# Circular arc

1. Aug 21, 2004

### Hypercase

The cross-section of a tunnel is a circular arc. The maximum height of the tunnel is 10 (units). A vertical strut 9 (units) high supports the roof of the tunnel from a point 27 (units) along the ground from the side. Calculate the width of the tunnel at ground level.

P.S:-I'd post this in HW help, but this isn't my home work, its just a sum thats bugging me.

2. Aug 22, 2004

### Galileo

I'm not sure I picture it correctly. Is it something like the picture I attached?

(My Paint skills are limited, so please bear with me)

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3. Aug 23, 2004

### Hypercase

hey sorry i tried attaching a diagram, it probably didnt load.
I'll draw another one in a moment.

-Cheers.

4. Aug 23, 2004

### Hypercase

Sorry, but i'm unable to load the pic.
What software did you use to draw that pic? I tried using paint but he file size always turns out greater than the limit.
Your pic is correct except that the base should be a minor chord and the 27 m is measured from the side closest to the 9 m high pillar.

5. Aug 23, 2004

### HallsofIvy

Set up a coordinate system so that the circle's center is at the origin. The equation of the circle is x2+ y2= R2. Take the "ground level" to be y= u (unknown). Since the maximum height of the tunnel is 10, we have u= 10- R.
One side of the tunnel is where x2+ u2= R2 or
x2+ (10-R)2= R2 or x2+ 100 - 20R= 0.
x= √(20R- 100) and the width of the tunnel is 2x= 2√(20R-100).

We are told that at 27 units from the edge (i.e. x= √(20R-100)- 27), the height is 9 units (i.e. y= 9+u= 19-R). That gives (√(20R-100)-27)2- (19-R)2= R2. Solve that equation for R and then find 2√(20R-100).

6. Aug 24, 2004

### K.J.Healey

Good answer, and you can SO tell that you've done a lot of classical mechanics... (unless im wrong)