Help needed for physic question

chifan SG

For the whole of this question, take acceleration of freeall as 10m/s2

A non-uniform metre rule pivoted on a stand at 50cm mark. The metre rule has a mass of 80g and is balanced by a bag of sand and 2 masses hanging on the 2cm mark and 98cm mark respectively. The weight of the bag of sand is 7.6N. The mass of the 2 masses are 0.25KG and 0.5KG.

Where is the C.G of the ruler?

Help appreciated :)

icytoad SG

This question can use priniciple of moments:

Pretend this is the ruler lol

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Sand ^ 2 masses

Weight of 2 masses= mg= (0.25+0.5)x10=0.75x10=7.5 N

Weight of ruler = mg=0.080(10)=0.8 N

Since the 2 masses lighter than sand, it is clear that the CG of the ruler is to the right of the pivot.

Let its distance from pivot be x.

Clockwise moments=anticlockwise moments.

7.5(48) + 0.8x= 7.6(48)

360+0.8x=364.8

0.8x=4.8

x=6

Hence cg at 56 cm mark

chifan SG

This question can use priniciple of moments:

Pretend this is the ruler lol

______________________________________

Sand ^ 2 masses

Weight of 2 masses= mg= (0.25+0.5)x10=0.75x10=7.5 N

Weight of ruler = mg=0.080(10)=0.8 N

Since the 2 masses lighter than sand, it is clear that the CG of the ruler is to the right of the pivot.

Let its distance from pivot be x.

Clockwise moments=anticlockwise moments.

7.5(48) + 0.8x= 7.6(48)

360+0.8x=364.8

0.8x=4.8

x=6

Hence cg at 56 cm mark

Thanks for the help! However, i do not really understand the part where u take the weight of the ruler to be the clockwise moment and mulitple it by x. Please enlighten me :) Thanks.

icytoad SG

Because the 2 masses are lighter than the sand, but yet the ruler balances in equilibrium, so the CG of the ruler must be on the side of the masses.

(this works because just nice the 2 masses and the sand are 48 cm away from the pivot)

This question the tricky part in my opinion is the units. The units are all mixed up in the question.