A uniform ruler is pivoted at its mid point. A weight of $50 g f$ is suspended at one end of it. Where should a weight of $100 g f$ be suspended, to keep the ruler horizontal?

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Solution

Let AB be the uniform metre rule, such that C is its midpoint. The metre rule is pivoted at C, such that $A C = 50 c m$ and $C B = 50 c m$ . 50 gf weight is suspended near A.
Let D be the point of suspension of 100 gf weight at a distance of x from C, so that the rule is horizontal.
Left hand side moment,
$L H M = 50 \times 50 = 2500 g f c m . . . . . (1)$
Right hand side moment,
$R H M = x \times 100 = 100 \times g f c m . . . . . (2)$
To keep the rule horizontal,
$L H M = R H M$
$2500 = 100 x \Rightarrow x = 25 c m$ .
$∴$ 100 gf is to be suspended at a distance of 25 cm from mid point (or) 75 cm from the end of the rule.

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A uniform ruler is pivoted at its mid point. A weight of 50 gf is suspended at one end of it. Where should a weight of 100 gf be suspended, to keep the ruler horizontal?

A uniform ruler is pivoted at its mid point. A weight of $50 g f$ is suspended at one end of it. Where should a weight of $100 g f$ be suspended, to keep the ruler horizontal?