Wikipedia:Reference desk/Archives/Mathematics/2010 September 27

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September 27

Elevator, Force/Mass question

an elevator has a mass of 490-kg and is carrying nine 65-kg passengers. The elevator has a maximum acceleration of 2.24 m/s^2. I'm supposed to find the tension of the cable if it accelerates downward at maximum acceleration. Since the tension + force due to gravity=totalmass times acceleration

I set up that T=totalmass(gravity+a) or that the tension equals 1075 kg(9.8m/s^2 + 2.24 m/s^2) ergo, the tension is 12943 N.

Are my maths right? Sometimes I screw up the signs of gravity.209.6.54.248 (talk) 00:14, 27 September 2010 (UTC)

Imagine you are holding a model elevator by a small cable. Will the cable be under more tension when you yank it upward, or when you let it drop? Does that help? -- ToE^T 00:43, 27 September 2010 (UTC)

Draw a free-body diagram. Is the tension pulling up or down? --COVIZAPIBETEFOKY (talk) 02:29, 27 September 2010 (UTC)

OP here, ah, so the acceleration is negative. so it should be 1075 kg(9.8 m/s^2 - 2.24 m/s^2) so the tension is 8127 N209.6.54.248 (talk) 02:54, 27 September 2010 (UTC)

You have a body of mass 1075 kg accelerating downwards at 2.24 m.s^-2. Newton's Second Law says that requires a resultant force of 2408 N downwards. The weight of the body is 10535 N and yet the resultant force on the body is only 2408 N downward. Clearly there is an upward force of (10535 - 2408) or 8127 N. What could that upward force be? Aha! it must be the tension in the cable, especially when you realize a flexible cable can't sustain a compressive force so if the body is hanging down on the cable the cable must be providing an upward force.

As you can see, it isn't necessary to assign a positive or negative sign to any of the forces. It is usually sufficient to recognise that upward and downward forces have different signs and therefore upward and downward forces must be subtracted, not added, when determining the resultant force. Dolphin (t) 06:20, 27 September 2010 (UTC)

Volume of a cone

Today in math I had a homework question that went something like explain why it is not possible to find the volume of a cone with radius r and height h by multiplying the area A of the right triangle formed by r, h, and the slant by 2πr (circumference. I said that when you do that you're actually finding the volume of a triangular prism with length 2πr, which is not the same because the inside points of the cone should not 'spin' as much as the outside points. But intuitively I think there should be some way of relating a cone's area to the triangle formed by r, h, and the slant. But how? THanks. —Preceding unsigned comment added by 24.92.78.167 (talk) 21:48, 27 September 2010 (UTC)

Take a look at the subsection volume of a cone. — Fly by Night (talk) 21:57, 27 September 2010 (UTC)

Yes i forgot to mention: i realize that the traditional way to integrate a cone for volume is through the center of the base to the tip, but I was wondering if there is another way. —Preceding unsigned comment added by 24.92.78.167 (talk) 22:03, 27 September 2010 (UTC)

You may wish to read our article on Pappus's centroid theorem. -- ToE^T 23:27, 27 September 2010 (UTC)

You have written I think there should be some way of relating a cone's area (volume?) to the triangle formed by r, h, and the slant. No, it is not possible to relate the volume of a cone to the area of the right triangle r, h and the slant. Think of two such triangles, both with the same area but with very different dimensions r and h. One triangle has a small base r but a large height h. The other triangle has the same area as the first, but with a large base r and a small height h. Clearly, the second triangle, the one with the large base r and small height h, produces the cone with the greater volume because the volume of the cone is related to r², but only to h, not h².

The volume of the cone is a function of r², and h. It is not a function of area of the triangle of dimensions r and h. Dolphin (t) 23:53, 27 September 2010 (UTC)

The OP never suggested the volume is a function of only the area A of that triangle. There is a relation, which is V = 2πrA/3. I think he/she is looking for a nice geometric way to intuitively understand that value. The best I can do is to cut the cone into small radial slices (like cutting a cone shaped cake) and noting that each one when laid on its side is nearly a tetrahedron with base A and all the heights summing up to 2πr. Unfortunately at that point you need to use the volume formula for a pyramid, but maybe someone has a nice geometric way of thinking about that one. Rckrone (talk) 00:15, 28 September 2010 (UTC)