A uniform rod AB has mass M and length 2a A particle of mass 2M is attached to the rod at the point C, where AC = 1.5a The rod rests with its end A on rough horizontal ground - Edexcel - A-Level Maths Mechanics - Question 4 - 2022 - Paper 1

To find the tension in the string, consider the moments about point A. The torque created by the weight of the particle at C, which is located at a distance of 1.5a, and the weight of the rod, acting at its center of mass, must balance the torque produced by the tension T at a distance of 2a. Therefore, we can set up the equation:

T imes 2a = (2M imes g) imes 1.5a imes rac{1}{2} + (M imes g) imes a \Rightarrow T = \frac{(2M imes g imes 1.5a)}{2a} = 2Mg \cos θ

Considering vertical forces, we have:

$R + T \sin θ = Mg + 2Mg \sin θ$

Substituting T from the previous step into this equation, we resolve the vertical components:

$R + 2Mg \cos θ \times \frac{3}{5} = Mg + 2Mg \times \frac{4}{5}$

This simplifies to:

$R = \frac{57Mg}{25}$

In limiting equilibrium, the frictional force F is given by:

$F = \mu R$

From our earlier discussions, we substitute:

$F = T \sin θ \Rightarrow 2Mg \cos θ \sin θ$

Substituting for R and simplifying gives us:

$\mu \cdot \frac{57Mg}{25} = 2Mg \cdot \frac{3}{5} \cdot \frac{4}{5} \Rightarrow \mu = \frac{8}{19}$.