You pull downward with a force of 27.3 N on a rope that passes over a disk-shaped pulley of mass 1.43 kg and radius 0.0792 m. Th
e other end of the rope is attached to a 0.700-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.
1 answer:
Answer:
The tension in the left side string = 17.21 N
The tension in the right side string = F = 27.3 N
Explanation:
Given that
F= 27.3 N
M= 1.43 kg ,r= 0.0792 m
Moment of inertia of disk ,I = 0.5 m r²
I = 0.5 x 1.43 x 0.0792² = 0.0044 kg.m²
m= 0.7 kg
Lets take linear acceleration of system is a m/s²
Lets take tension in left side string = T
From Newtons law
T- mg = ma
T- 0.7 x 10 = 0.7 a ----------1
(F - T) r = I α
α = Angular acceleration of disk
a = α r
(F - T) r = I α
(F - T) r² = I a
( 27.3 - T) x 0.0792² = 0.0044 a --------2
Form equation 1 and 2
a= 1.42 T - 10 m/s²
a = 1.42 ( 27.3 - T) m/s²
1.42 T - 10 = 38.9 - 1.42 T
T=17.21 N
The tension in the right side string = F = 27.3 N
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- radius of the hill:
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