3
$\begingroup$

I have a mechanical engineering question that I can seem to figure out.

Basically I have an object this is being pulled down by two forces. Each force is 10N in the downward direction. There is a nut welded to the bottom of a plate and a threaded rod runs through the plate and into the nut.

I can't figure out how much torque will be needed on the threaded rod to lift the the object. Let's ignore friction for the time being. A diagram is included to show what I've got.

It would be really helpful if I could get some insight into how to setup the problem. I did well in my college physics courses but I'm was an applied math major.. 8-) And they really didn't talk about how to convert rotational work into translational work.

enter image description here

$\endgroup$
1
  • $\begingroup$ the other answers are correct, but all you really needed to know is that a screw is just a simple inclined plane. $\endgroup$
    – Tiger Guy
    Mar 29 '20 at 6:35
2
$\begingroup$

Since we're not having to deal with friction, conservation of energy is our best friend. There's no need to "convert rotational work into translational work", merely observe that they must be equal in a stable object. So

$$W_F = W_T$$

where $W_F$ is the work done by the applied forces and $W_T$, by the torque.


A note before we continue: a screw's pitch is usually defined as the axial distance between two threads. However, it can actually be given in two different ways: metric screws usually give you the pitch as a distance (i.e. 1.5 mm), while inch-based screws tend to be given in terms of threads-per-inch.

enter image description here

For the derivation below I'll be using pitch as a distance.


So, if we think of a single rotation of the screw, the plate will move along the rod by a length equal to the screw's pitch $d$ (the linear distance between two threads).* Therefore, we know that

$$W_F = Fd$$

As for the rotational work, we can imagine it is created by a force $f$ applied tangent to the nut, rotating the screw once around the rod's spiral. We can save ourselves some work and consider the length of that spiral equal to the circumference of the rod.

$$\begin{align} W_T &= 2\pi rf \\ &= 2\pi T \end{align}$$

The last steps are now trivial:

$$\begin{align} W_F &= W_T \\ Fd &= 2\pi T \\ \therefore T &= \dfrac{Fd}{2\pi} \end{align}$$

If you instead have the pitch defined in threads-per-distance, the modification is trivial:

$$T = \dfrac{F}{2\pi d}$$

where the torque's distance unit is the distance unit in the pitch's threads-per-distance (usually threads-per-inch, therefore inches).


* Strictly speaking, the dimension $d$ to be used is the lead, not the pitch. The lead is the axial distance covered by one rotation around the screw, while the pitch is the distance between two threads. In most screws, these values will be the same (as in the left side of the image above) and pitch is the more common term, which is why I used it above. However, when using multi-start screws, the lead will be greater than the pitch (right side of the image), and $d$ should be adopted as the lead.

$\endgroup$
0
$\begingroup$

Let's call the diameter of the screw d and and its pitch N (number of threads/ meter).

If we ignore the friction the screw will act just like a ramp or slope of a ramp, and assuming the slope smaller than a few degrees we can say

sin (alpha) =alpha.

The torque you need has to be bigger than

T= 20/ (Npid/)Newton. m

$\endgroup$
1
  • 1
    $\begingroup$ The SI unit for distance is the meter, 'm'. There is no SI unit with a symbol 'M'. All SI units named after a person have their symbols capitalised but are lowercase when spelled out, so 'N' or 'newton'. $\endgroup$
    – Transistor
    Mar 28 '20 at 8:20

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.