2 simple Physics q's driving me nuts

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SUMMARY

The discussion addresses two physics problems related to rotational equilibrium. The first problem involves determining the pivot point on a seesaw with a 400 N child and a 300 N child, requiring the application of torque calculations. The second problem focuses on a uniform meterstick supported at the 30.0 cm mark with a 0.50 N weight at the 0 cm mark, where the weight of the meterstick must be calculated using torque principles. The key formula for both problems is the equilibrium condition where the sum of torques equals zero.

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kangta
This looks really simple , but my teacher didn;t explain how to do them so i have no clue how to approach them cna someone kindly help?


1. A 400 N child and a 300 N child sit on either end of a 2.00 m long seesaw. where along the seesaw should the pivot support be placed to ensure rotational equilibrium?

2. A uniform meterstick, supported at the 30.0 cm mark, is balanced when a .50 N weight is hung at the 0 cm mark. What is the weight of the meterstick?
 
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For static equilibrium, the sum of the Torques must equal 0.

Torque is a vector, and is equal to distance X force
 
so how would i solve these problems?
 
Like I said: sum the torques. set equal to zero. torque is force times distance.

400*x - 300*(2-x) = 0

There will be a better analysis in your book, I'm sure. Check for the chapter on rotational motion.
 
Last edited:
as for the second part of your question:

the torques in both direction have to be the same again
torque one arises from .5 N being placed at .3 m from the axis of rotation. the second torque is going to originate from the mass of the meter stick acting at the center of mass of the meter stick( in this case .5 m because the stick has a uniform density)
equate the torques and you have an equation with one unknown. solve for it.
 

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