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 need to resolve the forces parallel to the slope. The force component down the slope due to gravity is:

$F = 2.7g \sin(40°) - 15 \cos(50°)$

Calculating the values gives:

$F = 2.7 \times 9.81 \times \sin(40°) - 15 \cos(50°) \approx 17.0 \, \text{N}$

The frictional force can be expressed as:

$F = \mu R$

Setting it equal gives:

$\mu = \frac{F}{R} = \frac{2.7 \times 9.81 \times \sin(40°) - 15 \cos(50°)}{31.8} \approx 0.23 \text{ or } 0.232$

To determine if P moves, we compare the net force acting down the slope with the frictional force. The component of weight acting downhill is:

$F_{\text{max}} = 2.7 \times 9.81 \times \sin(40°) \approx 17.0 \, \text{N}$

Since 17.0 N exceeds the maximum frictional force (calculated from part b), it indicates that P will begin to slide down the slope, thereby justifying that P will move.