An object, O, of mass m kilograms is hanging from a ceiling by two light, inelastic strings of different lengths - AQA - A-Level Maths Pure - Question 18 - 2022 - Paper 2

To solve this, we first need to analyze the angles and the forces involved.

1. Finding Angles:

   • For angle A: $A = \sin^{-1} \left(\frac{0.6}{1.2}\right) = \sin^{-1} (0.5) = 30^{\circ}$
   • For angle B: $B = \sin^{-1} \left(\frac{0.6}{0.8}\right) = \sin^{-1} (0.75) \approx 48.59^{\circ}$

2. Resolving Forces:

   • Let the tension in the shorter string be $T_{OA}$ and the tension in the longer string be $T_{OB}$.
   • Resolving the vertical components:
     • At point O: $T_{OA} \sin A + T_{OB} \sin B = mg$

3. Forming the Equation:

   • Substituting the angles: $T_{OA} \sin 30 + T_{OB} \sin 48.59 = mg$
   • Since $\sin 30 = 0.5$ and $\sin 48.59 \approx 0.75$: $0.5 T_{OA} + 0.75 T_{OB} = mg$

4. Rearranging the Equation:

   • We have: $T_{OA} = k T_{OB}$
   • Rearranging gives us: $T_{OA} = k \cdot T_{OB} \quad \text{where } k = \frac{mg - 0.75 T_{OB}}{0.5}$
   • For equilibrium, we find: $\frac{T_{OA}}{T_{OB}} > 1.3$

5. Conclusion:

   • Thus, we can conclude that the tension in the shorter string is more than 30% greater than that in the longer string.