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Standard IX

Physics

Second Law of Motion

In a simple A...

Question

In a simple Atwood machine, two unequal masses $m_{1}$ and $m_{2}$ are connected by a string going over a clamped light smooth pulley. In a typical arrangement shown in the figure $m_{1} = 300 g$ and $m_{2} = 600 g$ . The system is released from rest. Find the tension in the string.

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Solution

$m_{1} = 0.3 k g, m_{2} = 0.6 k g$
$T - (m_{1} g + m_{2} a) = 0$ ... (i)
$\Rightarrow T = m_{1} g + m_{1} a$
$T + m_{2} a - m_{2} g = 0$ .... (ii)
$\Rightarrow T = m_{2} g - m_{2} a$
From equation (i) and equation (ii)
$m_{2} g + m_{1} a + m_{2} a - m_{2} g = 0$ , from $(i)$
$\Rightarrow a (m_{1} + m_{2}) = g (m_{2} - m_{1})$
$\Rightarrow a = f (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) = 9.8 (\frac{0.6 - 0.3}{0.6 + 0.3}) = 3.266 m s^{- 2}$
From (i) $T = m_{1} (g + a) = 0.3 (9.8 + 3.26) = 3.9 N$

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Figure

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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 shown in the figure m1=300g and m2=600g. The system is released from rest. Find the tension in the string.

m1=0.3 kg,m2=0.6 kgT−(m1g+m2a)=0 ... (i) ⇒T=m1g+m1aT+m2a−m2g=0 .... (ii) ⇒T=m2g−m2aFrom equation (i) and equation (ii)m2g+m1a+m2a−m2g=0, from (i)⇒a(m1+m2)=g(m2−m1)⇒a=f(m2−m1m1+m2)=9.8(0.6−0.30.6+0.3)=3.266 ms−2 From (i) T=m1(g+a)=0.3(9.8+3.26)=3.9 N

In a simple Atwood machine, two unequal masses $m_{1}$ and $m_{2}$ are connected by a string going over a clamped light smooth pulley. In a typical arrangement shown in the figure $m_{1} = 300 g$ and $m_{2} = 600 g$ . The system is released from rest. Find the tension in the string.