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 analyze the forces, consider the moments about point A. The tension T contributes a vertical component, which can be expressed as T sin θ. This will create a counterclockwise moment about the pivot point A, and must be balanced by the moments due to the weights of the rod and the mass attached. The condition for equilibrium gives us the equation, where net torque equals zero.

Taking moments about point A, we have:

T imes 2a imes rac{1}{2} = (Mg imes a) + (2Mg imes 1.5a)\cos θ
Simplifying this leads us to:

$T = \frac{(Mg + 3Mg)}{2a} imes 2a \cos θ = 2Mg cos θ$.

Using the value of cos θ, we substitute into the previous equation for tension found. The vertical force can be found using:

$R + T \sin θ = Mg + 2Mg$ where R is the vertical force exerted by the ground.
We know that:
$T = 2Mg \cdot \frac{3}{5}$
Thus, this allows us to express R as:

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

In limiting equilibrium, the frictional force can be expressed in terms of the normal force and the coefficient of friction μ. Given that frictional force equals the horizontal component of the tension:

$F = μR$
From our previous expressions, substituting values will yield:

$μR = T \cos θ = \frac{8}{19}$.