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Three blocks of masses m1 = 2 kg, m2 = 3 kg and m3 = 5 kg are suspended with string, which is passed over a pulley as shown in the figure. Calculate the tension T1, T2 and T3, when the whole system (a) is moving upward with an acceleration of 2 ms-2, and (b) is held stationary. |
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Answer» Here, m1 = 2 kg, m2 = 3 kg and m3 = 5 kg (a) When the system is moving upward with an acceleration 'a' where a = 2 ms-2. As per the fig., consider the free body diagram of the system of three blocks. Then, T1 - (m1 + m2 + m2) g = (m1 + m2 + m3) a or, T1 = (m1 + m2 + m3) (a + g) ...(i) or, T1 = (2 + 3 + 5) x (2 x 9.8) = 118 N Consider the free body diagram of blocks of masses m2 and m3. Then T2 - (m2 + m3)g = (m2 + m3) x a ....(ii) or, T2 = (m2 + m3)(a + g) ⇒ T2 = (3 + 5) x (2 x 9.8) = 94.4 N Consider the free body diagram of block of mass m3. Then, T3 - m3g = m3a ....(iii) or, T3 = m3(g + a) = 5 x (2 + 9.8) = 59 N (b) When the system is held stationary: Obviously, a = 0, setting a = 0, in the equations (i), (ii) and (iii), we have T1 = (m1 + m2 + m3) g = (2 + 3 + 5) x 9.8 = 98 N T2 = (m2 + m3) g = (3 + 5) x 9.8 = 78.4 N and T3 = m3g = 5 x 9.8 = 49 N |
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