Ten one-rupee coins are placed 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 $7^{\text{th}}$ coin (counted from the bottom) due to all the coins on its top.
$(b)$ The force on the $7^{\text{th}}$ coin by the $8^{\text{th}}$ coin.
$(c)$ The reaction of the $6^{\text{th}}$ coin on the $7^{\text{th}}$ coin.

(N/A) The force on the $7^{\text{th}}$ coin is exerted by the weight of the three coins $(8^{\text{th}}, 9^{\text{th}}, 10^{\text{th}})$ on its top.
Weight of one coin $= mg$.
Weight of three coins $= 3mg$.
Hence,the force exerted on the $7^{\text{th}}$ coin by the three coins on its top is $3mg$. This force acts vertically downward.
$(b)$ The force on the $7^{\text{th}}$ coin by the $8^{\text{th}}$ coin is equal to the weight of all the coins above the $7^{\text{th}}$ coin,which are the $8^{\text{th}}, 9^{\text{th}},$ and $10^{\text{th}}$ coins.
Total weight $= mg + mg + mg = 3mg$.
Hence,the force exerted by the $8^{\text{th}}$ coin on the $7^{\text{th}}$ coin is $3mg$ acting vertically downward.
$(c)$ The $6^{\text{th}}$ coin supports the weight of all coins above it,which are the $7^{\text{th}}, 8^{\text{th}}, 9^{\text{th}},$ and $10^{\text{th}}$ coins.
Total weight supported by the $6^{\text{th}}$ coin $= 4mg$.
According to Newton's third law of motion,the $6^{\text{th}}$ coin exerts an equal and opposite reaction force on the $7^{\text{th}}$ coin.
Therefore,the reaction force of the $6^{\text{th}}$ coin on the $7^{\text{th}}$ coin is $4mg$ acting vertically upward.