Frictional force?

Question

A 22.5 kg block is at rest on the incline. The coefficients of static and kinetic are .6 and .4 respectively. The angle is 25 degrees. What is the frictional force acting on the 22.5 kg mass ?

Answer 1

You don't need the coefficients of friction for this problem. All you need to do is recognize that there are 2 forces acting on the block parallel to the incline:

1) Parallel component of gravity: Fg(par) = m·g·sin(Θ) = 93.19 N
2) Frictional force (pointing UP the incline)

Since the block is at rest, these two forces must be equal and opposite:

F(fric) = 93.19 N (up the incline)

Answer 2

Since the block is at rest kinetic friction has no role to play in this till the block starts motion. Assuming the block is in a tendency to slide down, therefore, the frictional force acts in opposite direction. Resolving force components along x-axis(along the slope) and y-axis (perpendicular to the slope) we get F=(mu)N
F=(mu(s))mgsin(x) , where mu(s) is the coefficient of static friction.
Hence, it can be calculated from here.

Answer 3

The frictional force must balance the component of weight along the incline. Draw a little sketch and apply a little trig and you're home free.

Answer 4

You need to look at the forces that are acting on the block. Since the block is not moving you know that it is not accelerating and that the net force acting on the block is zero. The fricitonal force is one of the components that adds to zero. I recommend drawing a free-body diagram which illustrates the forces. HINT: There are 3 of them.

Answer 5

22.5*9.81*sin25
because, the maximum frictional force at rest is more than the force mgsin25, the frctional force coming in to action is only mgsin25. However if more force is applied fictional force increases to mgcos25*0.6 and then decreases to mgcos25*0.4

Answer 6

Yes