In leveraging weight, if the load and effort grow at the same distance from the fulcrum, will the effort grow in leverage, go down in leverage, or remain the same.

For instance:

If you have a 1 oz weight off set from the fulcrum one inch further than a 1 oz weight on the opposite side and the leveraging side has 14 grams of leverage. If you moved the leveraging weight 3 inches further from the fulcrum and the opposite 2 inches further from the fulcrum "so that the leveraging weight is still 1 inch further out than the opposite", will the leveraging weight gain leverage "meaning more than 14 grams of leverage", go down, or remain the same amount?

Thank you!

Not sure where you are getting "14 grams of leverage" from in your example. If you put two weights of equal mass m on the two sides of the fulcrum, with one mass placed at distance d1 from the fulcrium and the other mass at distance d2 from the fulcrum, then the sum of the torques about the fulcrum point is:


T = mgd_1 - mg d_2 = mg(d_1-d_2)


So as you can see the torque is a function of the difference between the two distances.

However, I suspect that you are trying to determine what forces the two masses "feel." That would be equal to the toque divided by the distance of the mass from the fulcrum. Thus for the closer in mass at distance d2, the force pushing it up is:


mg(d1/d2)


If d1 = d2+ 1, then this force is:


mg(1+\frac 1 {d_2})


You can see that as d2 increases then the upward force on that mass at d2 gets closer and closer to its weight. Hope this helps.

The leveraging advantage depends on the ratio of the two distances, not their absolute difference. A 1" difference is pretty significant if the leverage point is 3 inches from the fulcrum, and the load is 2" away. That advantage would be enormously diminished, however, if the leverage point was 1001" from the fulcum while the load was 1000" away. To maintain the same advantage as before, the 3:2 ratio would have to be maintained. The leverage point would have to be 1500" to maintain the same advantage over a load 1000" from the fulcrum (1500:1000 = 3:2).

So as you can see, as you move further and further away from the fulcrum, a constant 1" difference becomes less and less significant. So the answer is that your leverage weight gain would go down.

EDIT since seeing Ebaines's post: Note that this is in agreement with the formula that Ebaines gave you at the end of his post. As the distance d2 gets larger and larger, the amount that is added to the effective weight becomes less and less.

Jcaron2, please answer the following:

I asked ebains this same question, could you please give me your input about the same question? Thank you
I created a diagram so it could more easily be understood. I just need to know if a smaller weighted wheel will have less torque than a bigger weighted wheel.

The diagram can be seen here: http://bizidia.com/wheel.jpg

Thank you so very much for your time!

Sincerely,

JC

In that case, the torque would be the same in both cases if the weight was the same. Let's say both wheels had weights of w lbs. The torque on each wheel in the counter-clockwise direction (i.e. as a result from the weight that's right on the rim of the wheel) is \tau=\frac{2}{\pi}wr, where r is the wheel's radius. The factor of 2/pi is mostly irrelevant here and comes about from the fact that the weight is spread around the rim (for example, the portion of the weight at the exact top and exact bottom of the wheel will not be contributing to the torque whatsoever). Anyway, suffice it to say that the torque is proportional to wr (which stands to reason). Meanwhile, the clockwise torque on the wheel (i.e. the torque from the weight which is 1" further out from the center) is going to be proportional to w(r+1). Since the two torques are in opposite directions, that means the net torque is the difference between the two, which is

\tau_{net}=\frac{2}{\pi}w(r+1)-\frac{2}{\pi}wr=\frac{2}{\pi}w

Note that the net torque is independent of the radius. It would be the same for a 10" wheel as for a 20" wheel. The torque is, however, proportional to the weight. Thus in your example, the 20" wheel would be experiencing three times as much torque as the 10" wheel because the weights are three times as large.

Please note that I'm answering this purely in terms of the torque. Torque, by definition, is measured as the axle of the wheel. That's quite different from the tangential force that would be felt at the point you labeled "Torque measured here". The force would be torque/radius, and the equations would resort back to something quite similar to the equation Ebaines gave you. In that case, just like in our previous answers, the result would be that the force would decrease as the diameter got larger. (But in your specific example the doubled diameter would be more than compensated for by the tripled weight).

The short answer is, wheel two in your specific example would experience more torque AND more tangential force at the rim.

I hope that all makes sense. Let me know if you need me to clarify something.