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

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
Force on the seventh coin = weight of the three coins above it.
The mass of the above three coins (counted from the bottom) will be $30 g = 0.03 k g$ . Now the magnitude and direction of the force on the 7th coin (counted from the bottom) due to all the coins (three coins) above it $= 0.03 \times 10 N = 0.3 N$
Direction of force will be downward.

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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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