A beam AB has mass m and length 2a - Edexcel - A-Level Maths Mechanics - Question 3 - 2021 - Paper 1

To demonstrate the inequality, we first consider the moments about point A. The forces acting on the beam include the weight of the beam, mg, acting downwards at its center of mass, and the reaction force from the wall, F, acting horizontally at point B. The moments about A can be expressed as:

1. For moments about A: $mg \cos(\theta) \cdot a = F \cdot 2a \sin(\theta)$

2. Since F = μN, we substitute N = mg \cos(θ) into the moment equation: $mg \cos(\theta) \cdot a = μN \cdot 2a \sin(\theta)$ Thus, we rewrite it: $mg \cos(\theta) = μ(2mg \cos(\theta) \sin(\theta))$

3. Dividing both sides by mg \cos(\theta) (assuming mg \cos(\theta) ≠ 0), we get: $1 = 2μ \sin(\theta)$ Simplifying gives: $μ = \frac{1}{2} \cot(\theta)$

4. Therefore, since μ must be greater than this expression, we find: $μ > \frac{1}{2} \cot(\theta)$