Not a mastercam question, but....

[cadcam]

(and why would you hire a contractor that didnt have a tape measure???)

is this like why did the chicken cross the road?

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[altimu]

I agree that it is a trick question and here is why:

The Hyp. of each triangle needs to be greater than 6ft so the remainder of the 10ft ladder is less the 4ft then apply the principal of similar triangles that says the Hyp. of the 12ft ladder would need to be less than 4ft which we already stated needs to be greater than 6ft. Solution - not possible.

Was fun trying!

[Guest]

It would seem the solution would require a set of simultaneous equations and then we would need to solve for the variables - Hope this is some insight... The ladders are not required to intersect at their midpoints. Dont forget the solution where the ladders intersect at their endpoints.

[Allan Brown]

In the limit as the distance x approaches 0 and the ladders approach vertical, the intersection height h approaches roughly 5.454545... Would have been much more fun if there were a solution.

[Simtech]

Why fumble with numeric values then:

1st ladder: L1

2nd ladder: L2

Intersecting height: h

Distance between walls: x

Now, that's better!

Good luck to you all!

(I'll supply the answer if a few days!!!)

 

[Michael Reynolds]

If you were to break this problem down into two opposing right triangles, where the the intersection of the 12' ladder and the 6' height met, the opposite side of that triangle would be "W", then where the 10' ladder met the 6' height, the opposite side would be "Z", the equation would be the following:

Z+W=(12W)/(SQUARE ROOT OF 36+Wsquared)

Does THAT help.....

Mike R.

 

[yenkls]

The question is tricky since it is impossible. This situation can never exist. Here is why:

Assume that both ladders are 12 feet each, so they intersect exactly in the middle of X. Also each diagonal segment from the wall to the intersection point is 12/2 =6 feet. Since this 6 feet is the Hyp ot the triangle, the height will always be less than 6. But Simtech told us that the height is 6, so it is impossible.

Now if the second ladder is only 10 feet, than we know that the length from the bottom to the intersection is even less than 6, so the height has to be even less than that.

The question is either irrational or Simtech is illegal;-)

 

[Guest]

h = intersection point

H1 = height of first ladder

H2 = height of second ladder

L1 = length of first ladder

L2 = length of second ladder

x = distance between walls

can make several simultaneous equations from this :

(1) 1/(H1) + 1/(H2) = 1/(h)

(2) (H1)^2 + (x)^2 = (L1)^2

(3) (H2)^2 + (x)^2 = (L2)^2

subtract (2) from (3) gives

(4) (H1)^2 - (H2)^2 = (L1)^2 - (L2)^2

(1) manipulates into :

(5) (H1) = ((h)(H2)) / ((h)-(H2))

replace (H1) in equation (4) with equation (5) and there is only one variable left which can be solved. its turns into an ugly fourth power equation which is even worse because i didnt use any numbers but if you do the grunt work the problem can be solved.

btw: how do you import a drawing into the forum, for the drawing above

[This message has been edited by mbowling (edited 06-13-2001).]

[gstephens]

Where did you get Eq 1 from (Am I forgetting some trig identity?)? I tried splitting x into a and b and came up with:

a/h = x/H1 and b/h = x/H2 (Similar triangles), but ended up with 1 too many variables.

 

[Guest]

your right, eq (1) came from the similar triangles. split "x" at intersect point into "a" & "b". then (bh)/(x) = (h)/(H1) and (ah)/(x) = (h)/(H2). add the two equations and with alittle manipulation you get eq (1).

[Guest]

The distance x=d1+d2

Find the distance of d1 and d2 using trig and the relationship between the similar triangles. Simtech is laughing at the rest of us... The education system is better in Quebec than the rest of the plannet I guess!

Andrew

[Simtech]

Just came back to this post after a few years of absence ... here goes the solution.

 

a = height of first ladder (from left corner)

b = height of second ladder (from right corner)

d = distance between walls

x = horizontal point of intersecting ladders from left corner

y = vertical distance of intersecting ladders

 

Eq1: f(x)= (a/d)x #equation of first ladder

Eq2: g(x)= (-b/d)x + b #equation of second ladder

 

when does Eq1 = Eq2, so f(x)=g(x)

 

(a/d)x = (-b/d)x + b

(a/d)x + (b/d)x = b

x((a+B)/d) = b

x = bd/(a+

 

replacing x in Eq2 we get

y = ab/(a+

 

Example:

a = 38

b = 31

d = 45

Intersecting point will be:

x = 20.217

y = 17.072

 

My son solved the problem, he's in 12th grade highschool!!!

[Matthew Hajicek - Singularity]

Why not just draw it up in Solidworks?

[Philcott]

It's been a while since Chris B. left In-house and emastercam.

 

Hope you're doing well Chris.

[Bob W.]

quote:

---

He mesured the intersection height "h" of the two ladders from ground level to be 6 ft.

---

quote:

---

Intersecting point will be:

x = 20.217

y = 17.072

---

6' isn't exactly 17.012". It is impossible to have the ladders intersect at 6' with two vertical walls.

[Justin Beebe at Folsom Tool]

I happen to be enrolled in Algebra and Trigonometry this semester. 8(20+x+y) is an example of the "Distributive Property" that states a(b+c)= ab+ac. In our book back in chapter 1 we had similar problems that were presented as follows:

 

Use the Distributive Property to remove the parentheses:

 

8(20+x+y) = 160+8x+8y

 

or

 

(x+2)(x+4) = x²+6x+8

 

Without seeing how the problem was presented it's impossible to know for sure but it was most likely used to teach students how to use the Distributive Property which is very useful for solving all kinds of mathmatical equations.

[JDowe]

From geometry opposite inerior angles are equal

when striaght lines intersect - so the tringles

formed are proportional - so let a be the long

unknown and b be the short unknown

5a - 6b = 0

a + b = 6

 

5a - 6b = 0

6a + 6b = 36 Mulitply by 6 and add

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11a = 36 a = 3.2727 b= 2.7272