Idea I had since elementary school

[preoccupied]

I was referred to torque wrenches as a means to measure the mechanical advantage. Does anybody know where I can get a good deal on a torque wrench? What materials should I use to to make a right angle that can attach to a torque wrench? I will also have to make a straight rod to compare it to. I hypothesize that if my hypothesis is correct that the right angle will produce mechanical advantage when turned into the right angle.

I'm discouraged to test my hypothesis because the viewpoints so far state that the science already says I'm wrong but I had this experience with a car jack that backs my hypothesis so I'm also motivated to continue. To test my hypothesis I will need to see if one right angle on an axle produces any kind of mechanical advantage.

With the car jack the jack went way under the car and a long rod attached to the jack and another piece attached to the long rod. The piece that attached to the long rod flipped back and forth and went perpendicular and to the side a right angle, not a handle. I tried to turn the rod pushing out of the right angle and it was very difficult and I flipped it to the other side and tried pushing into the right angle and it was easy. So maybe pushing into the right angle might not produce mechanical advantage though I hope it does but pushing out on the right angle produces resistance somehow, from my personal experience with the car jack.

[nicbordeaux]

craiglist

[path_finder]

Dear preoccupied,
You have already under your eyes the right angle crank: your own arm holding the torque wrench. Adding a new bar will change nothing.

[preoccupied]

I am arranging an experiment to see if a right angle bar and a straight bar with the same radius from the axle have different torque. I guess the science says that the straight bar and the right angle bar would have the same torque but I had an experience with a car jack that makes me want to reconsider if that is true. If a right angle bar two feet up and two feet across had 200 pounds of force applied to it it should give torque equal to the radius which is sqrt(8) = 2.83 times 200 which is 566ft*lb. I want the right angle bar to give more than that. I might not use 200 pounds I might use like 10 pounds.

[Tarsier79]

Why don''t you arrange them opposite? Making the right angle will mean there is more material on that side and weigh more, so you will have to balance the arrangement before adding weights. If your right angle/ swastika produces more torque, that side will lift the other weight. Make sure the distance between the weight and the axis is equal.

[preoccupied]

Is "balance" balanced the way it appears in the picture? When I was using the car jack when I was pushing away from the right angle like in your picture if gravity were pulling down it was more difficult than when I pushed into the right angle. I was able to do either or because the piece flopped on either side. I hypothesize that your picture is not balanced and the straight side will go down. There has to be some reason why it was more difficult for me to turn the right angle out of the right angle than into the right angle unless I did something else that was some kind of user error.

[jim_mich]

I'm guessing you used a car jack that looked something like this picture where you inserted the lug wrench into the jack, either as shown in black or as shown in blue. Depending upon which way the ell shaped lug wrench was inserted, you will get different leverage, and thus it will be harder to jack with it inserted one way and easier when inserted the other way. This is due to the effective leverage length. Note the little purple circle is the lever pivot point.

[preoccupied]

I think the car jack was for a 1988 dodge neon. It did not resemble the jack you shared Jim_Mich. It had a right angle that was not a handle to turn but to pull and push. When I pushed towards the outside of the right angle bar going the correct direction it was difficult to turn and when I pushed towards the inside of the right angle bar going the correct direction it was easy. I was able to change whether I pulled towards the inside or outside of the right angle bar because it flipped back and forth. I hypothesize based on that experience that there is mechanical advantage towards the inside and resistance towards the outside of the right angle bar.

[Tarsier79]

Preoccupied, why would you think the straight lever would go down if you think a right angle produces a mechanical advantage? Gravity is pushing down equally on the weights, that are equidistant from the axis, so if a right angle produced a mechanical advantage, then the angled side should go down.

I was just giving you an alternative way to test your torque rather than with a wrench. In fact, in a perfect world, it will balance, which would show no mechanical advantage to either lever over the other.

PS"...user error." I suspect that to be the case.

[preoccupied]

I don't know why, tarsier79, maybe it was user error. I can't know why I had that experience from a scientific perspective because I haven't an education in physics. The situation reminds me of how it is more difficult to bend a bar further in than to bend it back out. There is more tension trying to bend a bar in than to bend it back out. But I don't know how tension works in physics so I can't say it is because of tension of some kind. I tried pushing on my own arms. I held my arm at a right angle and I pulled my arm in and had to use a lot of muscle to keep it from going in and when I pushed it out I didn't seem to have to use as much muscle but this could be like a placebo effect where I kind of think it's supposed to be that way so I think it happened that way. I look forward to when I can borrow a torque wrench and test the hypothesis. I decided borrowing a torque wrench was the cheapest way to test the hypothesis.

[path_finder]

Dear preoccupied,
In the same domain we can consider this kind of 'energy extractor', the cyclorameur (I don't know the translation).
When I was a kid (in the 50's) I had once a 'cyclorameur', where two arms must be pulled for a forward motion (see an example below).
And for the fun the 'Fête du Jouet Sportif' (Fair of the sportive toys) in Paris (October 3rd 1934):
http://video.google.fr/videoplay?docid= ... 7304&hl=fr

Also the rowing bike (see the animation below).

[preoccupied]

I'm probably wrong about the animation because people say the science says I'm wrong. Okay, so what I am told is that the radius determines the distance in the equation "Work Done = Force x Distance" on an axle. In the animation according to the science that I am told those animations should be in balance but I animated them out of balance to show my hypothesis.

I drew the right angle bar with inkscape and an equal radius straight bar on the other side of a pivot. Lets say the distance of the bars on the right angles are 2 ft and the distance of the straight bar is sqrt(8) or 2.83 ft. On the left the weight is pushing straight down on the right angle in a way that makes it look like it's only applying force to the first horizontal bar. I hypothesize that the right angle on the left animation only has a distance of 2 ft because of how it looks. I hypothesize that the straight bar on the left animation will pull down because it has more leverage. On the right animation I hypothesize that the right angle bar will pull down. On the right animation I think tension is applied to the vertical bar attached to the axle by the horizontal bar and that increases its distance.

[Fletcher]

Here's your answer I think.

Two examples - rhs V shaped - each lever 1 kg - balanced.

Note that combined CoM is vertically over the pivot so no torque.

lhs - right angle each 0.5 kg [ 2 x 0.5 = 1 kg]

Note the combined CoM for each side of pivot - then see the combined system CoM.

Note it is to the left of the pivot - result it will have torque to the left & rotate CCW.

Not exactly the same as your animation but demonstrates the principle of CoM & Leverage about a pivot.

-----------------------------------

By my estimation both your pairs in the animation are falling the opposite way to what would happen in a real build, assuming both masses each side of the pivot were equal.

N.B. To turn the right angle upside down [to lift the CoM] means doing Work to raise the Potential Energy [Joules] & Work Done = Force x Distance = Newton Meter's = Joules of Energy.

[daanopperman]

Hi all,
Fletcher,
maybe if you draw the left side pic in perspective the com on the angle will change to the pivot point, the two pieces should be the same length as the rh side of the drawing.

[preoccupied]

I look to wiki now. I might not understand what I read on wiki. wiki says the "right hand grip rule" can determine the direction of the torque: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque. In the left animation if the right hand holds the vertical bar on the right angle bar its torque would be zero because the thumb would be pointing to nothing but if the horizontal bar is held its torque would be pointing at the axle. I think that's one bar worth of distance. In the right animation if the right hand holds the horizontal bar on the right angle bar the thumb will point towards the elbow. I therefore think the elbow is the torque. If the right hand holds the vertical bar the thumb points towards the axle so I think the axle is the torque. I think it has two bars worth of torque.

On the left animation for Work done = force x distance, the right angle bar is 2kg x 2ft and the straight bar is 2kg x sqrt(8). For the right animation, the right angle bar is (1kg x 2ft) + (1kg x 2ft) and the straight bar 2kg x sqrt(8). Of course though, Fletcher, we are both probably wrong. The work done should be equal for equal radius from the axle according to what I have heard from people. But I stand by my hypothesis until I can test it.