Question

two masses m1 and m2 are attached to a spring balance S.if m1 is greater than m2 then the reading of spring balance will be

Answer 1

Given -

m1 > m2

Reading on spring balance = ?

Solution-

Since m1 > m2, there will be an acceleration in the system.

m1g-m2g = (m1+m2)a

a= (m1-m2)g/(m1+m2)

Restoring force,

R= T1 = T2

Thus, R = T1 = m1(g-a)

R = m1[g- (m1-m2)g/(m1+m2) ]

R = m1g (m1+m2-m1+m2)/(m1+m2)

R = 2m1m2g/(m1+m2)

The reading of the spring balance is nothing but restoring force.

Hence, reading on spring balance is 2m1m2g/(m1+m2).

Hope that was useful.

Answer 2

Since m1 > m2, there is an acceleration in the system

• m1g-m2g = (m1+m2)a
• a= (m1-m2)g/(m1+m2)

Restoring force,

• R= T1 = T2
• R = T1 = m1(g-a)
• R = m1[g- (m1-m2)g/(m1+m2) ]
• R = m1g (m1+m2-m1+m2)/(m1+m2)
• R = 2m1m2g/(m1+m2)

I hope that this was useful for you.