Answer:
Given::
force P = 32 N
mass of the pully M = 1.4kg
radius R = 0.075m
mass attached at the other end
=0.77kg
since the pully mass is not negligible
tensions in the rope on both sides not same
if a is the accelration of the mass m
and T₁ is the tension in the rope where the mass is attached then
ma = T₁ ₋mg1
T₁ = ma₋mg1 . . . . 1
and if T₂ the tension on the other side then
F=T₂ ..... 2
since the mass is moving up T₁ is less than T₂
So, (T₂ ₋ T₁ ) R =1 (a/R)
Putting the value of T₁ from equation 1
(T₂ ₋ ma₋mg1 ) R =1 (Ma/2)
a = (T₂ ₋ mg)/m+M/2
a =
![(32-0.77*9.8)/(0.77+1.4/2)](https://img.qammunity.org/2021/formulas/physics/college/uk2llh49362syuncq98znkyba7bprrpcx6.png)
= 24.454/1.47
= 16.6353741
So, T₁ = ma₋mg
= 0.77(16.6353741+9.8)
T₁ = 20.355
T2 = 32