Question

Find the reading of the spring balance shown in fig. The elevator is going up with an acceleration g/10, the pulley and the string are light and the pulley is smooth?( The mass of the blocks are 1.5 Kg and 3.5 Kg)

Answer 1

`4.4 Kg`

Explanation:

Here,we can think from non inertial frame of reference that as the lift is moving upward with an acceleration of `g/10`,so net force acting downward for the two blocks will be their weight+(g/10)their mass.

So,if the tension in the string is `T`,we can write for the larger block as, `3g +(3g)/10 -T =3a` (as it is going down)

similarly for the smaller block `T-1.5g -(1.5g)/10 =1.5a` (where, `a` is the acceleration of the blocks relative to the lift)

Solving both we get, `T=(11/5)g`

Now,the spring is attached to the pulley which is bearing the tension of the string from both the side,so the spring will show a reading of `2T=(22/5)g=43.12 N`

so,reading of the spring balance will be `((22g)/5)/g =4.4 Kg`

Answer 2

See below.

Explanation:

Let the lift acceleration be `gamma`

`x_1, m_1` coordinate and mass of left body.
`x_2,m_2` coordinate and mass of right body.
`x_0` the pulley center coordinate
`L` rope length.

For left and right bodies

`T-m_1 g = m_1 ddot x_1`
`T - m_2 g = m_2 ddot x_2`

The restrictions to the movement are

`(x_0-x_1) + (x_0-x_2) = L` or

`ddot x_0-ddot x_1+ddot x_0-ddot x_2 = 0` or

`ddot x_0 = 1/2(ddot x_1 + ddot x_2)`

Now solving

`{(T-m_1 g = m_1 ddot x_1),(T - m_2 g = m_2 ddot x_2),(ddot x_0 = 1/2(ddot x_1 + ddot x_2)):}`

we obtain

`{(T = 2(m_1m_2)/(m_1+m_2)(ddot x_0 + g)),(ddot x_1 =(2 ddot x_0 m_2+(m_2-m_1)g)/(m_1+m_2) ),(ddot x_2 = (2 ddot x_0 m_1+(m_1-m_2)g)/(m_1+m_2)):}`

but `ddot x_0 = gamma` then

`T = 2(m_1m_2)/(m_1+m_2)(gamma + g)`

So the spring balance will read

`P = 2T = 4.4 g/g = 4.4`