Simple physics problem

  • Thread starter Rasine
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  • #1
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Three forces act on a particle that moves with a constant velocity v = (2 m/s) x - (7 m/s) y. Two of the forces acting on this particle are F1 = (-9 N) x + (10 N) y + (2 N) z and F2 = (6 N) x + (1 N) y + (9 N) z. What is the magnitude of the third force?

so what i was trying to do is take the f1z component and the f2z compoment and finnding the magnitude by doing squroot(9^2+2^2)....but that is not how you do it...

can someone please point me in the right direction?
 

Answers and Replies

  • #2
Doc Al
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First find the components of the third force, considering that the net force on the particle must be zero.
 
  • #3
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so if the net force does not change the velocity then f1 and f2 should yeild 0 right?
 
Last edited:
  • #4
cristo
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No. You need to find the components of the third force; F3. Since the acceleration is zero in each direction, you know that the sum of the components of the forces in each direction must be zero; i.e. f1x+f2x+f3x=0... etc.
 
  • #5
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so how would i go about doing this.... f1 and f2 should be equal right? um...so the magnitude of f3 should make be the variable that makes f1 and f2 =0 right?
 
  • #6
cristo
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Don't think about the magnitude yet. f1 and f2 are not equal; you are given the values of them. Set up three equations, f1x+f2x+f3x=0, and the other similar two. Then you know the values for f1x and f2x, and so obtain -9+6+f3x=0. This will give you the x component of f3. Use similar equations for the y and z components.
 

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