A dynomometer D (a force meter ) is attached to two masses M = 10 kg and m = 1 kg . Force F = 2 kg f and f = 1 kgf are applied to the masses . What will applied to the masses . What will the dynamometer show if :
(a) F is applied to M and f ot m
(b) F is applied to m and f to M
(c) if M = m = 5 kg .

Explanation

In all the cases as net pulling force is `F-f` while the mass involved is `m+M`
Acceleration (a) `("force")/("mass")=(F-f)/(m+M)=1m/s^(2)` ...(i)
So, the whole system will move in the direction of larger force with acceleration given by Eqn.(i)
(a) if `T_(1)` is the reading of dynamometer then for motion of M and m respectively we have
`F-T_(1)=Ma and T_(1)-f=ma`
i.e., `T_(1)=F-Ma andT_(1)=f+ma`
Substituting a from Eqn. (i) in any of the above, we get
`T_(1)=(mF+Mf)/(m+M)~=f=10N` ...(ii)
(b) For this case, equations of motion for m and M will be respectively
`F-T_(2)=ma and T_(2)-f =Ma`
Either of these equations in the light of Eqn. (i) yields
`T_(2)=(MF+mf)/(m+M)~= F=20 N` ....(iii)
(c) Substituting `m=M` in Eqn. (ii) or (iii) we get
`T_(3)=(F+f)/2=(20+10)/2=15N`
The tension in the spring is independent of mass in this case.
Note: if `f=F, a=0 and T=F=f`
i.e., if two equal forces are applied to two differencet masses connected by a spring the system remains at rest and the tension in the spring is equal to either force. The stretch in the spring will be `(F//k)` as F=ky.