A smooth bead B is threaded on a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 3 - 2005 - Paper 1

To find the tension, we consider the horizontal components of the forces acting on bead B.

From equilibrium in the horizontal direction, we have:

$T \, \cos \alpha = 6 \text{ N}$

Given that ( \tan \alpha = \frac{2}{3} ), we can find ( \cos \alpha ) and ( \sin \alpha ) using:

$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} \Rightarrow \sin \alpha = 2k, \, \cos \alpha = 3k$

Where ( k ) is a scaling factor. Since ( \sin^2 \alpha + \cos^2 \alpha = 1 ), we get:

$(2k)^2 + (3k)^2 = 1 \Rightarrow 4k^2 + 9k^2 = 1 \Rightarrow 13k^2 = 1 \Rightarrow k^2 = \frac{1}{13} \Rightarrow k = \frac{1}{\sqrt{13}}$

Thus,:

$\cos \alpha = \frac{3}{\sqrt{13}} \text{ and } \sin \alpha = \frac{2}{\sqrt{13}}$

Substituting ( \cos \alpha ) back, we have:

$T \cdot \frac{3}{\sqrt{13}} = 6$

Solving for T:

$T = 6 \cdot \frac{\sqrt{13}}{3} = 2\sqrt{13} \text{ N} \approx 7.5 \text{ N}$