Thursday, May 1, 2008

Round 260

Date Released: May 1st, 2008.
Date Rewarded: May 8th, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every 3 metre increment, there is a spherical steel weight with a diameter of 30cm.

If the density of the steel is exactly 7.8 g/cm3, the steel can withstand a tensile stress of 400 MPa, and gravity is 9.81m/s2, what is the maximum possible length, in metres, that the bar can extend without breaking? Round down to the nearest metre, submit only a number (with no additional or extraneous information), and disregard any deformation of the bar or any other forces acting on the bar.


Steel Negg

Click to show/hide solution

Using the stress equation [1]:
rho = F/A

where rho = 400 MPa = 400,000,000 kg/m·s2 is the tensile stress that steel can withstand, F is the force that the steel has to withstand and A is the crossed area of the steel bar.

The crossed area is a regular pentagon can be calculated by the following equation [2]:
A = 5/4 t2 tan(54°) = 0.000688190960236 m2

where t = 2 cm = 0.02 m is the side of the pentagon.

Now, we can calculate the force that the steel bar can withstand:
F = rho * A = rho * 5/4 t2 tan(54°)

Using Newton's gravity law [3], we have:
F = m * g

Therefore, the mass that the steel bar can widthstand is:
m = F /g = rho * 5/4 t2 tan(54°) / g = 28060.793485650833 kg

where g = 9.81 m/s2 is the gravitational acceleration.

Don't round anything yet. The mass is contributed by 3 components: the big sphere m1, small spheres m2, and the steel bar m3. Since all of them are steel, they have the same density. We have:
m1 + int(h/3) * m2 + m3 = m
<=>[V1 + int(h/3) * V2 + A * h] * d = m

where h is the length of the steel bar, int is the floor function (every 3 meter, we have one small sphere), d = 7.8 g/cm3 = 7,800 kg/m3 is the density of steel, V1, V2 are the volumes of big and small spheres. The volume of the steel bar is the product of the crossed area with the length of the bar.

We can calculate V1, V2 by the sphere volume formula [4]:
V1 = 4/3 * pi * (r1)3 = 0.065449846949787 m3
V2 = 4/3 * pi * (r2)3 = 0.014137166941154 m3

where the radius r1 = d1/2 = 25 cm = 0.25 m, r2 = d2/2 = 15 cm = 0.15 m. Don't round both volume either. We already know A, so the equation now should be:
V1 + int(h/3) * V2 + A * h = m/d
<=>int(h/3) * V2 + A * h = m/d - V1

We know the right hand side of the equation but the left hand side is a little bit tricky. So, we can do a method of trial-and-error to test. Solve h by the following equation:
h/3 * V2 + A * h = m/d - V1
<=>(V2/3 + A) * h = m/d - V1
<=>h = (m/d - V1)/(V2/3 + A) = 654.0200901109209 m

Now round h down to an integer. We then have h = 654 m. Next, test h into the equation int(h/3)* V2 + A * h = m/d - V1. If the left hand side is smaller than the right hand side, the steel bar can withstand the gravitational forces. The largest h that is satisfied that the left hand side is smaller than the right hand side will be the answer.

Using h = 654 m, the left hand side: int(h/3)* V2 + A * h = 3.5320. The right hand side: m/d - V1 = 3.5321 (good). Now increase h by 1 (h = 655 m) to see if the steel bar can still withstand. The left hand side: int(h/3)* V2 + A * h = 3.5327. The right hand side: m/d - V1 = 3.5321. (not good). Therefore, the answer is h = 654 m.

Results: 986 people guessed the correct answer earning themselves 2029 NP each.

[1] Stress
[2] Regular pentagon area
[3] Newton's gravity law
[4] Sphere volume


vivek said...

hello im only 12 and i really have no idea what this whole page says. I have been trying for about 2 hours trying to figure out what it says. could u help me out a little more. i would like to figure it out but its really hard for me. my neopets account is mastervivek
plz help me. a hint or something

David said...

(Damit I should've done the puzzle earlier...)
I didn't get the d1 and d2... I didn't see 2 different diameters.
(I wonder if he will check back)
vivek, if you have a scientific calculator, use it, it helps so much (as long as you know what you're doing) but here's my hint: My answer adds up to 15. If your digits add up to 15, then you might be right. (You're only 12... uh... 143232=15, so the number is 143232 (NO IT'S NOT THE ANSWER))

LC Solver said...

sorry that you're 12. This problem is way beyond age 12. However, try to challenge and you can learn some

Most of the theory has been posted. You only need to put in number and calculate.

and yeah. My answer digits adds up to 15 but don't always count on me for right answer (I made mistakes sometimes)

Jennifer said...

took me two hours to do all the equations. what did you guys get for h? i got a four digit number. then for the answer i got a two digit number.

Anonymous said...

Dependable as always, aren't you? This really saves a lot of toil, since (as usual), I tend to overcomplicate things (see my first comment for the Round 258). The fact that the "V2" steel spheres were set 3 cm apart drove me crazy w/possible implications into the whole calculating process (e.g. strain effect, position issues, stuff you don't want to know). Graci.

P.S.: Do they usually put it up @ 12:00AM NST sharp?

LC Solver said...

My answer is a 3 digit number, adds up to 15 if you want to know.

I'm glad that you could solve it, L. They usually put up at 4-5 PM NST on Wednesday. I put the next date on my blog because I want to put on the date of the news. No fixed time, that's when they put up the news!

vivek said...

Thanks a lot guys. I dont know the answer yet but i am about to work on it. I have an older brother that helped me on round 258 but i still got it wrong. Well thanks for the hints. Thanks a lot. Hope i win a trophy

LC Solver said...

Good luck with your answer.

vivek said...
This comment has been removed by the author.
vivek said...

I got it. I think. I know i cant say the answer. My answer is even. it has all even numbers but one. Is that correct or wrong. It adds up to 15 so i think its right. It was very challenging, but after lots of work and help from my brother i got it.
I love this website. Thanks for all your help lc solver. By the way whats your neopet username.

LC Solver said...

look like you got the right one. I'm glad you make it. I don't tell people my neopets username (a lot of people like to report) but when I'm online, I can always help with hints...

Good luck!

Anonymous said...

Argh; that means my submission prbbly didn't make it to the top 250. They NEED to make better time adjustments. Anyway, thanks.

LC Solver said...

I believe it's pretty hard to post on the right time because it's hard to meet the time line in computer programming.

Still, it's not very good for those in different timezone (when they are still sleeping)

What's your timezone by the way?

Anonymous said...

GMT +8:00

JoNaH said...

i'm good at maths and i still don't understand it =(
ny help?

LC Solver said...

Ahh. Then, you must wake up early on Thursday to try to catch the problem early.

Jonah, the hints are supposed to help those who don't understand the problem, you only need to plug in the number and calculate.

JoNaH said...

yh but i don't get what numbers to plug in =/

vivek said...

lc solver. my username is chocoholikaddict
Just message me. so i can get ur username. I dont think anyone should report you. You didn't say the answer you just gave us hints to get it.

Anonymous said...

Question. An engineering friend o/mine seems to differ in opinion as to which area should be used in the rho=F/A stress equation. The rest of the concepts are fine; but on that particular point, he was emphatic that the area calculated (right there, not saying that the pentagons don't matter) should be that of the steel sphere due to its position (withholding all the vertical pressure). What do you think?

LC Solver said...

vivek, I would like to be safe than sorry. It's nice account you have there though.

jonah, the numbers are given on the problem. The equations/formulas are given in my hints. So I hope you can plug them in.

L, the stress equation said that F is the force acting on the area A (check the reference). What we concern here is where the steel bar can withstand the forces. So, F is the force acting on the steel bar and A is the cross-sectional area of the steel bar.

I would worry more about other equations but the stress equation, I'm pretty sure it's on the steel bar. Hope it helps :)

Tsuyoi said...

Does m = F/g = 400'000'000/9.81 ?

Tsuyoi said...

You wrote "r2 = d2/2 = 30 cm = 0.15 m"

I'm pretty sure that's wrong ;)

LC Solver said...

tsuyoi, thanks for your comments. It was a typo and it's fixed. The answer won't be affected by the typo.

m = F/g where F can be calculated by the above equation (F = rho * A), F is not 400,000,000.

Tsuyoi said...
This comment has been removed by the author.
LC Solver said...

like I said in the above comment, F is not 400,000,000 MPa. 400,000,000 MPa is the stress (rho). You made a mistake from there. Hope it helps :)

Tsuyoi said...

I got a 3 digit number thats adds up to 15 and its even odd even, is it right

LC Solver said...

look right to me :)

Anonymous said...

In that case (if you're right, which I hope you are), he'll have the chance to be pointed & laughed at (anticipation awaits); *Cackles evilly*...

LC Solver said...

well, let's wait for the answer and see how it goes. I'm confident about the method I used though.

JoNaH said...

i dont get where to plug the numbers in though =/
do i plug them in every equation or just 1?

matrix @ its best said...

hi dudes um... i still cant figure out the answer i think its 15 though can u plz help me to figure it out cause the matrix said so lol na im just messen around with ya but i seriously need help with the answer