Question

Calculate the distance from $O$ where a $100gf$ weight should be placed to balance the metre rule.

• A. $1cm$ on the left side of $O$.
• B. $1cm$ on the right side of $O$.
• C. $10cm$ on the left side of $O$.
• D. $10cm$ on the right side of $O$.

Answer 1

Answer: (D)

$10cm$ on the right side of $O$.
The turning effect of a force is known as the moment. It is the product of the force multiplied by the perpendicular distance from the line of action of the force to the pivot or point where the object will turn. That is, Moment = Force * distance.
If the moment of a force turns or rotates the body in anti-clockwise direction, then it is called anti-clockwise moment.
If the moment of a force turns or rotates the body in clockwise direction, then it is called clockwise moment.
In this case, the weight of 150 gf at the distance of 40 cm from the center point O, rotates the ruler in the anticlockwise direction and the weight of 250 gf at the distance of 20 cm from the center point O, rotates the ruler in the clockwise direction.
Therefore,
$Anticlockwisemoment=150gf\times 40cm=6000gf$
and the
$Clockwisemoment=250gf\times 20cm=5000gf$.
Here, the clockwise moment is lesser than the anticlockwise moment by 1000 gf.
The principle of moments states that when in equilibrium the total sum of the anti clockwise moment is equal to the total sum of the clockwise moment.
When a system is stable or balance it is said to be in equilibrium as all the forces acting on the system cancel each other out.
That is, In equilibrium, Total Anticlockwise Moment = Total Clockwise Moment. So, a 100gf weight should be placed to the right side of the point O to balance the metre rule.
Therefore, the expression is given as $1000gf=100gf\times x$ where x is the distance o the weight from the center O.
$x=\frac{1000gf}{100gf}=10cm$
Hence, the distance from where a 100gf weight should be placed to balance the metre rule is 10 cm on the right side of .