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A lift is moving upwards with an acceler...

A lift is moving upwards with an acceleration of `2m//sec^(2)`. Inside the lift a `4kg` block is kept on the floor. On the top of it , `3kg` block is placed and again a `2kg` block is kept on the `3kg` . Calculator `:`

`(i)` contact force between `2kg` block and the `3kg` block.
`(ii)` contact force between `4kg` block and floor of the lift.
Draw the free body diagrams properly & take `g=10m//sec^(2)`

Text Solution

Verified by Experts

The correct Answer is:
`(i) 24N (ii) 108N`
`(i) FBD` of `2kg`

`N_(23)-20=2(2)`
`N_(23)=24N`
`(ii) FBD` of `3kg`
`N_(34)-N_(23)-30=3(2)`

`N_(34)=N_(23)+30+6`
`N_(34)=24+30+6=60N`
`FBD` of `4kg`
`N_(G)-N_(34)-40=4(2)`

`N_(G)=N_(34)+40+8`
`N_(G)=60+40+8=108N`
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