2 simple Physics q's driving me nuts

In summary, the conversation discusses two problems related to rotational equilibrium involving torques and the placement of a pivot support. The first problem involves two children of different weights sitting on a seesaw and determining the placement of the pivot support for equilibrium. The second problem involves a meterstick being balanced with a weight hanging from one end and finding the weight of the meterstick. The solution to both problems involves setting the sum of the torques equal to zero and solving for the unknown variable. The conversation also suggests checking a textbook for more in-depth analysis on rotational motion.
  • #1
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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  • #2
For static equilibrium, the sum of the Torques must equal 0.

Torque is a vector, and is equal to distance X force
 
  • #3
so how would i solve these problems?
 
  • #4
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:
  • #5
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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