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 $g=10{\text{m/s}}^2$

• A. $0.3\text{N}$ downwards
• B. $0.3\text{N}$ upwards
• C. $0.7\text{N}$ downwards
• D. $0.7\text{N}$ upwards

Answer 1

The correct option is A $0.3\text{N}$ downwards

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