A small block is held in place against a rough wall by someone?

Question

A small block is held in place against a rough wall by someone pushing on it with a force directed at 22° above the horizontal. The coefficients of static and kinetic friction between the box and wall are 0.4 and 0.3 respectively. The box slides down unless the applied force has magnitude 13 N. The mass of the box is

Answer 1

forces in x direction are:
13 x cos22
and in y direction are:
-mg (weight)
Ff (friction force) which is equal to Normal force x coefficient of friction
13 x sin22

Your Normal force is = 13 x cos22

if you balance forces in y direction, since your box is not moving or accelerating you should make it equal to zero
-mg + Ff + 13 x sin22 = 0

-mg + N x cf + 13 x sin22 =0

-mx9.8 + 13 x cos22 x 0.4 + 13 x sin22 =0
-9.8 x m = -9.69

m=.99 kgs

please check the math

thanks for the correction, coefficient of friction should be static not kinetic, i corrected the solution
thanks

Answer 2

I looked at the solution of makemovie. His analysis is good. I'd like to say that I agree on his entire solution, with one exception: his using 0.3 as the coefficient of friction. I'd use 0.4 instead, which is the coefficient of static friction. As he also stated and I agree, the block is not moving, meaning that it's stationary. If it were moving then 0.3 would be applicable.

But please do the math as he suggested. That's something I didn't do.