If the coefficient of friction between all surface figure is `0.4` then find the minimum force `F` to have equilibrium of the system.

Options

A. `62.5 N`
B. `150 N`
C. `31.25 N`
D. `50 N`

Answer

c

Explanation

In the obsence of friction we can find that `15 kg` will acceleration downward and `25 kg` upward so various forces acting on these will be as shown

`N_(2)= N_(1) = F. f_(t_1) = mu N_(1) = 0.4 F. f_(t_2) =mu N_(2) = 0.4F`
For `15 kg T + f_(t_1)= 15kg rArr T + 0.4F = 15 kg`...(i)
For `25 kg 2T = f_(t_1) + f_(t_2)+25g`
`rArr 2T = 0.8F + 25g`...(ii)
Solve Eqs (i) and (ii) to get `F = 31.25 N`