Two questions about mechanics

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Q1: You have to note the conditions for the two proportionalities.
acceleration a is proportional to force F, when under a constant mass m.
Hence, F = K.a = (k).(m).a, where K is a constant of proportionality, and is expressed as another proportional constant k, i.e. K = k.m

Similarly, a is proportional to 1/m, under the condition that the applied force F is constant.
Thus, a = K'/m = k'.F/m, where K' and k' are constants
i.e. F = ma/k'
Multiplying the two equations together,
F^2 = [k/k'].(ma)^2
Taking square-root on both sides, F = K".ma
where K" = square-root(k/k').
Such equation is used to define the unit of force "Newton" by taking the constant K" equal to one.

Q2: The force diagram that you have drawn is wrong.
There are, in fact, three forces to keep the trolley in equilibrium. These are (i) the weight of the trolley that acts vertically downward, (ii) the frictional force that acts upward along the plane, and (iii) the normal reaction force that acts upward perpendicular to the plane surface. A correct force diagram should be drawn with all these three forces together. Just by taking the "weight" and "friction", but omitting the "normal reaction" would not form a closed force diagram representing the trolley in equilibrium.

If you draw out correctly the force diagram, it should be a right angle triangle, with the hypotenus representing the weight W, one side of the triangle representing the frictional force f, and the other side representing the normal reaction R, say. After finding the angles of the triangle, you could then easily see that,
sin(30) = f/W
i.e. f = W.sin(30)

[Note: just imagine if your force diagram is correct, the resultant of the two forces W and f would be a horizontal force acting to the right (representing by the dotted-line in your diagram). That is to say, the trolley would not be in equilibrium, but pushed up the slope.]

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