Finding the Center of Mass of a Triangle

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To find the center of mass of triangle ABC formed by a uniform wire, the x-coordinate can be calculated using the formula for center of mass, yielding a result of 12λ times the length of rod BC divided by 2. However, the y-coordinate remains unresolved, requiring further mathematical analysis. A visual representation of the triangle can aid in identifying the center of mass for each side. The symmetry of the triangle simplifies the calculation of one coordinate, but the other necessitates additional calculations. Completing the solution involves showing all working steps as per forum guidelines.
Acemave
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Homework Statement


a thin uniform wire is bent to form two equal sides AB and AC of triangle ABC , where AB=AC=5cm . the third side BC , of length 6cm is made from uniform wire of twice the density of the first . the distance of centre of mass from A is ?

Homework Equations


x centre of mass = m1R1+m2r2/ sum of masses(where r= centre of mass of particle in sys.)λ=m/l ( where m= mass of object , l=length ) uniform density formula

The Attempt at a Solution


i was able to calculate the X c.o.m which is = 12λ*lenght of rod BC/2
but was unable to calculate its y coordinate ?
 
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Acemave said:

Homework Statement


a thin uniform wire is bent to form two equal sides AB and AC of triangle ABC , where AB=AC=5cm . the third side BC , of length 6cm is made from uniform wire of twice the density of the first . the distance of centre of mass from A is ?

Homework Equations


x centre of mass = m1R1+m2r2/ sum of masses(where r= centre of mass of particle in sys.)λ=m/l ( where m= mass of object , l=length ) uniform density formula

The Attempt at a Solution


i was able to calculate the X c.o.m which is = 12λ*lenght of rod BC/2
but was unable to calculate its y coordinate ?
I don't think that's the solution this problem is looking for.

Once the wire is formed into the triangle, you are supposed to calculate the (x,y) coordinates of the center of mass from point A.
 
Make a drawing. Mark on it the centre of mass of each rod. One of the co-ordinates (say x) is trivial by symmetry. The other (say distance y from point A) needs some maths.

Acemave said:
but was unable to calculate its y coordinate ?

Show your working (forum rules).
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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