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, we again resolve the forces acting on the particle. The particle must remain in equilibrium, so the forces acting up the slope and down the slope must balance each other:

$P + F_{friction} = 120 \sin(\alpha)$

The normal reply reaction when $P$ is acting up the slope is:

$R = 120 \cos(\alpha) - F_{friction}$

With the frictional force given by $F_{friction} = \frac{1}{2}R$. Substituting for $R$:

$F_{friction} = \frac{1}{2}(120 \cos(\alpha))$

Substituting in the equation for forces acting along the slope:

$P + \frac{1}{2}(120 \cos(\alpha)) = 120 \sin(\alpha)$

Using values for $\cos(\alpha)$ and $\sin(\alpha)$ calculated previously, we can find:

$P + \frac{1}{2}(96) = 120 \cdot \frac{3}{5}$

Thus:

When $P = 30$ N, we set up the equation:

$30 + F = 120 \sin(\alpha)$

Substituting for $\sin(\alpha)$ gives:

$30 + F = 120 \cdot \frac{3}{5} = 72$

Solving for $F$ gives:

$F = 72 - 30 = 42 ext{ N}$

The direction of the frictional force is up the slope, as it opposes the motion or impending motion of the particle.