A particle P of mass 2.5 kg rests in equilibrium on a rough plane under the action of a force of magnitude X newtons acting up a line of greatest slope of the plane, as shown in Figure 3 - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

Using the limiting friction formula, we can express the force of friction as:

$F_f = ext{friction coefficient} imes R = 0.4 imes R$

We previously derived that:

$F_f = 0.4 imes 23 ext{ N} = 9.2 ext{ N}$

In equilibrium, the force X must equal the component of the weight down the incline plus the frictional force opposing it:

$X = W imes ext{sin}(20°) + F_f = 2.5 imes 9.81 imes ext{sin}(20°) + 9.2$

The calculated value of X will be approximately:

$X ext{ is about } 17.6 ext{ to } 18 ext{ N}.$

When the force X is removed, we need to analyze the forces again. The weight component acting down the plane is:

$F = W imes ext{sin}(20°) = 2.5 imes 9.81 imes ext{sin}(20°)$

Calculating this gives:

$F ext{ is approximately } 8.4 ext{ N}$

The maximum frictional force is given by:

$ext{max } F_f = ext{friction coefficient} imes R = 0.4 imes R$

Substituting R:

$ext{max } F_f = 0.4 imes 23 ext{ N} = 9.2 ext{ N}$

Since the force acting down the slope (8.4 N) is less than the maximum static friction (9.2 N), this indicates:

ightarrow ext{P remains in equilibrium on the plane}.$$