# Calculating the force required to lift a weight with a screw

I am trying to learn physics by myself as I do not have a good teacher at school. I've been reading Feynman Lectures on Physics and I can't figure out how he found out this number. Here is an excerpt of the book:

Let us now illustrate the energy principle with a more complicated problem, the screw jack shown in Fig. 4–5. A handle $20$ inches long is used to turn the screw, which has $10$ threads to the inch. We would like to know how much force would be needed at the handle to lift one ton ($2000$ pounds). If we want to lift the ton one inch, say, then we must turn the handle around ten times. When it goes around once it goes approximately $126$ inches. The handle must thus travel $1260$ inches, and if we used various pulleys, etc., we would be lifting our one ton with an unknown smaller weight $W$ applied to the end of the handle. So we find out that W is about $1.6$ pounds. This is a result of the conservation of energy.

If you divide $2000$ pounds by $1260$ in you get $\frac{100}{63}$, which is $1.58$, a number close enough to $1.6$ pounds, which according to Feynman is the weight that we need to apply to the end of the handle. I believe, however, that this calculation is wrong given a dimensional analysis of such operation.

Here is my dimensional analysis:

$\frac{2000 pounds * 1 in}{ 1260 inches }$ = $\frac{[M] * [L] }{[L]}$ = ${[M]}$ And this is wrong, as we are suppose to get a force, not a mass.

I found this website, where you can find what I think is an email from an student reporting the teacher that he believes he got the number diving $\frac{5}{\pi}$ . Can you also explain me this?

Finally, in a thread in a Physics Forum question,they concluded that $1.6$ is just the force to keep the system at equilibrium, and you need $\text{a force} > 1.6$ to actually lift the ton. After, someone said that Feynman would have said this answer is wrong. Why is that?

• He's ignoring friction, so it's just an energy balance. The energy put into lifting the weight is what? 2000 lb * 1 inch = 2000 inch-lbs. That handle must be pushed 1260 inches to get the same energy, and 1260 inches * 1.6 lb ~ 2000 inch-lbs. What's this dimensional analysis you're talking about? Dec 3, 2014 at 2:43
• @MikeDunlavey Reviewing my notebook it seems that I misunderstood what a pound was. English is not my native language, so when I think of pounds, I think in mass, not weight. Therefore, my dimensional analysis was completely wrong. Thanks for your answer. Dec 3, 2014 at 3:21
• @nonameable - Ah yes, the joys of the Imperial system, where you can have pound (mass) / poundal (force) or pound (force) / slug (mass). Metric really is simpler, even if you worry about cgs / mks. Dec 3, 2014 at 4:17
• I was also confused by the same thing Mike brought up. You were right on the verge of writing a really great question, and then you left out the most important part: the dimensional analysis that makes you think the calculation was wrong. (So I'd say your question is "only" good, not great!) Would you consider adding that into the question? (Don't mark it with "EDIT:...", just edit it right into the middle as if you'd written it there in the first place.) Dec 3, 2014 at 13:06
• The trouble with this problem is ignoring friction yields a way-wrong answer (set this up with anything like normal materials and the force required to break it is less than required to spin it down if applied where the load is) yet there is not enough information to determine the force required to overcome screw twist friction. Dec 3, 2014 at 16:24

Work is calculated as force times distance. $$W = Fd$$ The purpose of a simple machine like a screw jack is to lessen the force required. However, the work needed is still the same, so the distance over which you exert the force has to increase. Halving the force requires doubling the distance.

In this problem, you want to lift 2000 lbs a distance of 1 inch. So, the amount of work is 2000 inch-lbf (lbf = pound-force, or the amount of force required to lift one pound). To lift the weight with the screw requires 10 turns, so with a handle of 20 inches, this is a distance of $10*2\pi*(20 in) \approx 1256\,in$. Putting it all together: $$W = Fd$$ $$2000in \cdot lbf = F*(1256 in)$$ $$F = \frac{2000in \cdot lbs}{1256 in} = 1.6 lbs$$ Notice that this force works for any lifting distance. If you want to lift the weight 2 inches, then you have to do twice the work over twice the distance, which results in the same force.

Those who say that you need more than this force know that you need a short push to get the screw turning from a dead stop. This is similar to the fact that, even though you don't need any force to keep an object moving, you have to give it a push to get it moving in the first place. Once you've got the screw turning, you only need 1.6 lbs of force to keep turning it (ignoring friction).

For the $5/\pi$ answer, it amounts to the same thing, just doing the math in a different order: $$F=\frac{2000 in\cdot lbf}{10*2\pi*20in}=\frac{2000}{400\pi}\,lbf =\frac{5}{\pi}\,lbf = 1.59\,lbf$$

From the geometry, you can state that in order to move the screw by 1 unit of distance, you have to move the end of the handle by $10\times 20\times 2\pi$ units of distance. Let's call the unit of distance $[L]$ - in this case, an inch might be a good unit but we don't have to be explicit about that.

Conservation of energy says that work done on the system in one place must be equal to the work done by the system in another place - work done on moving the handle must equal work done to move the mass.

Mathematically, we can say

$$F_1d_1 = F_2d_2$$

so for a given force at the end of the handle, you know the force it can move on top of the screw is given by

$$F_{screw} = \frac{400 \pi [L]}{1[L]} F_{handle}$$

Now you can express the force in pounds, Newtons, dyne, kgf, kip, pond, poundal, or whatever unit of force takes your fancy. Since the ratio of distances becomes unitless (the inches, in this case, cancel out) you can see that the two forces will have the same units. Dimensional analysis, in other words, works (it usually does).

You can see by simple manipulation that if the force exerted by the screw is 2000 units, the force on the handle must be $\frac{5}{\pi}$[units] as was stated.

As for the question of equilibrium vs motion: if the system is frictionless, then you need an infinitesimally small additional amount of force to set the system in motion (to accelerate it from rest) after which you can reduce the force to be exactly equal to $\frac{5}{\pi}$[units] in order to keep moving. At the end you can reduce the force slightly and the screw will come to rest. The total work done to move the mass up is independent of how much the mass was accelerated as long as it it at rest at the beginning and end of the exercise - and thus the average force will be exactly as calculated.

As David Richerby pointed out, because there is no friction, you need to maintain the force at all times in order to prevent the screw from spinning down again - but no work is done (in the physics sense) for just holding it still.

• "no work is done (in the physics sense)" - I think your muscles would have different opinion :-) since they will expend energy for the isometric "non-work" as well... :-) Dec 3, 2014 at 14:59
• @Tomas that is why I added "in the physics sense". Mechanical work requires displacement - but maintaining the position with your muscles would require effort. Biomechanical effort. Dec 3, 2014 at 18:37