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

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
Coins are placed on top of one another.
Mass of each coin $= 10 g$
Number of coins above the seventh coin from the bottom $= 3$
Therefore,
Total mass over the seventh coin $= 10 \times 3 = 30 g$
Force on the seventh coin $= 30 \times 10^{- 3} \times 10$
$F = 0.3 N$

Since, there is no other force is acting on the coin except the force exerted by the other three coins over it, the net force will be $0.3 N$ Downward.

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(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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