Question

A uniform metre rule 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
rule horizontal ?
how did length be 50​

Answer 1

Answer:

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

Here we have moment of force of r×F.

After substituting we get moment of force=50×50gfcm.

Here this must be balanced by 100gf's moment of force.

Let us think that there would be x perpendicular distance.

On equating we get

50 × 50gfcm =100 × xgfcm

x=50 × 50/100 = 25cm.

Hence 100gf must be applied at a distance of 25cm.