Question

A 5.35-kg box is pulled up a ramp that is inclined at an angle of 33.0° with respect to the horizontal, as shown below. The coefficient of kinetic friction between the box and the ramp is 0.165, and the rope pulling the box is parallel to the ramp. If the box accelerates up the ramp at a rate of 2.09 m/s2, what must the tension FT in the rope be? Use g = 9.81 m/s2 for the acceleration due to gravity.

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the box = m = 5.35 kg

Angle of incline = = 33^o

Tension in the rope = T

Normal force on the box from the incline = N

Coefficient of kinetic friction between the box and the ramp = = 0.165

Friction force on the box = f

f = N

Acceleration of the box = a = 2.09 m/s²

From the free body diagram,

N = mgCos

f = mgCos

ma = T - mgSin - f

ma = T - mgSin - mgCos

(5.35)(2.09) = T - (5.35)(9.81)Sin(33) - (0.165)(5.35)(9.81)Cos(33)

T = 47.03 N

Tension in the rope = 47.03 N