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Two blocks m(1)=1kg, m(2)=2kg are placed...

Two blocks `m_(1)=1kg`, `m_(2)=2kg` are placed one upon the other. `m_(1)` is kept on `m_(2)`. The force of static friction between `m_(1)` and `m_(2)` is `0.2` and between `m_(2)` and the horizontal surface and the floor is `0.28`. Calculate the maximum force that can be applied on `m_(2)` so that `m_(1)` and `m_(2)` does not get separated.`(g=10ms^(-2))`

Text Solution

Verified by Experts

`(i) R=(m_(2)+m_(1))g`
`f_(2)=mu_(2)(m_(2)+m_(1))g=0.28(2+1)9.8=8.232N`
`f_(1)=mu_(1)m_(1)g=0.2xx1xx9.8=1.96N`
`F=(m_(1)+m_(2))a+mu_(2)(m_(1)+m_(2))g`
`F=(1+2)a+8.232`
but `f_(1)=m_(1)a`
`:.a=1.96//1=1.96Nkg^(-1)`
`:.F=3xx1.96+8.232=14.112N`
`F=m_(1)a=1.96N`
If `f gt 1.96N`, then `m_(1)` slip over `m_(2)`
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