A 12 kilogram object is suspended from a horizontal ceiling by two light, inextensible strings at angles of 30° and 45°, as shown in the diagram.
The acceleration due to gravity is \(g\) m s\(^{-2}\) and the tensions in the strings are \(T_1\) newtons and \(T_2\) newtons.
- Show that \(T_2=\sqrt{\dfrac{3}{2}} T_1\) (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
- Determine the tensions, in newtons, of \(T_1\) and \(T_2.\) (2 marks)
--- 8 WORK AREA LINES (style=lined) ---
Show Worked Solution
a. \(\text{Resolve forces into horizontal / vertical components:}\)
\(\text{Horizontal forces are equal.}\)
| \(T_1\cos 30^{\circ}\) | \(=T_2 \cos 45^{\circ}\) |
| \(T_1 \times \dfrac{\sqrt{3}}{2}\) | \(=T_2 \times \dfrac{1}{\sqrt{2}}\) |
| \(T_2\) | \(=\dfrac{\sqrt{3} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \, T_1=\sqrt{\dfrac{3}{2}}\, T_1\) |
b. \(\text{Vertical forces are equal.}\)
| \(T_1 \sin 30+T_2 \sin 45\) | \(=12 g\) | |
| \(T_1 \times \dfrac{1}{2}+T_2 \times \dfrac{1}{\sqrt{2}}\) | \(=12 g\) |
\(\text{Substitute} \ \ T_2=\sqrt{\dfrac{3}{2}}\, T_1:\)
| \(T_1 \times \dfrac{1}{2}+T_1 \times \sqrt{\dfrac{3}{2}} \times \dfrac{1}{\sqrt{2}}\) | \(=12 g\) |
| \(T_1\left(\dfrac{\sqrt{3}+1}{2}\right)\) | \(=12 g\) |
\(T_1=\dfrac{24g}{\sqrt{3}+1} \times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}=12(\sqrt{3}-1)g\)
| \(T_2\) | \(=\sqrt{\dfrac{3}{2}} \times 12(\sqrt{3}-1)g\) |
| \(=6 \sqrt{6}(\sqrt{3}-1)g\ \ \ (=6 \sqrt{2}(3-\sqrt{3}))\) |