Question

A lift is moving upwards with an acceleration of 2 m/sec2. Inside the lift a 4 kg block is kept on the floor. On the top of it, 3 kg block is placed and again a 2 kg block is kept on the 3 kg block. Find the contact force between 4 kg block and floor of the liſt. Take g = 10 m/sec?
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Answer 1

Answer:

Explanation:

Answer

(i)24N(ii)108N

Solution

(i)FBD of 2kg

N23−20=2(2)

N23=24N

(ii)FBD of 3kg

N34−N23−30=3(2)

N34=N23+30+6

N34=24+30+6=60N

FBD of 4kg

NG−N34−40=4(2)

NG=N34+40+8

NG=60+40+8=108N

Answer 2

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