Problem with the weigth of the moment arm

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SUMMARY

The discussion centers on calculating the total mass required for equilibrium in a system involving a rod, counterweight, and hanging mass. The rod has a mass of 400g, a counterweight of 1600g, and a length of 0.8m, with equilibrium achieved when the handle is positioned at 12.0cm. The calculations involve determining the moment arm and integrating the mass distribution of the rod to find the total mass of the hanging weight (m) necessary for balance. The final calculations yield a total moment of 0.112kg at the specified distance from the pivot.

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SwedishFred
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Hi
A rod comprises a counterweight (a) a handle and a hook which hangs mass (m), the rods mass is 400g counterweight mass (a) = 1600g and the rod is 0.8m long. Equilibrium exists when the handle is at position X = 12.0cm

How I did

1: calculated the length of the handle to (m) which was 0.68m
2: figured kg from 1.6 kg * 0.12 = 0.192 kg
3 0.192 = 0.68*m
but how do you do with the rod weighs 400g how do I calculate it??
The question is to calculate the total mass of (m) to obtain equilibrium
hope you understand the language and metrics, used google translate, did work out that great, but I´m loosing time

regards fredrik
 
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Consider a uniform horizontal rod length L pivoted at one end. Using calculus:
An element length dx at distance x from the pivot has mass ρdx. What is its moment about the pivot? Can you integrate that to get the total moment?
Or, without calculus, where is the mass centre of the rod?
 
Oki now I see, its in its center then You just calculated it to the handle..
0.8/2-0.12=0.28
0.4kg*0.28=0.112kg

Thanx for your help! Sent from my iPhone using Physics Forums
 

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