Question

A horizontal board of negligible mass supports a mass weighing $20\text{N}$ at a distance of $2\text{m}$ from the point $A$. Calculate the weight which must be placed at point $B$, which is $5\text{m}$ from point $A$, in order to keep the board in rotational equilibrium?

• A. $4\text{N}$
• B. $8\text{N}$
• C. $10\text{N}$
• D. $40\text{N}$

Answer 1

The correct option is B $8\text{N}$


Let $W_1=20\text{N},r_1=2\text{m},r_2=5\text{m},W_2=?$
As the system is in rotational equilibrium, thus the moment about point $A$ is zero,
$\therefore W_1\times r_1-W_2\times r_2=0$
or $20\times 2=W_2\times 5\Rightarrow W_2=8\text{N}$