Consider three blocks placed one over the other as shown in Fig. Let as now pull the blocks with the force of magnitudes `18 N, 100 N` and `15 N` Take `m_(1) = m_(2) = m_(3) = 10 kg` If the coefficient of static and kinetic friction between all contacting surface are `mu_(g) = 0.2` respectively find the

(a) Acceleration of the block s
(b) Friction at each surface

Explanation

Let us first calculate the limiting and kinetic friction force between various contact surface:
Between `m_(1)` and `m_(2)` :
`f_(t_1) = 0.3 xx 10 g = 30 N`
`f_(k_1) = 0.2 xx 10 g = 20 N`
Between `m_(2)` and `m_(3)` :
`f_(l_2) = 0.3 xx 20 g = 60 N`
`f_(k_2) = 0.2 xx 20 g = 40 N`
Between `m_(3)` ground :
`f_(l_3) = 0.3 xx 30 g = 90 N`
`f_(k_3) = 0.2 xx 30 g = 60 N`
Consider `FBD` of `m_(3)`

Maximum value of `f_(2)` is `f_(t_2) = 60 N` which is less then the maximum value of `15 + f_(3)` which `15 + 90 = 105 N` Hence, `m_(3)` will not acceleration so `a_(3) = 0`
Now assume that `m_(1)` and `m_(2)` move together with common acceleration a

`a= (100 - 18 - 40)/(20) = 2.1ms^(-2)`
Let us calculate how much friction is required between `m_(1)` and `m_(2)` for common acceleration a for this , consider `FBD` of `m_(1)`

`f_(1) - 18 = 10a rArr f_(1) - 18 = 10 xx 2.1 rArr f_(1) = 39 N`
This is not possible because the maximum value of `f_(l_(1)) =30 N`. Hence, slipping occurs between `m_(1)` and m`_(2)`

`a_(1) = (f_(k_1) - 18)/(10) = (20 - 18)/(10) = 0.2ms^(-2)`
`a_(2) =(100-f_(k_(1))-f_(k_(2)))/(10) (100 - 20 - 40)/(10) = 4 ms^(-2)`
a. So acceleration of `m_(1)` is `a_(1) = 0.2 ms^(-2)`
Acceleration of `m_(2)` is `a_(2) = 4 ms^(-2)`
Acceleration of `m_(3)` is `a_(3) = 0`
Since slipping occure between `m_(1)`and `m_(2)` Friction between `m_(1)`and `m_(2)` is `f_(k_1)= 20 N` Similarly . Slipping accures between `m_(1)`and `m_(2)` and `m_(2)` is `f_(k_2) = 40 N`

To find friction between `m_(3) ` and ground :
`f_(k_2) = 15 + f_(3) rArr 40 = 15 + f_(3) rArr f_(3) = 25 N`