A box at rest on a ramp is in equilibrium, as shown. What is the force of static friction acting on the box? Round your answer to the nearest whole number. N What is the normal force acting on the box? Round your answer to the nearest whole number. N

Answer
Ffs = 251 N
Fn = 691 N

Explanation
Take the y direction to be normal to the ramp and the x direction to be parallel to the ramp. The angle of the ramp is 20°, so the angle that the weight vector makes with the normal is also 20°.  Therefore:
Fgx = Fg sin 20°
Fgy = Fg cos 20°
Sum of the forces in the x direction (parallel to the ramp):
∑F = ma
Ffs − Fgx = 0
Ffs = Fgx
Ffs = Fg sin 20°
Ffs = 735 sin 20°
Ffs ≈ 251
Sum of the forces in the y direction (normal to the ramp):
∑F = ma
Fn − Fgy = 0
Fn = Fgy
Fn = Fg cos 20°
Fn = 735 cos 20°
Fn ≈ 691

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way