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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.)
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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.

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    $\begingroup$ Half the weight of the pipe... nice. Thank you for the clear and detailed answer. $\endgroup$
    – Cognitive Hazard
    Feb 18, 2015 at 2:30
  • $\begingroup$ Note that both the original premise of the questionand this answer only apply if the bar is horizontal. $\endgroup$
    – Peter Green
    Dec 8, 2015 at 5:37
  • $\begingroup$ @PeterGreen: or if the weight of the bar is negligible compared to the torque required to turn the bolt. It appears that OP understands this. $\endgroup$
    – Ross Millikan
    Dec 8, 2015 at 5:44

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