In the diagram, a mass of 12kg hanging from a light inextensible string is pulled aside by a horizontal force, R, such that the string is inclined at 45\(^o...

The 12 kg mass hangs from a string whose upper end is fixed at \(P\). A horizontal force \(R\) pulls the mass sideways so that the string makes \(45^{\circ}\) with the vertical. Three forces act at the junction (knot): the tension \(T\) along the string, the horizontal pull \(R\), and the weight \(W\) acting vertically downward.

Weight of the mass (taking \(g = 10\,\text{m s}^{-2}\)):

\[ W = mg = 12 \times 10 = 120\,\text{N} \]

Since the system is in equilibrium, resolve the tension into vertical and horizontal components. The string is \(45^{\circ}\) from the vertical, so the vertical component of \(T\) is \(T\cos 45^{\circ}\) and the horizontal component is \(T\sin 45^{\circ}\).

(a) Tension in the string

Resolving vertically (the vertical component of the tension supports the weight):

\[ T\cos 45^{\circ} = W \]\[ T \times \frac{\sqrt{2}}{2} = 120 \]\[ T = \frac{120}{\cos 45^{\circ}} = \frac{120}{0.7071} = 120\sqrt{2} \]\[ T \approx 169.7\,\text{N} \]

So the tension is about \(170\,\text{N}\).

(b) Value of \(R\)

Resolving horizontally (the horizontal force balances the horizontal component of the tension):

\[ R = T\sin 45^{\circ} = 120\sqrt{2} \times \frac{\sqrt{2}}{2} \]\[ R = 120\sqrt{2} \times 0.7071 = 120\,\text{N} \]

Hence \(R = 120\,\text{N}\).

Check: Because the string is at \(45^{\circ}\), the vertical and horizontal components of \(T\) are equal, so \(R\) equals the weight, \(120\,\text{N}\), which agrees with the result above. (Using \(g = 9.8\,\text{m s}^{-2}\) gives \(W = 117.6\,\text{N}\), \(T \approx 166.3\,\text{N}\) and \(R = 117.6\,\text{N}\).)