# Calculating the Width of a Triangle at Position x

1. Feb 20, 2012

### chris_avfc

1. The problem statement, all variables and given/known data
Given an isosceles triangle
- Length = L
- Uniform Density = ρ
- Width Varies from 0 at x = 0 to a at x = L

I attached a picture of it.

2. Relevant equations

Have to show the width at position x is given by

(a/L)x

3. The attempt at a solution

Now it is only two marks, but I don't have a clue where to even begin.
Could anybody help?

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2. Feb 20, 2012

### tiny-tim

hi chris!

hint: similar triangles?

3. Feb 20, 2012

### chris_avfc

Cheers for the answer Tim, but I still don't get it :(

4. Feb 20, 2012

### tiny-tim

have you done "ordinary" geometry, with similar triangles, congruence, etc?

5. Feb 20, 2012

### chris_avfc

In the past, yeah.

edit: wait! I get it.
God it is so simple.
I couldn't get the idea of integrals out of my head, I guess because they are used later in the question!

Cheers mate,
I'll probably be back asking for help in the next question haha.

Last edited: Feb 20, 2012
6. Feb 20, 2012

### chris_avfc

1. The problem statement, all variables and given/known data
Following on from this it asks you to find the mass ΔM contained in a strip of width w and length Δx.
In terms of x, δx, p, t, a and L

2. Relevant equations

Mass = Density x Volume

3. The attempt at a solution

My initial reaction was to go along the lines of

(ΔM/M) = (Area of strip/ Total Area)

And then substitute in for mass as ρV, then substitute in for V, but then I ended up with

Δm = Δx, which is clearly wrong.

Is that along the right lines at all?

Edit - This time I got Δm=wptΔx, which seems more right that before, but I'm still not sure.

Last edited: Feb 20, 2012
7. Feb 20, 2012

### tiny-tim

hi chris!
yes, the volume is w*t*∆x, so the mass is that times ρ

8. Feb 20, 2012

### chris_avfc

Glad I'm finally getting somewhere then.

So if I wanted total mass I'd have to then integrate that, which would be

M = ∫ ρ w t Δx, with the limits of L and 0.

Which would be

M = ρt ∫w Δx, with the same limits.

Right?

Thank you so much for the help by the way.

9. Feb 20, 2012

### tiny-tim

that's right

(and of course you write w as a function of x, from the first part)

10. Feb 20, 2012

### chris_avfc

Which would give

(pt a/L) ∫ x Δx ?

11. Feb 20, 2012

### tiny-tim

yup!

(oh, except that with an ∫ , we write dx not ∆x )

12. Feb 20, 2012

### chris_avfc

yeah I presumed so and ended up with

1/2 * ptaL ?

13. Feb 20, 2012

### tiny-tim

that's right

(and that agrees with area of a triangle = 1/2 aL)

14. Feb 20, 2012

### chris_avfc

Awesome, thanks so much.

Mind if I check one more answer with you?

I've just wanted to know if the centre of mass I found is correct.

Centre of mass is given by
C.O.M = ∫(x*dM)/M
= ∫(x wpt Δx)/M
= (pt/m) ∫ ( x w dx )

Substitute in for w, leading to
= (pta/mL) ∫ x^2 dx with limits L and 0

C.O.M = (1/3M) * L^2 pta

Would I have to substitute in for the total mass as well, or leave it as it is?

15. Feb 20, 2012

### tiny-tim

yes of course …

the result should just be a number times L, shouldn't it?

16. Feb 20, 2012

### chris_avfc

ah of course, which means

(1/3).* ( pta L^2 / wpt Δx)
(1/3).* ( a L^2 / w Δx )

Substitute in for w = (a/L)x

(1/3).* ( L^3 / x Δx)

How do I get rid of the x's?

17. Feb 20, 2012

### tiny-tim

where did all those x and ∆x come from??

what happened to your …
?

18. Feb 20, 2012

### chris_avfc

I substituted in
M = wρtΔx
So

C.O.M = (1/3)*( ρta L^2/ wρt Δx )

The ρt cancels, leaving

C.O.M = (1/3)*( a L^2/ w Δx )

I then substituted in
w = (a/l) x

So
C.O.M = (1/3)*( a L^2/ (ax/l) Δx)

The L on the bottom goes to the top as it is 1/L and the a cancels
Leaving

C.O.M = (1/3)*( L^2/ (x Δx)

19. Feb 20, 2012

### tiny-tim

i don't get it!

and, for the mass, M= …
now combine them

20. Feb 20, 2012

### chris_avfc

Argh no, I'm such an idiot, I was putting in the mass of a strip!

Now its really simple!

Now I get 2L/3