# Center of mass problem?

## Homework Statement

http://img9.imageshack.us/img9/2866/siyx.png [Broken]

## Homework Equations

Where is the COM of the figure? (width = 8, height = 8)

## The Attempt at a Solution

area of triangle: 0.5(8*8) = 32
32 - (pi(0.52) - (2.5*1) - (2.5*0.5) = 27.4646

I don't have an axes to locate the center of mass. How do I solve a problem like this?
Thanks.

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Equation for center of mass?
(equate it with c.o.m. in terms of - c.o.m. of whole triangle, c.o.m. circle, c.o.m. rectangle 1, c.o.m. rectangle 2.)

Can you please elaborate more or give me an example.

What is the formula for center of mass?

gneill
Mentor
Are there any vertical placement measurements for the cutouts? How do we know how high the circle is positioned, or rectangles? How long are the rectangles?

All I know is this formula: Mcmx = (m1x1 + m2x2 ... mnxn)/(m1+m2...mn) where x can be x,y, or z.
It does not help me that much in this problem.

Are there any vertical placement measurements for the cutouts? How do we know how high the circle is positioned, or rectangles? How long are the rectangles?

Yes I listed above.

A of circle = pi * 0.5^2
A of rectangle = 2.5*1
A of rectangle = 2.5*0.5

gneill
Mentor
Yes I listed above.

A of circle = pi * 0.5^2
A of rectangle = 2.5*1
A of rectangle = 2.5*0.5

Okay, but how high are these items from the bottom of the triangle? Their exact placement matters.

And it appears to me from the diagram that the two rectangular cutouts are the same width, so why are you giving one as 2.5*1 and the other 2.5*0.5?

All I know is this formula: Mcmx = (m1x1 + m2x2 ... mnxn)/(m1+m2...mn) where x can be x,y, or z.
It does not help me that much in this problem.

Okay, vector format would have been better but we can work with this. Just need to work out x and y coordinates separately.
1. Can you find out CM of the figure without any holes (geometrically)?
2. Can you find out the individual CM of the shapes that fit into the holes (geometrically)?
3. Can you after all this express the actual CM in terms of these? (Think in terms of the formula.)
Note that you will need all the values specifying their placement....like gneill just said.

Okay, but how high are these items from the bottom of the triangle? Their exact placement matters.

And it appears to me from the diagram that the two rectangular cutouts are the same width, so why are you giving one as 2.5*1 and the other 2.5*0.5?

I don't have the exact placement for the cutouts.
I am positive that the two rectangular cutouts are not the same. I agree with you that they look somewhat same size (the height is the same), but the with is for one is 1 and for the other is 0.5. You can verify that from the diagram provided above.

Okay, vector format would have been better but we can work with this. Just need to work out x and y coordinates separately.
1. Can you find out CM of the figure without any holes (geometrically)?
2. Can you find out the individual CM of the shapes that fit into the holes (geometrically)?
3. Can you after all this express the actual CM in terms of these? (Think in terms of the formula.)
Note that you will need all the values specifying their placement....like gneill just said.

[*]Can you find out CM of the figure without any holes (geometrically)? I think it is (x,y) should be (4,4) right?
[*]Can you find out the individual CM of the shapes that fit into the holes (geometrically)?
rectangle one is (0.5, 1.25)
rectangle tow is (0.25, 1.25)
[*]Can you after all this express the actual CM in terms of these? (Think in terms of the formula.)
I don't know?

mmm...I think 0.5 is the height at which rectangles are.
2.5 the length is given horizontally too, besides the question can't be solved if we don't know height we can't solve for CM...

1. Can you find out CM of the figure without any holes (geometrically)?
I think it is (x,y) should be (4,4) right?
2. Can you find out the individual CM of the shapes that fit into the holes (geometrically)?
3. rectangle one is (0.5, 1.25)
4. rectangle tow is (0.25, 1.25)
5. Can you after all this express the actual CM in terms of these? (Think in terms of the formula.)
6. I don't know?
Where is your origin? It should be same for all figures.
And the first one's wrong...

I took each piece and put it bottom left corner (x,y) (0,0) origin.

Chestermiller
Mentor
You should have listened to Gneill, so I'll say it again. You can't do this problem if you don't know how far above the base of the triangle the centers of the rectangular cutouts and the center of the circular cutout are located. Secondly, the two rectangular cutouts are identical. When the figure says "centered from both sides," it means that the figure is symmetrical about the vertical centerline of the triangle. You misread the figure when you thought that the base of one of those rectangles is 0.5. Look at the figure again. Both triangles have a base of 1.0, as required if the figure is symmetric.

gneill
Mentor
A question for vac: Were you given just the figure and a ruler to work with? In other words, are you expected to take your own measurements directly from the figure?

Chestermiller
Mentor
A question for vac: Were you given just the figure and a ruler to work with? In other words, are you expected to take your own measurements directly from the figure?
It doesn't seem like this could be the case. The figure implies that the altitude and base of the triangle are both 8 units. Yet, the figure as drawn looks like an equilateral triangle. So the figure could not be to scale.

Chet

gneill
Mentor
It doesn't seem like this could be the case. The figure implies that the altitude and base of the triangle are both 8 units. Yet, the figure as drawn looks like an equilateral triangle. So the figure could not be to scale.

Chet

I agree that the diagram is not very accurate. I find that if I cut and paste it into a drawing program and scale it so that the base is as close as I can estimate it to 8 cm, then the height of the triangle is just short of 8 cm: About 0.37 cm short. So about 5% off. Mind you, the circular cutout as drawn is then considerably bigger than a 0.5 cm radius indicated.

I think we need more details about the actual problem statement as it was presented to the OP.