Hi there!
[tex]\large\boxed{545N}[/tex]
Since the object is being pulled at a constant velocity, the forces must be balanced.
Since there is no movement vertically, we must take into account the horizontal forces. We can also assume a positive acceleration to be in the direction of motion.
The acceleration and force due to gravity on an incline is:
a = gsinФ
F = MgsinФ
∑F = -MgsinФ + T
Since it is getting pulled at a constant velocity, ∑F = 0. So:
0 = -MgsinФ + T
MgsinФ = T
Solve for T by plugging in values. Let g = 10 m/s²
T = (120)(10)sin(27) ≈ 545 N
