Finding Center of Mass of 800g Steel Plate Triangle

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Homework Help Overview

The problem involves finding the center of mass of an 800 g steel plate shaped like an isosceles triangle. Participants are tasked with determining the x and y coordinates of the center of mass based on the provided dimensions and properties of the triangle.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of integrals to find the center of mass, with some expressing confusion about the variables and equations involved. Questions arise regarding the meaning of certain terms and the appropriateness of using integrals for this specific shape.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the necessity of integration for determining the centroid of a simple geometric shape. Some guidance has been offered regarding the relationships between the variables, but no consensus has been reached.

Contextual Notes

There is a mention of a hint provided in the image, which suggests symmetry in the triangle, but the specifics of how to utilize this hint remain unclear among participants. Additionally, confusion exists around the definitions of variables and the setup of the problem.

Anthonyphy2013
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Homework Statement



An 800 g steel plate has the shape of the isosceles triangle shown in the figure. What are the x and y coordinates of the center of mass?
https://www.physicsforums.com/attachment.php?attachmentid=13559&d=1208292919

Homework Equations



x=1/M ∫ x dm

3. The Attempt at a Solution [/b
I have no idea how to start the question
 
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The image gives you a big ole hairy HINT. Did you try following it?
 
SteamKing said:
The image gives you a big ole hairy HINT. Did you try following it?

what hint is that ? the triangle is a smmetry and I put xcm=1/M∫xd(ρ/AH)
and the height is y=2/3 x as =mx+C. and should I put h = 2/3 x ?
 
Anthonyphy2013 said:
xcm=1/M∫xd(ρ/AH)
and the height is y=2/3 x as =mx+C. and should I put h = 2/3 x ?
That's a pretty confusing mix of variable names. What's AH, C, the m in mx+C? Why switch from y to h?
That said, y (=h) = 2x/3 looks right. What do you get for the integral?
 
xcm=1/M∫xd(ρ/AH) , A is the area and H is the height , I need to find h and I use h=mx+c to find the height and the slope is 2/3 . Question is to find the center of mass and only this integration could help me find the answer .
 
Anthonyphy2013 said:
xcm=1/M∫xd(ρ/AH) , A is the area and H is the height
ρ/AH is not going to give you mass. And... the area and height of what, exactly?
You want the mass, dm, of an element of width dx and height y = 2x/3. It's a 2-dimensional lamina, so the density is mass/area. So what is dm equal to?
 
More importantly, you have a plate with a simple shape composed of a material which has a constant density. Do you really need an integral to determine the location of the centroid?
 

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