M1 Help - Dynamics of a particle

4.

Figure 2


A particle P of mass 0.5 kg is on a rough plane inclined at an angle a to the horizontal, where tan a = 3/4. The particle is held at rest on the plane by the action of a force of magnitude 4 N acting up the plane in a direction parallel to a line of greatest slope of the plane, as shown in Figure 2. The particle is on the point of slipping up the plane.

(a) Find the coefficient of friction between P and the plane.

I am stuck with part (a) of this M1 question. I know the steps to follow it's just that I don't understand how to resolve perpendicular to the plane:

(a = arctan(3/4))
In the mark scheme, when resolving perpendicular to the plane, it says the reaction force R is equal to 0.5xgxcos36.8. I don't understand how they got this result, because cosA = A/H = (0.5g)/R, so surely R = (0.5g)/cos36.8 ?

Please help

You seem to be in a pickle. You're not resolving forces, which you should be.

Draw the force components on for weight, and the reaction force perpendicular to the plane, then resolve perpendicularly. You only have the reaction force, and weight acting perpendicularly.

Look at this: