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

To find the value of X,

1. Analyze the forces acting along the plane. The force X is opposed by the component of the weight acting down the slope: $F = W imes ext{sin}(20°) = 2.5 imes 9.81 imes ext{sin}(20°)$.

2. The frictional force f can be calculated using: $f = ext{μ}R = 0.4R$.

3. Set up the equilibrium condition:

   • The forces along the slope give: $X = f + F$ Substitute the known values and calculate: $X = 0.4 imes R + 2.5 imes 9.81 imes ext{sin}(20°)$.

4. Calculate:

   • After substituting R, we get the value of X to be approximately: $X \\approx 17.6 ext{ or } 18 ext{ N}$.

To show that P remains in equilibrium after the removal of force X,

1. Evaluate the forces when the force X is removed. The remaining forces acting on P are the weight, normal reaction R, and frictional force f:

   • The weight W acts down the slope, while the frictional force opposes motion: $F = W imes ext{sin}(20°) = 2.5 imes 9.81 imes ext{sin}(20°)$.

2. Substitute values and calculate:

   • The weight component gives: $F \\approx 8.38 ext{ or } 8.4 ext{ N}$.

3. Calculate the maximum frictional force: $ext{μ}R = 0.4 imes R ext{ where } R \\approx 23.0 ext{ N}$

   • This gives: $ext{μ}R \\approx 9.21 ext{ or } 9.2 ext{ N}$.

4. Compare forces:

   • Since $F < μR$, we confirm that P remains in equilibrium. Thus, we have shown that: $F < μR ext{ implies equilibrium.}$