Question

Ten coins are placed on top of each other on a horizontal table. If the mass of each coin is $10\text{gm}$, 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 $g=10m/s^2$

• A. 0.3 N downwards
• B. 0.3 N upwards
• C. 0.7 N downwards
• D. 0.7 N upwards

Answer 1

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, $3\text{mg}=3(10\times {10}^{-3})\left(10\right)=0.3\text{N}$ (in downward direction)