The Triangle's Puzzle

[CANGAS]

OK.

You say yo idea dont work.

I'm to dumb to to figure out why it should work, or, should not work.

Spain why it does , or, dont , work.


CANGAS35

[gmbajszar]

a^2 + b^2 = c^2

If we can fit 10 weights on the left, we can fit 14.14 weights on the right on a 45 degree slope.
What balances is 10 weights vertically, or 20 weights on a 45 degree slope.
(Yet we only fit 14.14 weights on the right side, not 20.)

It was shown that British were extremely bad in math internationally, even math teachers couldn't pass an international math test when all Russian math teachers scored 97 percent on it. I am not implying that there is a UK/US imperialism disorder so people can't figure out high school first grade level math.

How many weights do we need on the 45 degree slope to balance out 10 weights?
Ok, 20. But 20 weights extend lower than the height of 10 weights.

And still, this machine does not work, why. The answer may be Ph.D math level.

George

[Staffman]

George,


The math behind this is trigonometry. It is really needed to understand a lot of physics problems.




The force to move an object up an incline plane is equal to its Mass (m), gravity (g), and the sine of the angle (theta).


          f=m*g*sin(theta)




[For ease in demonstrating the math, we need to agree on the mass of the beads and their distance apart. I will say that the beads weigh 1 Lb each, and are 1 meter apart. This makes the math a bit easier to lay out, although we all know that a real beaded necklace would be lighter and the beads closer together. Also I will be ignoring friction to make things simpler.]


On the vertical side of the triangle, two beads weighing 1 Lb each, the Force in SI units is 8.8964432 Newtons

           F = 8.896 Newtons (Vertical)




If the height of the 45 degree triangle is 2 (meters), the length of the slope is 2 * sqrt(2) (exactly)... or approx 2.828 (meters). [I use meters, the SI unit, to prevent goof ups]


Now to calculate the force needed to move beads up the incline plane of 45 degrees. There are 2.828 beads on the incline plane (the third bead isn't all the way on the incline plane). The force needed to move the beads is:


           2.828 Lb = (1.283 Kg * 9.81 m/s/s) = 12.5795706848 Newtons


           F = m*g*sin(theta)


           F = 12.580 Newtons * Sin(45 Degrees) = 8.895 Newtons (up 45 Degrees)




Here we show that the force need to move the mass up the 45 degree incline is approximately equal to the force need to lift the mass vertically.




Quick recap....
               Each bead is about 1 Lb, 1 meter apart.  (1 Lb = 4.4482216 Newtons = 0.453437472 Kg * 9.81 m/s/s)


               Calculate lift force. (If you know the weight in Lbs, just convert to Newtons. Stay in SI units, and you'll have no problems.)


               Calculate force up incline plane. (you have to calculate the lengths of the triangle using trig if you use anything other than 45 degree triangle with the rule 1 Lb, 1 meter apart)
                                  F = m*g*sin(theta)




Now to see what happens when you go to a 30 degree triangle.


            Lift force is the same.


                     F = 8.896 Newtons (two one Lb beads 1 meter apart = 2 Lb * 4.4482216 = 8.896 Newtons)


            Because the vertical length is the same, we need to know the length of the hypotenuse (aka the number of beads on the hypotenuse).


                    Hypotenuse Length = Vertical Length / Sin (30 Degrees)


                    H = 2 / Sin(30) = 4


                    So there are 4 Beads on the hypotenuse. (4 Lb * 4.4482216 = 17.792 Newtons)


            The force needed to move the beads up the incline plane of 30 degrees is:


                   F = m*g* sin(theta)        (Note that m*g is weight in Newtons)


                   F = 17.792 Newtons * sin(30) = 8.896 Newtons    (Note that the force needed to move an object up a 30 degree incline plane is 1/2 the vertical force.)


                   F = 8.896 Newtons


Again we show that the lift force and the force needed to move the mass up the 30 degree incline plane is exactly the same.




No matter what the angle of the incline plane, using evenly distributed masses on a string, the lift force will always equal the force need to move the mass up the incline plane.




I hope this helps...

[gmbajszar]

What level math is this? Third grade high school?

Interesting that logically the mind assumes that on a 45 degree slope it is twice as easy to pull weight up.

I have a good similar puzzle for you, sir.


It cannot generate free energy, but what I want it to do is to move a spaceship in space.

I attached a picture below. We set up an experiment that begins with being in space in a long rocket. An astronaut throws a heavy weight downward from the top of the rocket, which then makes the rocket move upward until the weight hits the bottom of the rocket when the rocket stops moving. But for a few seconds until the weight flew from the top of the rocket to the bottom, the rocket was in motion upward.

Now, if we analyze the picture below, we use a shotgun to fire two weights downward. As the weights are fired, the rocket moves upward. But then the weights make a 180 turn, which should reverse the momentum of the moving rocket to a full stop. Only when the weights hit the top of the rocket again, I believe we generate motion for the rocket.

Question: Is this a valid theory?

George

[Staffman]

George,


This is first year physics, either high school or college. Normally, a course in trigonometry is required before taking the class. It's the trig part that can trick you up if your not careful. AND, making sure you stay in SI units... it's really easy to mess things up. A good place to learn is Khan Academy. Plenty of video's to get you rolling fast. You don't need to create an account, that is if you don't want progress saved. And, did I mention it's free?


Hope this helps. Good Luck!!!