Question

two massed 5 kg and 3 kg are suspended from the ends of an inextensible light string passing over a frictionless Pulley. when the mass are released, force exerted by the string on the pulley is

Answer 1

Lets do an Free body diagram of m₁=5kg
T is acting upwards whereas m₁g is acting downwards.
m₁g>T                (m₁>m₂)
Free body diagram of m₂=3kg
T is acting upwards whereas m₂g acts downwards.
m₂g<T
m₁g-T=m₁a             (i)
T₂-m₂g=m₂a            (ii)
add (i) and(ii)
m₁g-m₂g=m₁a+m₂a
g(m₁-m₂)=a(m₁+m₂)
10(2)=a(8)
20=8 a
20/8=a
a=2.5 m/s²
For finding T
T-m₂g=m₂a
T-30=7.5
T=37.5N
Therefore Force on pulley by rope =2×Tension 
Force on pulley= 37.5N×2
force=75N

Answer 2

Answer:

7.5kgf or 75N

Explanation:

hope it will help u.....