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Angle between a force and radius?

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data
    A force, F=(6i-2j)[N], moves a particle through a distance deltaR=(3i+1j)[m]

    A. Find the work done on the particle

    B. Find the angle between F and deltaR


    2. Relevant equations
    W=F*D


    3. The attempt at a solution

    A. F=Root(6^2+2^2)=6.32N

    R=Root(3^2+1^2)=3.16m

    W=FD=(6.32*3.16)=20.00N/M

    I can't figure out B though. By the way, is my A correct? Thanks!
     
  2. jcsd
  3. Dec 12, 2007 #2

    rl.bhat

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    W=FD=(6.32*3.16)=20.00N/M It is true if they are in the same direction.
    W= F.deltaR = (6i-2j).(3i+1j) = 18 - 2 = 16J
    And angle is given by cos(theta) = F.d/FD
     
  4. Dec 12, 2007 #3
    I'm kind of confused. Did you foil those out? I was taught to use pythag. theorum for component vectors. I'm also kind of confused on F.D/FD. What does this mean?
     
  5. Dec 12, 2007 #4
    I think B can be found if you find the dot product of the two vectors...
     
  6. Dec 12, 2007 #5

    rl.bhat

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    In vector you can perform scalar product and vector product. If Fand D are the force and dispalcement vectors then the work done = F.D and angle between the vectors =F.D/FD Bold letters indicatre the vectors.
     
  7. Dec 12, 2007 #6
    Oh, so the dot product is called the scalar product around here...
     
  8. Dec 12, 2007 #7
    Oh I think I see what you're saying. So the scalar, or dot product divided by FD is equal to the angle?
     
  9. Dec 12, 2007 #8

    rl.bhat

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  10. Dec 12, 2007 #9
    I think I just confused myself. You said W=F.D, or F multiplied by D. Then, you say that the angle is F.D/FD. If F.D is F multiplied by D, then wouldn't F.D/FD be 1? Since top and bottom would be the same? Or is the top the dot product and and bottom FxD?

    Also, the dot product is:

    lAl*lBlcosx correct?
     
  11. Dec 13, 2007 #10

    rl.bhat

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    F.D = 6i-2j).(3i+1j) = 18 - 2 = 16J It is dot product.
    FD = FD=(6.32*3.16)=20.00N/M
    And cosx = 18/20
     
  12. Jul 22, 2009 #11
    radius

    the diameter of the large and the small piston of a hydraulic lift

    If the ratio of the diameter of the small piston to that of the large piston is 1:2, how does the forco on the small piston compare with that bon the large piston?

    a hydraulic is to be constructed so that a force of 10 newton can lift a load of 6250 newton. if the small piston has a radious of 4 cm, what must be the radious of the large piston?
     
  13. Jul 22, 2009 #12
    radius

    can please somebody help me solve these fallowing problems????

    the area of the small and large pistons of a hydraulic lift are 10cm and 500cm, respectively. What load can be lifted on the large piston if a force of 40 newton is applied in the small piston?

    the diameter of the large and the small piston of a hydraulic lift are 6cm and 75cm, respectively. what force must be applied on the piston if a load of 4000 newton must be overcome by the large piston?

    If the ratio of the diameter of the small piston to that of the large piston is 1:2, how does the forco on the small piston compare with that bon the large piston?

    a hydraulic is to be constructed so that a force of 10 newton can lift a load of 6250 newton. if the small piston has a radious of 4 cm, what must be the radious of the large piston?
     
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