Question

At wood machine is attached below the left hand pan of a physical balance while equal weight `(M+m)` g is placed in the right hand pan. The beam is horizontal when the masses M and m on the left hand side are braked What will happen if the masses M and m are unbraked on the left hand side?

Answer 1

Initially the system is in equilibium, so
`F_(L)=F_(R)=(m+M)g` ....(i)
When the L.H.S is unbraked, mass M will move downward while m upwards with acceleration
`a=((M-m))/(M+m)g`
So, the tension in the string,
`T=m(g+a) [or M(g-a)]=(2mMg)/((M+m))`
This in turn implies that force on L.H.S when the masses are in motion will be

Now from Eqns (i) and (ii) it is clear that
`F_(L)-F_(L)=(m+M)g-(4mMg)/((M+m))`
i.e., `F_(L)-F_(L)=((M-m)^(2))/((M+m))g gt 0`
i.e., `F_(L) gt F_(L) or F_(L) lt F_(L)`
i.e., the force on the left side decreases when the masses are unbraked and hence L.H.S will move up while R.H.S. down
Note: Physical reasoning When the masses of L.H.S are unbraked, M will move down while m up with same the weight of M will decrease (as accelerated down) while of m will increase (as accelerated up) As `M gt m` decrease in weight will be more than increase So net weight on L.H.S will decrease and hence, it will move up.