First slide

Equilibrium of a particle

Question

A block of mass m is connected to a peg at point D through a light and inextensible string of length 18 m. The length of section AB is 3 m. The point C of string is pulled down by applying a force F of 128 N. Block A remains stationary. If the mass of block A is $\frac{16}{\sqrt{n}}$ kg, the value of n is ________

Moderate

Solution

From the figure,
$\begin{aligned} \begin{aligned} E C = \sqrt{(7.5)^{2} - (4.5)^{2}} = 6 m \\ \sin θ = \frac{4.5}{7.5} = \frac{3}{5} \Rightarrow \cos θ = \frac{6.0}{7.5} = \frac{4}{5} \\ 2 F^{'} \cos θ = 128 \\ \Rightarrow 2 F^{'} \times \frac{4}{5} = 128 \Rightarrow F^{'} = \frac{128 \times 5}{8} = 80 N \end{aligned} \end{aligned}$
$\begin{array}{l} For block at A, F = m g \sin 45^{\circ} \\ \begin{aligned} 80 = m \times 10 \times \frac{1}{\sqrt{2}} \Rightarrow m = 8 \sqrt{2} k \\ ∴ & \frac{16}{\sqrt{n}} = 8 \sqrt{2} \\ Or \sqrt{n} & = \frac{16}{8 \sqrt{2}} = \sqrt{2} \\ ∴ n = 2 \end{aligned} \end{array}$

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