Finding the work done using force vectors in 3 dimensions

  • Thread starter marthkiki
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  • #1
marthkiki
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Homework Statement


Find the work done by a force of 3 newtons acting in the direction 2i + j + 2k in moving an object 2 meters from (0, 0, 0) to (0, 2, 0).


Homework Equations


Commonly known vector operations.


The Attempt at a Solution



I found the displacement vector to be 2j.
My problem is figuring out how to represent the direction and magnitude of force as an actual vector.
The answer is (6*SQRT 5)/5 Joules. I don't understand how.
 

Answers and Replies

  • #2
Amok
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Are you sure that is the right answer?
 
  • #3
marthkiki
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Are you sure that is the right answer?

Positive. This really doesn't make any sense to me.
 
  • #4
LCKurtz
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Homework Statement


Find the work done by a force of 3 newtons acting in the direction 2i + j + 2k in moving an object 2 meters from (0, 0, 0) to (0, 2, 0).


Homework Equations


Commonly known vector operations.


The Attempt at a Solution



I found the displacement vector to be 2j.
My problem is figuring out how to represent the direction and magnitude of force as an actual vector.
The answer is (6*SQRT 5)/5 Joules. I don't understand how.

I think that "answer" was gotten using sqrt(2+1+2) to normalize the vector instead of correctly using sqrt(22+12+22). Typo in your answer book.
 
  • #5
marthkiki
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I think that "answer" was gotten using sqrt(2+1+2) to normalize the vector instead of correctly using sqrt(22+12+22). Typo in your answer book.

That may well be true. If so, is the correct answer 2?
The displacement vector is 2j.

(2i + j + 2k) dot product 2j = 0 + 2j + 0 = 2j. So the correct answer is 2 Joules, correct?
 
  • #6
Amok
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That's what I got.
 
  • #7
marthkiki
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Wait where does the magnitude of 3 N come into play? Shouldn't it be multiplied to 2j to get the total work? So you get

W = F*AB
AB = 2j
F = 3(2i + j + 2k) = 6i + 3j + 6k

F dot product AB = 0 + 6j + 0 = 6j. So the work should in fact be 6 Joules, correct?
 
  • #8
Amok
256
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No, you only have to take into account the magnitude of the force in the (0,1,0) direction, because that is that component that is going to be doing all the work! Remember that:

WORK = FORCE dot product DISPLACEMENT
 
  • #9
LCKurtz
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Wait where does the magnitude of 3 N come into play? Shouldn't it be multiplied to 2j to get the total work? So you get

W = F*AB
AB = 2j
F = 3(2i + j + 2k) = 6i + 3j + 6k

F dot product AB = 0 + 6j + 0 = 6j. So the work should in fact be 6 Joules, correct?

No. A force of 3N is represented by a vector of length 3. Your vector specifying the direction of the force just happens to have length 3 already. That is just a lucky break for this problem. Normally what you would do is multiply a unit vector in the direction of the force by 3.
 

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