0
$\begingroup$

Page 4-5 of The Feynman Lectures, under the topic of conversation of energy, writes this, while it tries to give a feel about conservation principle.

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.

Unfortunately, I'm unable to upload the image here, but you can refer to this shortened link (not for commercial purpose) to my drive https://bit.ly/2OZTXLj

Now, what the thing is, I get that when it goes around once, the tip covers approximately 2πl ≈ 126 inches. And that 10 times is 1260 inches, but what I don't get is, how do you go from there (using any arbitrary number of pulley or any other system) to getting W = 1.6 pounds?

Any help would be appreciated. Thank you.

$\endgroup$
2

1 Answer 1

1
$\begingroup$

If you lift one ton one inch the energy you need is $U=mgh$ = 1 ton * $g$ * 1 inch, where of course $g$ is the gravity. This force has to be balanced by the work $L$ done by the force $F$ pushing the tip and that is $L=$F*1260inches. We can "convert" the force in weight by dividng it by $g$, i.e. $W=F/g$ ($W$ being the weight "equivalent" to the force $F$ in earth's gravitational field...).

By that ($U=L$) we find 1 ton * $g$ * 1 = $F$ *1260 so that $W=F$/$g$=1 ton/1260 which is 1.6 pounds. (In a decimal system this would have been much more straightforward, btw)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.