A lamp of weight 20 N hangs from two strings

Scalars & Vectors - OCR A-Level Physics

Worked Example

Lamp hanging from two strings
A lamp of weight $W = 20 \text{ N}$ hangs in equilibrium from two strings attached to a ceiling. String A makes $30^\circ$ with the horizontal; string B makes $50^\circ$. Find the tension $T_A$ in string A and $T_B$ in string B.

Diagram pending

Lamp hanging from two strings attached to a horizontal ceiling. String A on the left, making $30^\circ$ with the horizontal. String B on the right, making $50^\circ$. Three force arrows from the lamp: tension $T_A$ along string A (up-left), tension $T_B$ along string B (up-right), weight $W = 20 \text{ N}$ straight down. Dashed components $T_A\cos 30^\circ$, $T_A\sin 30^\circ$, $T_B\cos 50^\circ$, $T_B\sin 50^\circ$ shown faintly.

Will be replaced with a GeoGebra SVG in stream 2.

Step 1. Draw a free body diagramAll forces on a single object, drawn from a single point.. Three forces act on the lamp: weight $W = 20 \text{ N}$ downward, tension $T_A$ along string A, tension $T_B$ along string B.
Step 2. Resolve horizontally. The two string tensions pull in opposite horizontal directions; the weight has no horizontal component. For equilibrium, horizontal components must cancel.

T_A \cos 30^\circ = T_B \cos 50^\circ

Step 3. Resolve vertically. The two upward tension components together support the weight.

T_A \sin 30^\circ + T_B \sin 50^\circ = 20

Step 4. Rearrange the horizontal equation to express $T_A$ in terms of $T_B$:

T_A = T_B \times \frac{\cos 50^\circ}{\cos 30^\circ} = T_B \times \frac{0.643}{0.866} = 0.742\, T_B

Step 5. Substitute into the vertical equation: $0.742\, T_B \times \sin 30^\circ + T_B \times \sin 50^\circ = 20$.
$0.742\, T_B \times 0.500 + T_B \times 0.766 = 20$.
$0.371\, T_B + 0.766\, T_B = 20$, so $1.137\, T_B = 20$.
$T_B = 17.6 \text{ N}$ and $T_A = 0.742 \times 17.6 = 13.1 \text{ N}$.

T_A = 13.1 \text{ N}, \quad T_B = 17.6 \text{ N}

Common Mistake MEDIUM

Wrong: Assuming $T_A = T_B$ because "both strings hold the lamp". Or writing $T_A + T_B = 20 \text{ N}$ as if tensions add arithmetically.
Right: The tensions are different because the angles are different. You must resolve in two perpendicular directions and solve simultaneous equations. Only the vertical components of the tensions sum to balance the weight.

Examiner Tips and Tricks

Sense check: the string closer to vertical (B, at $50^\circ$) supports more weight ($T_B = 17.6 \text{ N} > T_A = 13.1 \text{ N}$), which matches intuition.
Verify: $T_A \sin 30^\circ + T_B \sin 50^\circ = 13.1 \times 0.500 + 17.6 \times 0.766 = 6.55 + 13.5 = 20.1 \text{ N}$, equal to $W$ within rounding.
If the question gives angles from the vertical instead, swap $\sin$ and $\cos$ in the equations.