A Level Maths 9709: Two Strings Supporting a Hanging Lamp

A decorative lamp of weight $20\text{ N}$ hangs in equilibrium, held by two light inextensible strings attached to two fixed points on a horizontal ceiling. The strings are on opposite sides of the lamp: one string makes an angle of $50^\circ$ with the ceiling and has tension $T_1$ , and the other makes an angle of $35^\circ$ with the ceiling and has tension $T_2$ .

(a) By resolving forces horizontally, write down an equation connecting $T_1$ and $T_2$ . [2]

(b) By resolving forces vertically, write down a second equation connecting $T_1$ and $T_2$ . [2]

(c) Solve your two equations simultaneously to find $T_1$ and $T_2$ , giving each answer correct to $3$ significant figures. [3]

Setting up the forces

Three forces act at the point where the two strings meet the lamp: the tension $T_1$ (at $50^\circ$ to the ceiling, i.e. to the horizontal), the tension $T_2$ (at $35^\circ$ to the horizontal, on the opposite side), and the weight of the lamp, $20\text{ N}$ , acting vertically downwards. Since the ceiling is horizontal, each string makes the same angle with the horizontal at both ends.

Part (a): Horizontal equilibrium

The two tensions pull in opposite horizontal directions (that is why the strings are attached on opposite sides). For the lamp not to accelerate horizontally, these horizontal components must be equal in magnitude: $T_1\cos50^\circ = T_2\cos35^\circ$

Part (b): Vertical equilibrium

Both tensions have an upward vertical component (since both strings run up to the ceiling), and together they must support the full weight of the lamp: $T_1\sin50^\circ + T_2\sin35^\circ = 20$

Part (c): Solving simultaneously

From part (a): $T_1 = T_2 \times \frac{\cos35^\circ}{\cos50^\circ} = T_2 \times \frac{0.8192\ldots}{0.6428\ldots} \approx 1.2744\,T_2$

Substitute into the equation from part (b): $1.2744\,T_2 \times \sin50^\circ + T_2\sin35^\circ = 20$ $1.2744\,T_2 \times 0.7660\ldots + T_2 \times 0.5736\ldots = 20$ $0.9762\,T_2 + 0.5736\,T_2 = 20$ $1.5498\,T_2 = 20$ $T_2 = \frac{20}{1.5498\ldots} \approx 12.9\text{ N}$

Substituting back: $T_1 \approx 1.2744 \times 12.9\ldots \approx 16.4\text{ N}$

Check: $T_1\cos50^\circ \approx 16.45 \times 0.6428 \approx 10.57$ , and $T_2\cos35^\circ \approx 12.90 \times 0.8192 \approx 10.57$ , the horizontal components match, confirming the solution.

Final answers

(a) $T_1\cos50^\circ = T_2\cos35^\circ$
(b) $T_1\sin50^\circ + T_2\sin35^\circ = 20$
(c) $T_1 \approx \boxed{16.4}\text{ N}$ , $T_2 \approx \boxed{12.9}\text{ N}$