m1 question

A box of weight 100 N rests in equilibrium on a plane inclined at an angle α to the horizontal. It is given that sin α = 0.28 and cos α = 0.96. A force of magnitude P N acts on the box parallel to the plane in the upwards direction (see diagram). The coefficient of friction between the box and the plane is 0.25.

(i) Find the magnitude of the normal force acting on the box. = 96N
(ii) Given that the equilibrium is limiting, show that the magnitude of the frictional force acting on the box is 24 N.
=96*0.25 = 24N
(iii) Find the value of P for which the box is on the point of slipping:
(a) down the plane.

The mark scheme states the answer is P=4, from P = 100×0.28 – 24.Why is the answer less than 24N, which is the frictional force acting on the box to stop it from slipping?