You pull downward with a force of 26.0 N on a rope that passes over a disk-shaped pulley of mass 1.31 kg and radius 0.0791 m. The other end of the rope is attached to a 0.601-kg mass. Calculate the tension in the rope on both sides of the pulley. Enter tension for the part of the rope that you are pulling on first. Then enter the tension for the part of the rope with the mass.

Tension for the part of the rope that was pulled on first, T1 = 26N

Tension of the other side of the rope, T2 = 9.4N

Explanation:

The first rope creates a tension, T1

Tension T1 = 26N

pulley of mass, M = 1.31 kg

radius, r = 0.0791 m

Mass of rope, m = 0.601 kg

Tension from downward pull on the other side of the rope gives an equation:

T2-m*g =m*a

a = (T2-m*g)/m ...equation 1

Where m = mass attached to the rope

Tension from pulley system also gives an equation:

(T1-T2) = 1/2 *M*a

Where M = mass of the pulley

a= 2(T1-T2)/M ...equation 2

Equating equation 1 and equation 2

a = a

(T2-m*g)/m = 2(T1-T2)/M

Let g= 9.81m/s^2

[T2 -(0.601×9.81)]/0.601 = 2(26-T2)/1.31

(T2 - 5.89581)/0.601 = 1.5267(26-T2)

T2 - 5.89581 = 0.9175(26-T2)

T2 - 5.89581 = 23.855 - 0.9175T2

T2 + 0.9175T2 = 23.855 - 5.89581

1.9175T2 = 17.95919

T2 = 17.95919/1.9175

T2 = 9.3659N

Tension from the other side of the pull, T2 is approximately 9.4N

conservation of mass

mass before something happens = mass afterwards