# How to simulate a torque wrench?

Two-part question here, concerning the real-world application of torque.

I have a bolt on my vehicle that I have to tighten to about 200 ft-lbs. But my torque wrench is only calibrated for up to 150 ft-lbs.

If I sheathe the end of my regular ratchet with a 4-foot "cheater pipe", and then hang a 50 lb weight at the end of that 4-foot moment until the bolt stops turning, I believe the torque will then be greater than 200 ft-lbs (because the 4-foot galvanized steel pipe itself has significant weight.)

My two-part question is:

1. Is my intuition for accurately simulating a torque wrench correct?
2. How might I account for the weight of the (straight, uniform) pipe in a manner that is physically correct? (This seems like possibly an integration problem.)

Yes your intuition is correct. Absent the weight of the pipe and your ratchet, you will produce 200 ft-lbs this way. I suspect the weight of the pipe is a small error compared to others in the system, but if we model the pipe as a uniform beam we can check. Let the mass of the pipe be $m$ lbs. The mass of a small length $dx$ is then $\frac m4 \ dx$ lbs. The torque applied is then $\int_0^4 \frac m4gx\ dx=mg\frac{x^2}8|_0^4=2mg$ As your english scale reads $mg$, this is twice the reading on your scale. If the pipe weighs $2$ lbs, it is only a $2\%$ error. You can reduce your $50$ lb weight to $50-\frac m2$ lb to compensate, if you wish.