Block B of mass 2 kg rests on block A of mass 10 kg. All surfaces are rough with the value of coefficient of friction as shown in the figure. Find the minimum force F that should ber appplied on blocks A to cause relative motion between A and B.`(g=10(m)/(s^2)`)

Answer

48 N

Explanation

FBD of B B FBD B
`(a_(B))_(max) = (f_(max))/(m_(B)) = mu sg = 2.5 m//s_(2)`
FBD of combined system

`f_(x)=0.15 (2+10)g =18N`
`F_(max)-f_(k)=(m_(A)-m_(B))(a_(B))_(max)`
`rArr F_(max)=f_(k)+12 xx 2.5 - 48 N`.