Mechanics - centre of mass lamina

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SUMMARY

The discussion focuses on calculating the center of mass of a uniform lamina with a mass per unit area of 0.12 grams/cm², shaped as a square with a side length of 80 cm, from which a triangular section has been removed. The coordinates of the square's corners are defined, and a particle of mass 0.5 kg is attached at corner C(80, 80). The solution involves applying the principle of superposition to find the center of mass by treating the removed triangle as having negative mass. This method effectively combines the mass distributions of the square, triangle, and attached particle.

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kolomsg
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hi, I'm stuck and I can't seem to find a simple solution to questions regarding how to find the centre of mass of a lamina.

for example:
A uniform lamina with mass/unit area 0.12grams/cm2 consists of a square of
side 80cm, with one corner at the origin O(0, 0) and other corners at B(80, 0),
C(80, 80) and D(0, 80). A triangle of material has been removed from one corner
by cutting a straight line from E(50, 0) to F(80, 40), and a particle of mass 0.5kg
is attached at C. Find the coordinates of the centre of mass.
If anyone could help me, that'd be much appreciated. thanks
 
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Welcome to PF!

Hi kolomsg! Welcome to PF! :smile:

Do you know how to find the coordinates of the centre of mass of several bodies, given that you know the coordinates of the centre of mass of each?

Apply that to the square, the triangle, and the particle, but with the triangle having negative mass! :wink:
 
hi, thanks a lot. it helped!
 

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