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 demonstrate that the tension in the shorter string is more than 30% greater than that in the longer string, we start by applying the sine rule and geometry.

The angles for the strings are derived using trigonometry. Let:

• Angle at A, $A = ext{sin}^{-1}\left(\frac{0.6}{1.2}\right) = 48.59^{\circ}$
• Angle at B, $B = ext{sin}^{-1}\left(\frac{0.6}{0.8}\right) = 30^{\circ}$

Using equilibrium conditions, we can set up the horizontal component equations:

$T_{OA} \cos A = T_{OB} \cos B$ Thus, $T_{OA} = T_{OB} \frac{\cos B}{\cos A}$

Substituting the angles:

$T_{OA} = T_{OB} \frac{\cos(30^{\circ})}{\cos(48.59^{\circ})}$

Calculating:

• $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$
• $\cos(48.59^{\circ}) = 0.669$

Therefore:

$T_{OA} \approx T_{OB} \cdot 1.309$

This indicates that:

$T_{OA} > 1.3 T_{OB}$

Since 1.3 is greater than 1.3, we conclude that the tension in the shorter string on average is more than 30% more than that in the longer string.