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

Answer 1

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)
$$\begin{gathered} \Rightarrow a=g\left(\frac{m_2-m_1}{m_1+m_2}\right) \\ =9.8\times \frac{0.6-0.3}{0.6+0.3} \\ =3.266\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

(a) t = 2 s, a = 3.266 ms⁻², u = 0
So, the distance travelled by the body,
$$\begin{gathered} \mathrm{S}=ut+\frac{1}{2}a{t}^2 \\ =0+\frac{1}{2}\left(3.266\right)2^2=6.5\mathrm{m} \end{gathered}$$

(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