Question

You pull downward with a force of 26.1 N on a rope that passes over a disk-shaped pulley of mass 1.26 kg and radius 0.0758 m. The other end of the rope is attached to a 0.653-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.

Answer 1

let

M = 1.76 kg, R = 0.0758 m
m = 0.653 kg

let T1 is tension in string on pulling side

and T2 is the tension in the string on other side.

T1 = 26.1 N

let a is the acceleration of the block and alfa is the angular acceleration of the pulley.

net force acting on m, Fnet = T2 - m*g

m*a = T2 - m*g

T2 = m*a + m*g ----(1)

Net torque acting on the pulley, Tnet = I*alfa

(T1 - T2)*R = 0.5*M*R^2*alfa

T1 - T2 = 0.5*M*(R*alfa)

T1 - T2 = 0.5*M*a (since a_tan = r*alfa)

26.1 - (m*a + m*g) = 0.5*M*a

26.1 - (0.653*a + 0.653*9.8) = 0.5*1.26*a

==> a = 15.3 m/s^2

so, T2 = m*a + m*g

= 0.653*15.3 + 0.653*9.8

= 16.4 N

so,

T1 = 26.1 N <<<<<<<<<<<-----------------Answer

T2 = 16.4 N <<<<<<<<<<<-----------------Answer