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 (In the following figure), m_{1} = 300 g and m_{2} = 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.

#### Solution

The masses of the blocks are m_{1} = 0.3 kg and m_{2} = 0.6 kg

The free-body diagrams of both the masses are shown below:

For mass m_{1},

T − m_{1}g = m_{1}a ...(i)

For mass m_{2},

m_{2}g − T= m_{2}a ...(ii)

Adding equations (i) and (ii), we get:

g(m_{2} − m_{1}) = a(m_{1} + m_{2})

\[\Rightarrow a = g{\left( \frac{m_2 - m_1}{m_1 + m_2} \right)}$\]

\[ = 9 . 8 \times \frac{0 . 6 - 0 . 3}{0 . 6 + 0 . 3}\]

\[ = 3 . 266 m/ s^2\]

(a) t = 2 s, a = 3.266 ms^{−2}, u = 0

So, the distance travelled by the body,

\[S = ut + \frac{1}{2}a t^2 \]

\[ = 0 + \frac{1}{2}\left( 3 . 266 \right) 2^2 = 6 . 5 m\]

(b) From equation (i),

T = m_{1} (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