Two bodies A and B of mass 5 kg and 10 kg contact with each other rest on a table against a rigid The coefficient of friction between the bodies and the table is 0.15 A force of 200 N is applied horizontal on A What are (a) the reaction of thee wall (b) the the action , reaction forces between A & B ? What happens when the wall is removed ? Does the answer to (b) Change, when the bodies are in motion ? Ignore difference between `mu_(s)` and `mu_(k)`
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Explanation

(i) When the wall exists and blocks A and B are pushing the wall, there can't be any motion i.e., blocks are at rest . Hence, (a) reaction of the partition = - ( force applied on A) = 200 N towards left. ltbRgt (b) action reaction forces between A and B are 200 N each. A presses B towards right with an action force 200 N and B exerts a reaction force on A towards left having magnitude 200 N .
(ii) When the wall is removed, motion can take place such that net pushing force provides the acceleration to the block system. Hence, taking kinetic friction into account, we have
200 - `mu` `(m_1 + m_2)`g = (`m_1 + m_2`) a
`implies` a = 200 - `mu(m_1 + m _2`)g / (`m_1 + m_2`) = `200 - 0.15 xx (5 + 10 ) xx 10 `/ ( 5 + 10 )
= 200 - 22.5 / 15 = 177.5 / 15 = 11.8 `ms^(-2)`
`therefore` If force excerted by A and B be `F_BA` ,then considering equilibrium ( or free body diagram ) of only block A ,we have
200 - `f_1` = `m_1`a + `F_BA` or 200 - `(mu)(m_1`g = `m_1`g + `F_BA`
`implies` `F_BA` = 200 - `mu``m_1`g - `m_1`a = 200 - (0.15 `xx 5 xx 10)` - `(5 xx 11.8)`
= 200 - 7.5 -59
= 200 - 66.5 = 133.5 N = 1.3 `xx 10^(2)`N towards right
~therefore` Force excerted on A by B `F _ AB` = `-F_BA` = 1.3 xx 10 ^(2)` N towards left.