Physics moments question

A uniform metre ruler, AB, freely pivoted at its centre of mass.

(c) A 1.0 N weight is placed on the ruler 0.30 m from the middle of the ruler towards A.


(ii) Calculate the distance the pivot needs to be moved to restore equilibrium when the weight of the ruler is 0.50 N.

Apparently the answer is 0.2M but how did they get to that? Can you explain where you get the figures you're using from

The two downward forces involved are 1.0N and 0.5 N and the these are at a distance of 0.3 m from each other
let the pivot be moved by a distance x towards A reducing the distance of the pivot and the 1.0N force to 0.3-x
Now apply the principe of moments
1.0 x (3.0 -x)=0.5 X ( x )
Simplifying 3.0 =1.0x +0.5x
That gives x = 0.2 m ANS Ok ?

You need to balance the rod equating the moments on each side.
Anticlockwise moments = Clockwise moments

Now the weight of 1N is 0.3m from the center towards A, so we need to move the pivot towards A so the rod's weight acts in the other direction. The new distance of the pivot can be called d. The distance of the weight from the pivot is now (0.5-d) m. I hope you can visualise it or draw it to understand it more clearly.
Now applying the principle of moments:
acw moments = cw moments
(1)(0.3-d) = (0.5)(d)
0.3-d = 0.5d
0.3 = 1.5d
d = 0.3/1.5
d=0.2m

There. I hope it helped. Best answer?