In the co-efficinet of friction between the floor and the body `B` is 0.1. The co-efficient of friction beteen the bodies `B` and `A` is 0.2 A fore `F` is applied as shown `B` The mass of `A` is `m//2` and of `B` is m Which of the following statements are ture ?
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Options

A. The bodies will move together if `F =0.25 mg` .
B. The body A will slip with respect to `B` if `F =0.5 mg` .
C. The bodies will move together if `F =0.5 mg` .
D. The bodies will be at rest if `F =0.1 mg` .

Answer

A, B, D

Explanation

Consider the adjacent diagram Frictonal force on `B(f_(1)` and frictional force on A `(t_(2))` will be as shown.
Let `A` and `B` are moving toghether
`a_("common")=(F-f_(1))/(m_(A)+m_(B)) =(F-f_(1))/((m//2)+m)=(2(F-f_(1)))/(3m)`
Peseudo force `A = (m_(A))xxa_(common)`
`=m_(A)xx(2(F-f_(1)))/(3m)=(m)/(2)xx(2(F-f_(1)))/(3m)=((F-f_(1)))/(3)`
The force (F) will be maximum when
Pseudo force on A =Frictional force on A
`rArr(F_(max)-f_(1))/(3)=mum_(A)g`
`rArr (F_(max)-f_(1))/(3) = 0.2 xx (m)/(2) x g = 0.1mg`
`rArrF_(max) = 0.3 mg + f_(1)`
`= 0.3 mg + (0.1) (3)/(2) mg = 0.04mg`
`rArr` Hence, maximum force upto which bodies will move togther is `F_(max) = 0.45mg`
(a) Hence for `F =0.25mg gt F_(max)` bodies will move together
(b) For `F =0.5mg gt F_(max)` body A will slip with respect to `B`
(c ) For `F = 0.5mg gt F_(max)` bodies slip
`(f_(1))_(max) = mu m_(B)g = (0.1) xx (3)/(2) m xx g = 0.15mg`
`(f_(2))_(max) = mu m_(A)g =(0.2) xx ((m)/(2)) (g) = 0.1mg`
Hence minimum force required for movement of the system `(A+B)`
`f_(min) = (f_(1))_(max) + (f_(2))_(max)`
`= 0.15mg + 0.1mg = 0.25mg`
(d) Ginve, force `F =0.1mg lt F_(min)`
Hence the bodies will be aat rest
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