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August 28

Formula

The above question might not make clear I'm looking for a formula that would include inclination as a factor regulating the force required. 71.100.13.184 (talk) 00:49, 28 August 2008 (UTC)[reply]

Someone else can probably help more than I can, but I can give you an idea for working out an approximate answer. The force required is going to be something close to the force required to balance the ball on the tip of the chock. In that situation, there will be three forces affecting the ball, weight, reaction from chock, and you pushing it. Weight will be straight down, the other two will point directly towards the centre of the ball from wherever the chock, or you, is touching it. You can then resolve those forces into horizontal and vertical components. The ball is stationary, so all the forces will cancel you, giving you two equations in three unknowns (the force you're pushing with, the angle you're pushing at, and the reactive force) which you can try and solve (the solution won't be unique, and will depend on the exact position of the ball on the chock). Good luck! --Tango (talk) 01:45, 28 August 2008 (UTC)[reply]

By the way ft lb is a unit of energy, not force. Also I am not clear from the picture what is "I"? --Spoon! (talk) 07:50, 28 August 2008 (UTC)[reply]

You may offer a formula which uses weight as the dependent variable since weight is a force, but since weight is suggested as an independent variable along with height of the chock or its momentary incline (I) dependent upon wheel diameter the force in ft lbs should be interpreted as ft lbs force, i.e. a unit of energy. —Preceding unsigned comment added by 71.100.13.184 (talk) 09:12, 28 August 2008 (UTC)[reply]

You don't specify where the force is applied to the wheel or in which direction it acts. Without this information, the question is incomplete.

Let's assume, for the sake of definitness, that the force is always horizontal and is always applied along a line through the centre of the wheel. If you take moments about the top right hand corner of the chock, you can ignore the normal force from the chock because this acts through that corner. You have a clockwise moment from the wheel's load (let's assume the load acts through the centre of the wheel too), and an anticlockwise moment from the applied force. You want the anticlockwise moment to exceed the clockwise moment, so this gives you a minimum value for the force required to start the wheel turning. (I'll let you work out the formula for yourself).

Once the motion starts, the point of contact between the wheel and the chock moves down relative to the wheel's centre. So the lever arm of the wheel's load decreases and the lever arm of the force increases, so you never need a force greater than that at the beginning of the motion.

I'm assuming that the sloping face of the chock is sufficiently steep that the wheel will pivot about a corner of the chock. If the sloping face is shallow enough then the wheel will roll up the face of the chock instead, which requires a slightly different calculation - the point of contact of the wheel and the chock starts out lower, but stays at a constant distance below the centre of the wheel as the wheel rolls. So the minimum required force is initially less than in the pivoting case, but stays constant until the wheel reaches the top of the slope and pivots. Gandalf61 (talk) 09:27, 28 August 2008 (UTC)[reply]

In those terms you could think of the chock corner as a pivot point or hinge, attached to a door with the edge of the door secured to the center of the wheel. In this scenario the wheel can be replaced entirely by the weight of the door (e.g., tailgate). The angle the door must swing up to the perpendicular corresponds to both the height of the chock and the radius or diameter of the forgotten wheel. Bigger wheel smaller angle, lower height smaller angle. We are looking for the energy required to lift the door to the perpendicular through this angle whatever it might be if I understand your reasoning correctly. —Preceding unsigned comment added by 71.100.13.184 (talk) 10:11, 28 August 2008 (UTC)[reply]

Well now, if you really want the energy required, and not the force, then the problem becomes much simpler. Think potential energy (the size of the wheel and the incline of the chock are irrelevant). Gandalf61 (talk) 10:24, 28 August 2008 (UTC)[reply]

That seems to make force easy to figure as well since if the wheel weighs 8 lbs it will take at least eight lbs to move it and 2 ft lbs if it is moved 1/4 of a ft. So the question is how is lifting 8 lbs, straight up .25 ft different from lifting an 8 lb tailgate through an angle which corresponds to a vertical height of .25 ft? In terms of lifting a weight straight up the force required will remain 8 lbs throughout the lift but with the tailgate the force will be less to start with and will diminish during the lift as well as the vertical distance traveled but there will also be an angular or horizontal distance component which I suppose can be vectorized to show the same 2 ft lbs of energy being required. —Preceding unsigned comment added by 71.100.13.184 (talk) 12:07, 28 August 2008 (UTC)[reply]

I think you need to review the difference between force and energy. Whichever route the wheel takes to the top of the chock - lifting, pivoting, or rolling up a slope - the net energy required is always the same (if we neglect any dissipative forces such as friction). This is because energy is conserved - if it weren't, we could take a low-energy route up and a high-energy route down, resulting in a net energy gain and a perpetual motion machine. But the force required depends on how the wheel is lifted. For example, you can lift an 8lb weight through 2ft vertically with a constant force of only 4lb if you use a lever, an inclined plane, a pulley or a screw jack - these are all mechanisms that use various forms of mechanical advantage. The energy required is always the same because energy = force x distance; instead of 8lb force acting over a distance of 2ft, you can have 4lb force acting over 4ft, or even 1lb force acting over 16ft. Gandalf61 (talk) 12:54, 28 August 2008 (UTC)[reply]

May be he wants to figure out the kinetic energy as well as the rotational energy of the wheel. Then the size of the wheel is very much relevant.(Igny (talk) 04:22, 30 August 2008 (UTC))[reply]

Hint: Use the notion of a torque and apply some formulas of rotational motion.

Topology Expert (talk) 03:45, 29 August 2008 (UTC)[reply]

Graphing inaccuracy on TI-84 Plus SE

N.B. The same problem appears on TI-83 (Plus and normal), TI-89, and TI-86. In math class, we were graphing log(x-4), and noticed that the calculator didn't draw the line all the way down the x=4 asymptote, it stopped at about y=-3. I realise that at some point it will stop because it can't handle numbers smaller than about 1e-12, but pressing "Trace" and inputting values did display the flashing crosshairs at the point, although no line was graphed through that point. I also found a site explaining that the TI graphs by assigning x and y values to pixels, and if an asymptote is directly on a pixel, it has problems displaying it. We did play with the window, though, to no avail. If anyone has any ideas about why it does this, or, even better, how to fix it, that would be great. Thanks! —Preceding unsigned comment added by 76.234.137.207 (talk) 01:14, 28 August 2008 (UTC)[reply]

This is a little different issue that handling smaller numbers like 1e-12. You're dead on that the issue is with how it draws each pixel. When drawing a graph the calculater chooses a large list of points to draw. In this case your calculater has choosen to draw a point at x=4 (y=undefined) and x=4.001 (y=-3)... Looks like your current window has a point approximatly every .001, which means that if, for example, your viewing window goes from 3 to 5 that is about 2000 points your calculater had to calculate (3,3.001,3.002...). If you wanted to get to y=-4 where x=4.0001, you calculater would either have to graph 10X more points or you would have to zoom in by a factor of 10 (going from a window of 2 to a window of .2 giving you from 3.9 to 4.1). So you can quickly see it won't go all the way down, even if you zoom in to 3.999 to 4.001, because that would only display y=-6. Anythingapplied (talk) 15:16, 28 August 2008 (UTC)[reply]

Usually calculators don't really draw asymptotes. They graph functions by choosing a bunch of x-values, plotting the points at the corresponding y-values, and then joining the points with straight lines. So the vertical line you usually see where an asymptote is supposed to be is really just a line joining some (x,y) point off the top of the screen to some (x,y) point off the bottom of the screen. The calculator isn't smart enough to realize there's actually an asymptote there—it's just connecting the dots. —Bkell (talk) 05:21, 30 August 2008 (UTC)[reply]

And if you're graphing a function like

\log(x-4)

, it won't draw a line at all, because the function isn't defined for

x\leq 4

—so the first point is at 4.001 or whatever (the first pixel with an x-coordinate for which the function is defined), and that's where the calculator starts its connect-the-dots thing. —Bkell (talk) 05:27, 30 August 2008 (UTC)[reply]

Coincidence

I noticed today that the numbers of possible sides to a regular polyhedron made of triangles, including degenerate cases (2, 4, 8, 20, inf) is exactly the solutions to Conway's Soldiers, if you include Tatham's solution for the last case. Think I'm on to something? Black Carrot (talk) 03:04, 28 August 2008 (UTC)[reply]

Human height

What is the real human's height? When he's out of bed or hours after he's awake? Magnus Armstrong (talk) 19:29, 28 August 2008 (UTC)[reply]

There are over 6 billion real humans, each has a different height. Wikiant (talk) 21:19, 28 August 2008 (UTC)[reply]

That isn't a maths question. If the question you are asking is "what is a human's real height..." then there is no answer. You just qualify it as you wish. Having said that, practically all height measurements are done several hours into the day rather than first thing so most statistics will be the "hours after" answer. -- SGBailey (talk) 21:33, 28 August 2008 (UTC)[reply]

The OP has evidently noted that the height of a human is larger in the morning than it is in the evening (due to compression of the spinal column). The difference (from memory) is something like half an inch. The question is: Which of those two heights is considered to be their "real" height? It's still not a math question - and I'm not sure there is an answer. If the question were (for example): "Which height should I put on my passport application?" then maybe there might be an answer - but even then, probably not. Personally, I think you should use the average height during the day. SteveBaker (talk) 01:19, 29 August 2008 (UTC)[reply]

I agree. However, I am quite sure that the human height has a standard deviation of up to 1 inch during the whole day (including the morning).

Topology Expert (talk) 03:43, 29 August 2008 (UTC)[reply]