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)
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"How to prove this ?
#2|k^2# then #2|k# for some #k\inZZ#"
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.
![enter image source here]()
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#
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#
