A particle of weight 120 N is placed on a fixed rough plane which is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac{3}{4}$ - Edexcel - A-Level Maths Mechanics - Question 6 - 2011 - Paper 1

When the force P is applied up the slope, we resolve the forces again, considering the equilibrium:

1. Perpendicular to the inclined plane: $R = 120 \cos \alpha$ Where, $R$ is the normal reaction, hence: $R = 120 \left(\frac{4}{5}\right) = 96$

2. Now from frictional force: $F_{max} = \frac{1}{2} R = \frac{1}{2} \cdot 96 = 48$

3. Resolving forces along the plane: In equilibrium we use: $P_{max} = F_{max} + 120 \sin \alpha$ Where $\sin \alpha = \frac{3}{5}$, hence: $P_{max} = 48 + 120 \left(\frac{3}{5}\right) = 48 + 72 = 120$

Thus, the greatest possible value of P is 120 N.

Again, we resolve forces for when P is 30 N.

1. Using the condition for force equilibrium: $30 + F = 120 \sin \alpha$ Given $\sin \alpha = \frac{3}{5}$, we substitute: $30 + F = 120 \cdot \frac{3}{5} = 72$

Now solving for F: $F = 72 - 30 = 42$

Thus, the magnitude of the frictional force is 42 N acting up the plane.