A man pulls 50 kg box at constant speed and applies a 200 N force at angle of 30 degrees?

Question

A) Sum the forces in x-direction. What is the value if the frictional force opposing the motion?

B) Sum the forces in y-direction. What is the value of the normal force?

Answer 1

A) Sum the forces in x-direction
Fcos(30) - f=0
f=Fcos(30)=173 N

B) Sum the forces in y-direction
N=mg - Fsin(30)
N= 50x 9.81 - 200x sin(30)=
N=390 N