Finding the Radius (word problem)

[Miike012]

Homework Statement

If a pipe is between two blocks... one block being 1 inch tall and the other block being 2 inches tall. Between the two blocks is an arc that is 6 inches...
What is the radius?


Can some one please help me... I have no Idea what to do..

thank you.

 

[tiny-tim]

Hi Miike012!

(I assume the pipe is touching the ground?)

Hint: draw the radii.

 

[LCKurtz]

Here's how to get started. Label the center of the circle O, the point where the circle hits the short block A, and the point where it hits the taller block B, and the point where the circle hits the ground as G. Let alpha = angle AOG, beta = angle BOG and call the radius is r.

Using some trig functions, you should be able to write three equations in the 3 unknowns alpha, beta, and r. They will have to be solved numerically.

 

[Femme_physics]

This is not my exercise, but I've taken it seriously upon myself to try and solve it.

I've asked my math teacher (BA in math)-- She didn't know how but thought that using the formula of the radius will give the solution. I've asked the smartest guy in my class, who also didn't see how to solve it. I myself am rather lost. You can our sketch of it here (3 people doodled with that sketch...my teacher, the guy, and me)...but we just thought a lot, we didn't actually solve it.

Can the homework helpers please provide more steps? This is very perplexing.

 

[LCKurtz]

Here's how to get started. Label the center of the circle O, the point where the circle hits the short block A, and the point where it hits the taller block B, and the point where the circle hits the ground as G. Let alpha = angle AOG, beta = angle BOG and call the radius is r.

Using some trig functions, you should be able to write three equations in the 3 unknowns alpha, beta, and r. They will have to be solved numerically.

This is not my exercise, but I've taken it seriously upon myself to try and solve it.

I've asked my math teacher (BA in math)-- She didn't know how but thought that using the formula of the radius will give the solution. I've asked the smartest guy in my class, who also didn't see how to solve it. I myself am rather lost. You can our sketch of it here (3 people doodled with that sketch...my teacher, the guy, and me)...but we just thought a lot, we didn't actually solve it.

Can the homework helpers please provide more steps? This is very perplexing.

OK, since this isn't a homework question I will give you the three equations I alluded to above. See if you can see where they come from:

r cos(α) + 1 = r
r cos(β) + 2 = r
r(α +β) = 6

Three equations, three unknowns.

 

[MacLaddy]

Like Femme_physics I am also currently in a Trigonometry class, and this question has definitely piqued my interests.

I have flipped this circle every which way, and created multiple triangles.. I somewhat see what LCKurtz is trying to say, but is there no way to completely solve for the angle?

 

[Miike012]

Here's how to get started. Label the center of the circle O, the point where the circle hits the short block A, and the point where it hits the taller block B, and the point where the circle hits the ground as G. Let alpha = angle AOG, beta = angle BOG and call the radius is r.

Using some trig functions, you should be able to write three equations in the 3 unknowns alpha, beta, and r. They will have to be solved numerically.

I labeled the circle with your description... Is this correct?
I made lines in both green and red... I am thinking it is the one in green, am I correct?

 

[LCKurtz]

Here's how to get started. Label the center of the circle O, the point where the circle hits the short block A, and the point where it hits the taller block B, and the point where the circle hits the ground as G. Let alpha = angle AOG, beta = angle BOG and call the radius is r.

Using some trig functions, you should be able to write three equations in the 3 unknowns alpha, beta, and r. They will have to be solved numerically.

I labeled the circle with your description... Is this correct?
I made lines in both green and red... I am thinking it is the one in green, am I correct?

Your diagram is not as I described. (emphasis added). The circle doesn't touch the blocks at the ground. So it would be the green lines but you need to change the angle labels and the points A and B.

 

[Miike012]

I know that is why I also added the green line that connects @ the top of the blocks.. is that one incorrect to?

 

[LCKurtz]

I know that is why I also added the green line that connects @ the top of the blocks.. is that one incorrect to?

Apparently I edited my post at the same time you replied to it. I hope it is clear now.

 

[Miike012]

I am unfamilliar with the some of your expressions for instance.. r cos(α) + 1 = r
I am not sure why you multiplied Cos(alpha) by r then added one...
I am only familliar with using trig function with right triangles... is this similar?

 

[LCKurtz]

I am unfamilliar with the some of your expressions for instance.. r cos(α) + 1 = r
I am not sure why you multiplied Cos(alpha) by r then added one...
I am only familliar with using trig function with right triangles... is this similar?

Hint: Can you make a right triangle in the figure that would have r cos(α) as one of its legs? Similarly for the second equation.

 

[Femme_physics]

OK, since this isn't a homework question I will give you the three equations I alluded to above. See if you can see where they come from:

r cos(α) + 1 = r
r cos(β) + 2 = r
r(α +β) = 6

Three equations, three unknowns.

Perhaps it's my lack of understanding in solving trigonometric equations that's speaking (I did try learning the subject from the grounds up today but only got so far...)-- but shouldn't all the equations contain all the unknowns? Only the third one does. I will say that this thread/question inspired me to take up some more complex trigo-geo understanding.

 

[Femme_physics]

Question.

If I pass a line from the top of the 2[m] block to the lowest point on the 1 meter block to create a triangle. Would my hypotenuse be equal to 6?

i.e. does the green line equal to the yellow line in my attached pic? If so, it just takes a bit of Pythagoras and divided by 2 to get the radius (2.8).

 

[tiny-tim]

If I pass a line from the top of the 2[m] block to the lowest point on the 1 meter block to create a triangle. Would my hypotenuse be equal to 6?

Nooo.

Hint: draw the radii! ​

 

[Femme_physics]

In my first sketch I posted here I have 3 radii! Not sure how it helps... Regardless, I was told that the solution for this problem can only be an approximation and that there is "no elementary solution". I'm not sure what to believe.

 

[tiny-tim]

Sorry, I didn't notice that sketch.

If you now draw the horizontal lines from the tops of the pegs, to make two right-angled triangles of angles θ and φ, you'll get three equations for r θ and φ, of which two involve cosθ and cosφ, but the third is simply r(θ + φ) = 6 (no cos) …

so yes, there's "no elementary solution".

 

[Femme_physics]

Can you elaborate what "no elementary solution" means?

 

[LCKurtz]

Can you elaborate what "no elementary solution" means?

It means that you won't be able to solve it exactly like, for example, you could solve a quadratic equation with the quadratic formula. It has to be solved numerically.

With all the hints combined, have you figured out from the picture where the three equations come from?

 

[Miike012]

Wow I am pleased to see a lot of people are trying to figure this one out... I didnt think this problem would be this popular lol ... After we figure this one out I have others to that I can post that are similar in terms of difficulty...??

 

[Miike012]

Figured it out!

 

[Femme_physics]

Wow I am pleased to see a lot of people are trying to figure this one out... I didnt think this problem would be this popular lol ... After we figure this one out I have others to that I can post that are similar in terms of difficulty...??

*bolts!*

 

[Miike012]

Inorder to solve for R... the book says a calculator should be used... Is there another way to solve without a calc?

 

[eumyang]

Inorder to solve for R... the book says a calculator should be used... Is there another way to solve without a calc?

I'm guessing that you didn't read the previous posts:

so yes, there's "no elementary solution".

Can you elaborate what "no elementary solution" means?

It means that you won't be able to solve it exactly like, for example, you could solve a quadratic equation with the quadratic formula. It has to be solved numerically.

 

[Miike012]

How would I put it into my calculator and solve for R?

And if anyone would like me to send them the answer to this problem from the back of my trig book just msg me and I will send it to you...

I also have a couple harder trig problems that I could post aswell... just let me know if you are interested...

 

[LCKurtz]

How would I put it into my calculator and solve for R?

And if anyone would like me to send them the answer to this problem from the back of my trig book just msg me and I will send it to you...

I also have a couple harder trig problems that I could post aswell... just let me know if you are interested...

OK eveyone. Here's a picture drawn to scale:

Only two questions left:

1. Does everyone understand where the equations come from?
2. Were any of you able to solve the system with a calculator or computer?

 

[Miike012]

OK eveyone. Here's a picture drawn to scale:

Only two questions left:

1. Does everyone understand where the equations come from?
2. Were any of you able to solve the system with a calculator or computer?

Yes.. this is what I came up with... But like I said I don't know how to use my calc to solve for R...

 

[Miike012]

This is my picture...
Then...
Segment DB = 1; Segment BO = R - 1
Segment DC = 2; Segment CO = R - 1

Thus Cos Alpha @ angle AOB is (R-1)/R
And Cos Beta is R-2/R

then from there is simple algebra...

 

[LCKurtz]

This is my picture...
Then...
Segment DB = 1; Segment BO = R - 1
Segment DC = 2; Segment CO = R - 1

Thus Cos Alpha @ angle AOB is (R-1)/R
And Cos Beta is R-2/R

then from there is simple algebra...

I don't agree it is "simple algebra" from there. You need a nonlinear system solver to get a numerical answer. If your school has access to programs like Maple, Mathematica, or maybe Matlab, they can easily solve it numerically for you. I'm not familiar enough with the latest hand calculators to tell you whether or not your calculator has the capability to solve it. I'm guessing not.

Just for a checkpoint, you should get r = 2.768 approximately.

 

[Miike012]

I don't agree it is "simple algebra" from there. You need a nonlinear system solver to get a numerical answer. If your school has access to programs like Maple, Mathematica, or maybe Matlab, they can easily solve it numerically for you. I'm not familiar enough with the latest hand calculators to tell you whether or not your calculator has the capability to solve it. I'm guessing not.

Just for a checkpoint, you should get r = 2.768 approximately.

Sorry, I was meaning to say that it is simply algebra to arrive to r cos(α) + 1 = r;
r cos(β) + 2 = r; r(α +β) = 6...
But yes that is the answer in the back of my trig book... How did you get that?
And I have a TI-89 calc..

 

[LCKurtz]

But yes that is the answer in the back of my trig book... How did you get that?
And I have a TI-89 calc..

I used Maple. With a program like that, there is little work to do. Here's the steps you type to solve it with Maple. I simplified it slightly for clarity:

> equation1 := r*cos(alpha)+1 = r;
> equation2 := r*cos(beta)+2 = r;
> equation3 := r*(alpha+beta) = 6;

> fsolve({equation1,equation2,equation3}, {r, alpha, beta});

{r = 2.768080732, alpha = 0.8779382494, beta = 1.289628589}

> alphadegrees := evalf(180*alpha/Pi); 50.30215635

> betadegrees := evalf(180*beta/Pi); 73.89027526

Takes all the pain and suffering out of it, eh?

 

[Miike012]

I have another problem...

Six Identical pipes, each with radius of 1 foot are ties tightly together with a metal band... Find the length of the metal band...

I posted a picture...

My strategy is dividing the figure in half
Then finding the measure of the two arcs in red ( which should be equal)
Then finding the measure of the yellow line..
Once I find that... I can multiply it by three...
I am just not sure of how to do it...

 

[tiny-tim]

erm … page 3 ? …

for a new problem, best to start a new thread

 

[Miike012]

okay