Calculating Tunnel Width at Ground Level

[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.

Please help me solve this.


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

 

[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)

 

[Hypercase]

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

-Cheers.

 

[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.

 

[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).

 

[K.J.Healey]

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).

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