Question

a uniform metre rule is pivoted at its midpoint.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?

Answer 1

Hi.

Here is your answer----

   Diagram of your question is----
            Units are in cm                                      
            -----------------------------------50cm---------------------------------------
                                                weight is 50 gf                       
According to the principle of moments----
     Sum of clockwise moments = Sum of Anticlockwise moments
                                   50 x 50 = d x 100
                                        d = 5 x 5
                                        d = 25 cm
Thus, the weight must be suspended on the 25 cm mark to keep the meter rule in equilibrium.

Hope it will help u.

Have a nice day.

Answer 2

Explanation:

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.

hop it helps