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

**(a)** the force on the 7^{th} coin (counted from the bottom) due to all coins above it.

**(b)** the force on the 7^{th} coin by the eighth coin and

**(c)** the reaction of the sixth coin on the seventh coin.

Asked by Abhisek | 1 year ago | 76

**(a)** The force on the 7^{th} coin is due to the weight of the three coins kept above it. Weight of one coin is mg. So the weight of three coins is 3mg

Force exerted on the 7^{th} coin is (3mg)N and the force acts vertically downwards.

**(b) **The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence the force on the 7^{th} coin due to the 8th coin will be the same as the force on the 7^{th} coin by the three coins above it.

Therefore, the force on the 7^{th} coin by the 8th coin is (3mg)N and the force act vertically downwards.

**(c)** The sixth coin is under the weight of four coins above it and experiences a downward force due to the four coins.

The total downward force on the 6^{th} coin is 4mg

Applying Newton’s third law of motion, the 6th coin will exert a reaction force upwards. Therefore, the force exerted by the 6^{th} coin on the 7^{th} coin is equal to 4mg and acts in the upward direction.

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