given: Force Vector $F = 2i - 3j + k$

force acts on the point $(1,5,2)$

line $\frac{x}{2} = y = \frac{z}{-2}$

Known: $T = n * (r X F)$

'n' is a unit vector in the direction of the given line

r is the position vector

vectors, so this is a dot and cross product respectively

Find: torque about the given line

I'm not used to dealing with lines in 3D space defined in this manner, so I'm a bit stumped on how to find a unit vector on the given line.

  • $\begingroup$ How do you know which of two orientations to use for $n$? If you’ve only interested in the magnitude of the torque, then it doesn’t make any difference, but torque also has a direction. $\endgroup$ – amd Feb 15 '19 at 0:59

If your line is defined as $y=x/2=-z/2$, then moving one unit along $y$ must take you 2 in the $x$ direction and $-2$ in the $z$ direction. So, $(2,1,-2)$ is a vector parallel to the line. The magnitude of that vector is $\sqrt{4 + 1 + 4} = 3$, so a unit length vector in the same direction would be, $$ \frac{1}{3} (2,1,-2) $$

  • $\begingroup$ cool, thanks. another problem now: when I use this unit vector, I get a torque of 5sqrt(6), but the answer is 17 units of torque. r is (1,5,2) rXF is (11,3,-13) and using this unit vector, n*(rXF) is 5sqrt(6). Where am I going wrong? $\endgroup$ – physicsTom Feb 14 '19 at 22:33
  • $\begingroup$ @physicsTom You’re going wrong by using the incorrectly-computed unit direction vector from this answer. $\endgroup$ – amd Feb 15 '19 at 1:14
  • $\begingroup$ Since the line passes through the origin, if $(1/2,1,-1/2)$ is indeed its direction vector, then that point must lie on the line. However, $x/2=1/4$ and $y=1$, so that test fails. $\endgroup$ – amd Feb 15 '19 at 1:16
  • $\begingroup$ @amd Thanks for catching me. I have corrected my answer, although I'm just as sleepy now as I was then so maybe it's still wrong. ;) $\endgroup$ – Display Name Feb 15 '19 at 16:44
  • $\begingroup$ Yay! Thank you guys. It's been a while since I worked with vectors, so I'm trying to get the wheels turning again. Really appreciate the help:) $\endgroup$ – physicsTom Feb 15 '19 at 22:02

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