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Normal Force

A wooden box ...

Question

A wooden box is placed on a table. The normal force on the box from the table is $N_{1}$ . Now another identical box is kept on first box and the normal force on lower block due to upper block is $N_{2}$ and normal force on lower block by the table is $N_{3}$ . For this situation, mark out the correct statement(s).

A

N_{1} = N_{2} = N_{3}

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B

N_{1} < N_{2} = N_{3}

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C

N_{1} = N_{2} < N_{3}

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D

N_{1} = N_{2} > N_{3}

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Solution

The correct option is C $N_{1} = N_{2} < N_{3}$
R.E.F image
The FBD for the box of mass m when put on the table is,
The normal force $N_{1}$ is equal to the weight of the box.
$N_{1} = m g$
The FBD of the first box when an identical box of mass m is placed above it is,
The normal force $N_{2}$ is equal to the weight of the second box.
$N_{2} = m g$
The normal force $N_{3}$ is equal to the combined weight of the boxes.
$N_{3} = 2 m g$
Clearly, $N_{1} = N_{2} < N_{3}$
Hence, option C is correct.

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