Dear Long Live Yorke

[aadams_22 3]

I'm currently studying for the Professional Engineer exam, so I'm on a math kick at the moment. I already know the answer to this, but I felt like challenging you for no real reason...procrastination maybe.

[David_Nicoson 1]

http://en.wikipedia.org/wiki/Two_envelopes_problem

[LongLiveYorke 38]

I'm currently studying for the Professional Engineer exam, so I'm on a math kick at the moment. I already know the answer to this, but I felt like challenging you for no real reason...procrastination maybe.

I have no idea what the hell this means. I have to do what with the load? What is the load? What do the pounds per foot indicate?

[aadams_22 3]

I have no idea what the hell this means. I have to do what with the load? What is the load? What do the pounds per foot indicate?

They are wanting you to break the total distributed load down into it's components with respect to the centroid and find the magnitude of each.

[LongLiveYorke 38]

They are wanting you to break the total distributed load down into it's components with respect to the centroid and find the magnitude of each.

Oh, I think I get it.Hold on.

[LongLiveYorke 38]

Partial answer:Mass = 4750Center of mass position = 8.42How am I doing so far?

[timwakefield 68]

How am I doing so far?

Nerdy.Zing!

[aadams_22 3]

Partial answer:Mass = 4750Center of mass position = 8.42How am I doing so far?

you're on the right track, but here's a few hintsFrom the two loads given you can decipher that it's a x=y² parabola. As with an x=y² parabola you can break it down into a smaller parabolic shape, a rectangle, and a triangle. You will need to find the centroid of each shape. Once you find the centroids of all three shapes you can then determine the magnitude of each force with respect to each shape. Also, if you wanted to take it a step further you can find the overall resultant force by summing the y-components, and find it's magnitude by taking a moment about an arbitrary point with respect to the resultant force.

[Shimmering Wang 1]

the question wasn't until like the tenth posthe reads every thread and every post including all the I don't think Daniel is concentrating enough on his game threads

LOL shutup

[Ouch-8s 4]

The amount of water displaced is the amount of water whose mass is equal to the mass of the boat. This is Archimedes' principle. If I have an ice cube in water, it will displace an amount of water equal to the mass of the ice cube. So, the volume of water displaced is equal to the mass of the ice cube divided by the density of water:V = m/R, where R is the density of water and m is the mass OF THE ICE CUBE. When the ice cube melts, it will have the density of water and therefore have the same volume V as the displaced water.So, if we imagine that we can freeze time and remove the ice cube from the water, there will be a rectangular void in the top of the water where the ice cube was. The volume of shape is equal to the volume of the water displaced, which is V. When the ice cube melts, it will as water have volume V, and so we can imagine that it forms just enough water to perfectly fill this void, and therefore the overall height of the water doesn't change.

this was very nicely worded, you should consider going into teaching; i hear there's big money in that.**I'm not very good at math, you should probably look into that "big money" thing for yourself.

[speedz99 145]

...that's what she said.

Obviously this calls for a

[runthemover 39]

this was very nicely worded, you should consider going into teaching; i hear there's big money in that.**I'm not very good at math, you should probably look into that "big money" thing for yourself.

I'd say it's more important to watch out for whammies

[El Guapo 8]

I'd say it's more important to watch out for whammies

No.No Whammies.Stop.

[Nikki_N 17]

I'd say it's more important to watch out for whammies

No.No Whammies.Stop.

Classic.I'm going to have to read the rest of this thread later. ISAM.

[qyayqi 11]

is melting ice not a special case in the displacement example?

[LongLiveYorke 38]

is melting ice not a special case in the displacement example?

All objects displace water that has mass equal to the object's mass. Ice has the nice property that when it melts the overall level of the water won't rise. This comes from the fact that when ice melts, it will have the same density of water. If I were to melt a boat, it could potentially raise the overall level of the water since there would be no relation between the density of melted boat and the density of melted water (which I guess is just water).

[qyayqi 11]

All objects displace water that has mass equal to the object's mass. Ice has the nice property that when it melts the overall level of the water won't rise. This comes from the fact that when ice melts, it will have the same density of water. If I were to melt a boat, it could potentially raise the overall level of the water since there would be no relation between the density of melted boat and the density of melted water (which I guess is just water).

hmm. i thought the level would go down as the ice melted, because solid water was somewhat unique in how it occupies a greater volume than liquid water.

[LongLiveYorke 38]

hmm. i thought the level would go down as the ice melted, because solid water was somewhat unique in how it occupies a greater volume than liquid water.

See my earlier post in this thread on this topic.

[hblask 1]

<br />All objects displace water that has mass equal to the object's mass. Ice has the nice property that when it melts the overall level of the water won't rise. This comes from the fact that when ice melts, it will have the same density of water. If I were to melt a boat, it could potentially raise the overall level of the water since there would be no relation between the density of melted boat and the density of melted water (which I guess is just water).

What if I made a boat out of ice? :)Actually, when you think about it, it would make a difference if there were trapped air pockets in the ice cube, right?

[Ouch-8s 4]

What if I made a boat out of ice? :)Actually, when you think about it, it would make a difference if there were trapped air pockets in the ice cube, right?

only if the ice was forced underwater, i think

[LongLiveYorke 38]

What if I made a boat out of ice? :)Actually, when you think about it, it would make a difference if there were trapped air pockets in the ice cube, right?

I don't think so. At no part in my argument did I need to consider the density of ice itself. The presence of air bubbles only effectively decreases the density of ice, so their presence shouldn't change the final answer.

[Ouch-8s 4]

I don't think so. At no part in my argument did I need to consider the density of ice itself. The presence of air bubbles only effectively decreases the density of ice, so their presence shouldn't change the final answer.

yeah. a glass FULL of icewater will have it's level drop, but a single cube floating on top won't.

[aadams_22 3]

I'm currently studying for the Professional Engineer exam, so I'm on a math kick at the moment. I already know the answer to this, but I felt like challenging you for no real reason...procrastination maybe.

Partial answer:Mass = 4750Center of mass position = 8.42How am I doing so far?

you're on the right track, but here's a few hintsFrom the two loads given you can decipher that it's a x=y² parabola. As with an x=y² parabola you can break it down into a smaller parabolic shape, a rectangle, and a triangle. You will need to find the centroid of each shape. Once you find the centroids of all three shapes you can then determine the magnitude of each force with respect to each shape. Also, if you wanted to take it a step further you can find the overall resultant force by summing the y-components, and find it's magnitude by taking a moment about an arbitrary point with respect to the resultant force.

It's been a few days so here is the solution, and please excuse my apparent misspelling of "concentrated." I'm a Civil Engineer, not an English professor.

[LongLiveYorke 38]

Okay, I think I finally understand the question :)What is the advantage of dividing the figure up into three shapes, aside from the fact that it's the question asked? When I calculated it, I just considered two shapes, a parabola and a trapezoid. And I guess you know a few tricks about calculating the parabola's centroid that I don't.To find the function for the parabola, I used the two points given and then used the condition that the slope at the highest point must match the slope of the linear region, giving me three conditions which is enough to specify a parabola. And, I found my mistake which gave me the wrong value of the overall center of mass. This time I got:6.557

[aadams_22 3]

Okay, I think I finally understand the question :)What is the advantage of dividing the figure up into three shapes, aside from the fact that it's the question asked? When I calculated it, I just considered two shapes, a parabola and a trapezoid. And I guess you know a few tricks about calculating the parabola's centroid that I don't.To find the function for the parabola, I used the two points given and then used the condition that the slope at the highest point must match the slope of the linear region, giving me three conditions which is enough to specify a parabola. And, I found my mistake which gave me the wrong value of the overall center of mass. This time I got:6.557

For this problem the fact that it asks for three loads is the reason you want to break it down into three separate shapes. That being said I personally find it easier to deal with more familiar shapes like rectangles and triangles. As for the 6.557 that's pretty close, and it shows that this problem isn't unlike the many other math problems out there...ie...they can be worked in many different ways to come up with the same answer.