Ten coins are placed on top of each other on a horizontal table. If the mass of each coin is $10 gm$ , what is the magnitude and direction of the force on the 7th coin (counted from the bottom) due to all the coins above it? Take $g = 10 m / s^{2}$

0.3 N downwards

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0.3 N upwards

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0.7 N downwards

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0.7 N upwards

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Solution

The correct option is A 0.3 N downwards
The normal reaction that the 7th coin applies on 8th will balance the weight of 3 coins above it. Therefore, force from $3$ coins above it will be, $3 mg = 3 (10 \times 10^{- 3}) (10) = 0.3 N$ (in downward direction)

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Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of

(a) the force on the 7^th coin (counted from the bottom) due to all the coins on its top,

(b) the force on the 7^th coin by the eighth coin,

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