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

The next step is to find the frictional force acting on P. We first resolve the forces parallel to the slope:

$F = mg imes ext{sin}(40^ ext{o}) - 15 imes ext{sin}(50^ ext{o})$

From our earlier calculations, we have: $F = 2.7 imes 9.81 imes ext{sin}(40^ ext{o}) - 15 imes ext{sin}(50^ ext{o}) \ ext{ (approximately) } = 7.366 ext{ N}$

Now, using the relationship between frictional force and normal reaction:

u R $$ Substituting the values to find the coefficient of friction (μ):

u = rac{F}{R} = rac{7.366}{31.8} ext{ (approximately) } = 0.232 $$. Thus, the coefficient of friction between P and the plane is approximately 0.23.

To determine whether P moves, we compare the component of the weight parallel to the slope with the maximum frictional force. The component of weight parallel to the slope can be calculated as:

$F_{ ext{max}} = mg imes ext{sin}(40^ ext{o}) \ ext{ (approximately) } = 17.0 ext{ N}$

With the force of 15 N pulling upward and the frictional force calculated as:

u R = 0.232 imes 31.8 \ ext{ (approximately) } = 7.366 ext{ N} $$ Since the component of weight (17.0 N) is greater than the combined forces acting on P (15 N upward - 7.366 N friction downward), the particle P will begin to slide down the plane.