A particle P of mass 2.7 kg lies on a rough plane inclined at 40° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2014 - Paper 2

Next, we resolve the forces acting parallel to the slope to find the friction coefficient.

The weight component parallel is:

$F = 2.7g \sin(40^\circ) - 15 \sin(50^\circ)$

Using the equation ( F = \mu R ):

$\mu = \frac{2.7g \sin(40^\circ) - 15 \sin(50^\circ)}{R}$

Substituting the values:

$\mu \approx 0.23 \text{ or } 0.232$

To determine if P moves, we need to compare the forces. The weight component acting down the slope is:

$F_{\text{down}} = 2.7g \sin(40^\circ)$

Calculating this, we find:

$F_{\text{down}} \approx 17.0 N$

Since this value (17.0 N) exceeds the maximum static friction force, which is ( \mu R \approx 32 \times 0.23 \approx 7.36 N ), the particle P will start to move down the slope.