Question

A box of mass ?=20.5 kg is pulled up a ramp that is inclined at an angle ?=19.0∘ angle with respect to the horizontal. The coefficient of kinetic friction between the box and the ramp is ?k=0.305 , and the rope pulling the box is parallel to the ramp. If the box accelerates up the ramp at a rate of ?=2.89 m/s2 , calculate the tension ?T in the rope. Use ?=9.81 m/s2 for the acceleration due to gravity.

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the box = m = 20.5 kg

Angle of incline = = 19^o

Normal force on the block from the ramp = N

Coefficient of kinetic friction between the box and the ramp = _k = 0.305

Friction force on the box = f

f = _kN

Tension in the rope = T

Acceleration of the rope = a = 2.89 m/s²

From the free body diagram,

N = mgCos

f = _kmgCos

ma = T - mgSin - f

ma = T - mgSin - _kmgCos

T = m(a + gSin + _kgCos)

T = (20.5)[2.89 + (9.81)Sin(19) + (0.305)(9.81)Cos(19)]

T = 182.71 N

Tension in the rope = 182.71 N