If force F is increasing with time and at t=0, F = 0, where will slipping first start?

Options

A. between `3 kg and 2 kg`
B. between `3 kg and 1 kg`
C. between `1 kg and ground`
D. Both `(a) and ( b)`

Answer

C

Explanation

`f_(1) rarr` force of friction between `2 kg` and `3 kg`
`(f_(1))_(max) = 0.5 xx 3 xx 10 = 15 N`
`f_(2) rarr` force of friction between `2 kg` and `1kg`
`(f_(2))_(max) = 0.3 xx 5 xx 10 = 15 N`
`f_(3) rarr` force of friction between `1 kg` and ground
`(f_(3))_(max) = 0.1 xx 6 xx 10 = 6 N`
When `F gt 6 N` system will start moving with `a` common acceleration
`a = (F - 6)/(3 + 2 +1) = ((F)/(6) - 1) m//s^(2)`

`f_(1) - 6 = (3) a = (F)/(2) - 3`
`:. f_(1) = (6 + (F)/(2) - 3)`
`= ((F)/(2) + 3)`
Since `F` is slightly greater than `6 N`
`:. f_(1) lt 15 N` or `lt (f_(1))_(max)`
`:.` No slipping will occur here

`f_(2) - 6 = (1) (a) = (F)/(6) - 1`
`:. f_(2) = (F)/(6) + 5`
Again `f_(2) lt (f_(2))_(max)`. So no slip will take place here also.