2 simple Physics q's driving me nuts

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Homework Help Overview

The discussion revolves around two physics problems related to static equilibrium and torque. The first problem involves determining the pivot point on a seesaw with two children of different weights, while the second problem concerns finding the weight of a uniform meterstick balanced with an additional weight.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the concept of torque and its role in achieving static equilibrium. There are attempts to set up equations based on the torques produced by the weights involved in both problems.

Discussion Status

Some participants have provided guidance on how to approach the problems by suggesting the use of torque equations. There is an ongoing exploration of the necessary conditions for equilibrium, but no consensus has been reached on the final solutions.

Contextual Notes

The original poster expresses confusion due to a lack of explanation from their teacher, indicating a potential gap in understanding the foundational concepts of torque and equilibrium.

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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