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

1. Identify Forces:

   • The weight of the beam acts vertically downwards at its center of mass.
   • Normal force (N) acts perpendicular to the surface at point A.
   • Frictional force (F) acts horizontally opposite to any applied force (at point B).

2. Apply Moments about A:

   • The sum of moments about point A must be zero for equilibrium:

   $mga \cos \theta - F \cdot 2a \sin \theta = 0$

   Rearranging gives:

   $F = \frac{mga \cos \theta}{2a \sin \theta} = \frac{mg \cos \theta}{2 \sin \theta}$

3. Frictional Force Relation:

   • The frictional force can also be expressed as:

   $F = \mu N$

   where N = mg (since the vertical forces also must balance).

4. Combine the Equations:

   • Substituting from the equation for F gives:

   $\mu mg = \frac{mg \cos \theta}{2 \sin \theta}$

5. Rearranging gives the inequality:

   • Dividing both sides by mg and rearranging gives:

   $\mu > \frac{1}{2} \cot \theta$

Thus, we have shown that ( \mu > \frac{1}{2} \cot \theta ).