Question

A uniform metre rule is pivoted at its mid point. A weight of 40gf is suspended at one end of it. Where should a weight of 100gf be suspended so as to maintain tha metre rule perfectly horizontal?​

Answer 1

Given:-

• A weight of 40 gf is suspended at one end of meter rule.

To Find:-

• Where should be 100 gf weight suspended so that it maintain the meter rule perfectly horizontal ?

Solution:-

Here, Let the 40 gf weight produce anticlockwise moment about the middle point of the meter.

As we have one meter rule pivoted at midpoint there will be a perpendicular distance of 50 cm from the contact of force of 40 gf.

Now, the weight of 100 gf produce clockwise moment about the middle point.

Let, its distance of 100 gf from the middle be d cm. Then,

According to principle of moment,

Anticlockwise moment = Clockwise moment

40 gf × 50 = 100 gf × d

$\sf{d = \dfrac{40\times 50}{100}}$

d = 20 cm

Hence, the 100 gf weight should be placed 20 cm apart from the middle of the meter rule.