I don't understand how you got 26.96 N. The tension on the side being pulled (T1) is 27N. There is less tension on the other side of the pulley (T2). The difference in tension provides the torque to turn and accelerate the pulley. Call the angular acceleration 'alpha'. It equals the acceleration of the mass (a), divided by R.
Let m = pulley mass = 1.3 kg
M = object mass = 0.61 kg
R = pulley radius
(T1-T2)* R = I*alpha
= (1/2) m R^2 * alpha
= (1/2) m R a
T1 - T2 = m a/2
T2 = M a
T1 = [M + (m/2)]a = 27 N
a = 27/1.26 = 21.4 m/s^2
T2 = T1 - m a/2 = 27 - 13.9 = 13.1 N
You pull downward with a force of 27 N on a rope that passes over a disk-shaped pulley of mass 1.3 kg and radius 0.075 m. The other end of the rope is attached to a 0.61 kg mass. Find the tension in the rope on both sides of the pulley.
I found that the tension on the side you are pulling is 26.96 and that the tension on the side with the weight is less than the side you are pulling but I still can't find the actual tension.
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