Rotational problems

  • #1

Homework Statement



1) A force F = 1.5 i + 2.4 j N is applied at the point x = 3.0 m, y = 0 m. Find the torque about the point x = -1.3 m, y = 2.4 m.

I believe the lever arm should be the distance between the two points. but then the force that acts perpendicular to this is where i cannot get the right answer. How do i do this problem?
 

Answers and Replies

  • #2
Doc Al
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Use the cross product definition of torque:

[tex]\tau = \vec{r} \times \vec{F}[/tex]
 
  • #3
ok so F = 1.5 i + 2.4 j and x = -1.3 m, y = 2.4
i get -3.6 K and -1.95K so is it.. -5.55 k?
 
  • #4
Doc Al
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ok so F = 1.5 i + 2.4 j
OK.
and x = -1.3 m, y = 2.4
No, that's just the point about which you are finding the torque. [itex]\vec{r}[/itex] is the displacement from that point: r = (x2 - x1) i + (y2 - y1) j
 
  • #5
F = 1.5 i + 2.4 j N is applied at the point x = 3.0 m, y = 0 m.
x = -1.3 m, y = 2.4

so the r displacement would be 3-(-1.3) i + 0-2.4 j
so r would be sqrt( 4.3^2 + 2.4^2) ?
 
  • #6
Doc Al
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so the r displacement would be 3-(-1.3) i + 0-2.4 j
Right.
so r would be sqrt( 4.3^2 + 2.4^2) ?
That would be the magnitude of r.
 
  • #7
ok so since i have 4.3 i - 2.4 j how do i begin to find the torque and such?
 
  • #8
Doc Al
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ok so since i have 4.3 i - 2.4 j how do i begin to find the torque and such?
Take the cross-product, as I indicated in post #2.
 
  • #9
so i get 13.92 k is that right?
 
  • #10
Doc Al
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so i get 13.92 k is that right?
Sounds good to me. (Don't neglect the units.)
 
  • #11
ok thanks a lot!

i ask if it was right because i am unable to check my answer.
 

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