Two bodies A,B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall. The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. (i) What is the reaction of the partition? (ii) The actionreaction forces between A,B? (iii) What happens when the wall is removed ? Does the answer to (ii) change when bodies are in motion ? [ Ignore the difference between `mu_(s)` and `mu_(k)`]

Explanation

We take the round off value g = 10 `"m.s"^(-2)`.
Normal reaction on A,
`N_(A) `=its weight (mg)
`= 5xx10` = 50 N
Normal reaction on B, `N_(B)=10xx10`=100 N
So the limiting friction on A,
`f_(A)= mu N_(A)= 0.5xx50=7.5 N`
and limiting friction on B,
`f_(B) = muN_(B)=0.15xx100=15` N
[ As the applied force , 200 N, is higher the force of friction attains its limiting value ]
(i) We take the combination of A and B as a single body. It is at rest due to the presence of the wall. If R be the reaction of the wall, the force equation in the horizontal direction is,
`200-f_(A)-f_(B)-R=0`
or, R = 200 -7.5 -15 = 177.5 N
(ii) Let `F_(AB)` = action force of A on B, `F_(BA)` = reaction force of B on A, `F_(AB)` and `F_(BA)` are equal and opposite. From the equilibrium of A, we get
200-`f_(A) -F_(BA)`=0
or, `F_(BA)` = 200 -7.5=192.5 N
So, `F_(AB)=- 192.5 N`
(iii) Now if the wall is removed, the reaction force R is absent due to the applied force the combination of A and B would move forward with an acceleration a (say). Then the force equation in the horizontal direction would be,
`200-f_(A)-f_(B)= (m_(A)+m_(B))a`
or, `a = (200-7.5-15)/(5+10)= (177.5)/(15)=11.83 "m.s"^(-2)`
Now, the force equation for the body A is,
`200-f_(A)-F_(BA)=m_(A)a`
`[ F_(BA)= ` new value of force by B on A]
or, `F_(BA)=200-7.5-5xx11.83=133.3` N
Similarly, `F_(AB)` =-133.3 N
So due to the accelerated motion the action-reaction pair changes to a much reduced value.