A box is being pulled along a rough horizontal surface by a rope which has a tension of 10N at an angle of 60 degrees with the horizontal. If the box moves with constant velocity find the frictional force and the normal reaction force given the weight of the box is 100 N

Answer 1

Take upwards and direction of travel as positive.
Horizontally:
Fa + Ff = 0
10cos(60) N + Ff = 0
Ff = -5 N
Vertically:
Fa + Fg + Fn = 0
10sin(60) N + (-100) N + Fn = 0
Fn = 100 - 5sqrt(3) N

Answer 2

M*g = 100.
M = 100/g = 100/9.8 = 10.2kg = Mass of the box.

Fn = 100 - 10*sin60 = 91.34 N. = Normal force

Fap-Fk = M*a.
10*Cos60-Fk = 10.2*0,
Fk = 10*Cos60 = 5 N. = Force of kinetic friction.