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 1

The frictional force F_f can be calculated from the equilibrium condition in the direction parallel to the plane:

1. The sum of the forces acting parallel to the plane must equal zero:

   $F_{ ext{friction}} = W_{ ext{parallel}} - T imes ext{sin}(50)$

   Where:

   $W_{ ext{parallel}} = mg imes ext{sin}(40) = 2.7 imes 9.8 imes 0.6428 \\ \approx 17.00 N$

Thus, using the normal force from part (a):

$F_{ ext{friction}} = ext{μ}R$

Substituting values:

$ext{μ} = \frac{17.00 - 9.64}{R} = \frac{7.36}{10.85} \\ \approx 0.68$

When the 15 N force is removed, the only forces acting are the weight and the normal reaction. The particle is now influenced only by the weight component along the plane.

Using the components identified in previous parts:

$W_{ ext{parallel}} = 2.7 imes 9.8 imes ext{sin}(40) \\ \approx 17.00 N$

Since the friction was counteracting this component, and is given as:

$F_{ ext{friction}} = ext{μ}R \\ \approx 10.85 imes 0.68 \\ \approx 7.37 N$

Now, we can see:

$17.00 > 7.37$

Thus, the particle will move down the incline.

To determine if P moves after the removal of the 15 N force, we compare the forces acting down the plane to the maximum static friction:

Previously, we established that:

• Weight component down the plane = 17.00 N
• Max static friction = 7.37 N

Since 17.00 N (down the plane) is greater than 7.37 N (friction), it implies there is a net force down the incline. Therefore, P will begin to slide down the plane.