A light inextensible string is connected at each end to horizontal ceiling, as shown in the diagram. A mass of \(m\) kilograms hangs from a smooth ring on the string.
A horizontal force of \(F\) newtons is applied to the string until the tension in the string equals \(T\) and is constant across the whole system. At one end, the string makes an angle \(\theta\) with the ceiling and at the other end it makes an angle of \(2\theta\).
- Resolve the vertical forces to show that \(T=\dfrac{mg}{sin\,\theta+\sin\,2\theta}\) (1 mark)
- Hence, or otherwise, show \(F=mg\Bigg( \dfrac{1-\cos\,\theta}{\sin\,\theta} \Bigg) \). (3 marks)
Show Worked Solution
a.
\(\text{Resolving forces vertically:}\)
| \(T \sin \theta+T \sin 2 \theta\) | \(=mg\) |
| \(T(\sin \theta+\sin 2 \theta)\) | \(=mg\) |
| \(T\) | \(=\dfrac{m g}{\sin \theta+\sin 2 \theta}\) |
b. \(\text{Resolving forces horizontally: }\)
| \(F+T \cos 2 \theta\) | \(=T \cos \theta\) |
| \(F\) | \(=T(\cos \theta-\cos 2 \theta)\) |
| \(=\dfrac{m g\left(\cos \theta-2 \cos ^2 \theta+1\right)}{\sin \theta+2 \sin \theta \cos \theta}\) | |
| \(=\dfrac{m g(1-\cos \theta)(1+2 \cos \theta)}{\sin \theta(1+2 \cos \theta)}\) | |
| \(=mg\left(\dfrac{1-\cos \theta}{\sin \theta}\right)\) |
