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Question

# In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure 5−E7), m1 = 300 g and m2 = 600 g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds; (b) find the tension in the string; (c) find the force exerted by the clamp on the pulley. Figure

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Solution

## The masses of the blocks are m1 = 0.3 kg and m2 = 0.6 kg The free-body diagrams of both the masses are shown below: For mass m1, T − m1g = m1a ...(i) For mass m2, m2g − T= m2a ...(ii) Adding equations (i) and (ii), we get: g(m2 − m1) = a(m1 + m2) $⇒a=g\left(\frac{{m}_{2}-{m}_{1}}{{m}_{1}+{m}_{2}}\right)\phantom{\rule{0ex}{0ex}}=9.8×\frac{0.6-0.3}{0.6+0.3}\phantom{\rule{0ex}{0ex}}=3.266\mathrm{m}/{\mathrm{s}}^{2}$ (a) t = 2 s, a = 3.266 ms−2, u = 0 So, the distance travelled by the body, $\mathrm{S}=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=0+\frac{1}{2}\left(3.266\right){2}^{2}=6.5\mathrm{m}$ (b) From equation (i), T = m1 (g + a) = 0.3 (3.8 + 3.26) = 3.9 N (c)The force exerted by the clamp on the pulley, F = 2T = 2 × 3.9 = 7.8 N

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