What is the frictional force keeping the Tv in place

Question

A 22 kg TV sits at rest on a ramp inclined at an angle of 15°

Normal force= 208.3 N

Frictional force=?

Coefficient of Friction= ?

I was able to get Normal force, but I'm really stuck of Frictional force and Coefficient of Friction

Answer 1

Considering the question setup, there are only three forces acting on the TV: The weight (gravitational force), the normal force, and friction. The gravitational force acts directly downwards, whereas the frictional and normal forces act parallel and perpendicular to the surface, respectively. We have to break the gravitational force into x and y components in order to observe the contribution to each direction. Since the TV is at rest, we can set the sum of the forces in each direction equal to zero. With this, we can relate the y-component of mg to the normal force (N - mgcosθ = 0), as you did to get the answer of 208.3 N, and we can relate the x-component of mg to the frictional force (mgsinθ - F_f = 0) since it also acts in the x-direction.

The frictional force is directly related to the normal force, as given by the equation F_f =µ_sN, where µ_s is the coefficient of static friction. (If the object begins moving, this is replaced with the coefficient of kinetic friction, which is solved the same way.) Once you have solved for the normal and frictional forces, all you have to do is plug the values into this equation to solve for the coefficient of static friction (μ_s = F_f / N = tanθ).

Hope this helps!