Calculating the Width of a Triangle at Position x

[chris_avfc]

Homework Statement

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.

Homework Equations

Have to show the width at position x is given by

(a/L)x

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?

 

[tiny-tim]

hi chris!

hint: similar triangles?

 

[chris_avfc]

hi chris!

hint: similar triangles?

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

 

[tiny-tim]

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

 

[chris_avfc]

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

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.

 

[chris_avfc]

1. Homework Statement
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. Homework 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.

 

[tiny-tim]

hi chris!

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

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

 

[chris_avfc]

hi chris!


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

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.

 

[tiny-tim]

M = ρt ∫w Δx, with the same limits.

Right?

that's right

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

 

[chris_avfc]

that's right

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

Which would give

(pt a/L) ∫ x Δx ?

 

[tiny-tim]

yup!

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

 

[chris_avfc]

yup!

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

yeah I presumed so and ended up with

1/2 * ptaL ?

 

[tiny-tim]

that's right

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

 

[chris_avfc]

that's right

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

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

Leading to final answer

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?

 

[tiny-tim]

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?

yes of course …

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

 

[chris_avfc]

yes of course …

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

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?

 

[tiny-tim]

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

what happened to your …

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

?

 

[chris_avfc]

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

what happened to your …


?

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)

 

[tiny-tim]

i don't get it!

you had …

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

and, for the mass, M= …

1/2 * ptaL

now combine them

 

[chris_avfc]

i don't get it!

you had …


and, for the mass, M= …


now combine them

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

 

[tiny-tim]

Now I get 2L/3

at last!

 

[chris_avfc]

at last!

haha, turns out the whole of this question was pretty simple really, but I do have one more question.

It asks you to verify the centre of mass corresponds with the answer you would get with an equilateral triangle when you use symmetry.

So I know the centre of mass is where the lines of symmetry cross.

But how do I verify this on paper?

 

[tiny-tim]

draw it: you should see a 30°,60°,90° triangle

 

[chris_avfc]

draw it: you should see a 30°,60°,90° triangle

And then use trig?

 

[tiny-tim]

get on with it!

 

[chris_avfc]

get on with it!

You are actually going to hate me but I don't get it.

We're talking about the small triangles made my lines of symmetry right, so there are six of them?

 

[tiny-tim]

hi chris!

(just got up :zzz:)

yes, six identical triangles …

and you know the angles, so you can find the ratios of the lengths