Question

A steel ball of mass $m = 10 \, kg$ is suspended from the ceiling of a moving carriage by two inextensible strings making $60^\circ$ with the horizontal as shown. The carriage has an acceleration $a$ such that the tension in the string on the right is double the tension in the string on the left. Take $g = 10 \, m/s^2$. The acceleration $a$ (in $m/s^2$) is ............. (rounded off to one decimal place).

Hint:

Always confirm whether angles are given with horizontal or vertical. This changes sine and cosine usage in equilibrium equations.

Answer 1

Correct Answer: 1.8

Step 1: Define tensions.
Let tension in left string = $T$, tension in right string = $2T$.
Step 2: Vertical equilibrium.
Both strings make $60^\circ$ with the horizontal $\Rightarrow$ angle with vertical = $30^\circ$. Vertical components: \[ T \cos 30^\circ + 2T \cos 30^\circ = mg \] \[ 3T . \frac{\sqrt{3}}{2} = 10 . 10 \] \[ T = \frac{100}{3 . 0.866} \approx 38.5 \, N \]
Step 3: Horizontal components (net force).
Right string horizontal: $2T \sin 30^\circ = 2T . 0.5 = T$ Left string horizontal: $T \sin 30^\circ = 0.5T$ Net horizontal force: \[ F_{net} = T - 0.5T = 0.5T \] \[ F_{net} = 0.5 \times 38.5 = 19.25 \, N \]
Step 4: Acceleration.
\[ a = \frac{F_{net}}{m} = \frac{19.25}{10} \approx 1.9 \, m/s^2 \] But recheck: If strings are $60^\circ$ from horizontal (not from vertical), then angle with vertical is $60^\circ$. Correcting: Vertical equilibrium: \[ T \sin 60^\circ + 2T \sin 60^\circ = mg \] \[ 3T . 0.866 = 100 \] \[ T = 38.5 \, N \] Horizontal force: \[ F_{net} = 2T \cos 60^\circ - T \cos 60^\circ = T . 0.5 = 19.25 \, N \] \[ a = \frac{19.25}{10} \approx 1.9 \, m/s^2 \] Final Answer: \[ \boxed{1.9 \, m/s^2} \]