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

Explanation

Mass of block `A m_(1) = 5 kg`
Mass of block B `m_(2) =10 kg`
Co- efficient of friction between block an surface of the table `mu= 0.15 `
Horizontal force on block A F = 200 N
(a) Frictional force will be in opposite directin of external force
`F= mu R`
`= mu(m_(1) + m_(2)) g`
` =22.05 N`
Net force in rigth direction F= f - t
By Netwtons third law of motion rection on the wall = force acting on wall
( b) Actin= rectin force between A and B = friction force acting on block A

`F_(1) = mu_(1)`
`=0.15 xx 5 xx 9.8`
` F_(1) = F= f_(1)`
= 192 .65 N `to ` in right directin
Resultant force applied by A on B= 192.65 N

By Netwon third law of motion force applied by B on A = rection =`F_(1)`
(c ) When block are in motion ( assume wall is not present )
`:. a= 11.86 ms^(-2)`
Let common acceleration of system is a (d ) If `m_(1)` (block) and `m_(2)` both are in motion when block are in motion force applied by A on B
`=F-F_(1) - m_(1) a`
`=192 .65 -59 .3`
`=133.35N`
=Force applid by A on B
Reaction of B= 133.75 N `to` in left direction