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