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Cosine theta in work

  1. Aug 2, 2011 #1
    Hello, I was wondering if the cosine of theta, in the work equation, related to the angle of the applied force?
     
  2. jcsd
  3. Aug 2, 2011 #2

    Doc Al

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    Staff: Mentor

    The angle theta is the angle between the force and the displacement. Examples: If they are in the same direction, the angle is 0 and the cosine is 1; If they are opposite, the angle is 180 degrees and the cosine is -1.
     
  4. Aug 2, 2011 #3
    If you understand dot products then I find it easier to understand that form of the equation for work. That way you understand why there is a cosine in the expression and what angle they are referring to.
     
  5. Aug 2, 2011 #4
    "It is important to note that work is an energy transfer; if energy is transferred to the system (object),W is positive; if energy is transferred from the system, W is negative."

    I understand the transfer of energy to a system (object), but what about the transfer of energy from a system? Could someone give me an example. Thank you
     
  6. Aug 2, 2011 #5

    Doc Al

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    Imagine a block sliding along a surface. There is friction slowing it down. The work done by friction on the object is negative, since the displacement and the force are in opposite direction. Since the work is negative, energy is being removed from the car. The kinetic energy of the car is being transformed into thermal energy via the friction.
     
  7. Aug 3, 2011 #6
    Ive seen this relation cos(theta+90) in Work problems. Ive seen two examples that show a perpendicular force (like the force of gravity) and the object is moving at an angle theta. can someone explain this relation?
     
  8. Aug 3, 2011 #7
    Well you want the force and distance to be on the same axis of movement. Thats where the cosine comes from because it would give you the vector's component that you want.

    If i understand you completely then you have theta being the angle between the vector and ground rather than between the two vectors. If so then you'd want the vertical component of the displacement vector. In this case that would be equal to sin (theta) which is the same as cos( theta + 90).
     
  9. Aug 3, 2011 #8
    Chunkysalsa: Its like having the displacement vector at an angle theta being protected on the force vector?
     
  10. Aug 3, 2011 #9
    sin(theta) does not always equal cos(theta+90). Depends on theta.
     
    Last edited: Aug 3, 2011
  11. Aug 4, 2011 #10
    I do not know if this is right or if it will help the discussion between you to, but I believe that cosine theta, like all you have been saying, is the angle between the two vectors. But I think I have figured it out: the force component moves along the angle of inclination. (Is that right?)
     
  12. Aug 4, 2011 #11
    Bashyboy: The problem say that and object of mass m is moving at a 45 degree angle and the force acting on it is the force of gravity. Find the work.

    W=mg*d*cos45 <---this is what i understand, because the angle between them is 45 but is wrong!

    W=mg*d*cos(45+90)<--- this is what is correct and i can't figure out were does de +90 comes from.
     
  13. Aug 4, 2011 #12
    Theta is the angle between the force vector and the displacement vector. The angle give is the angle between the x-axis and the displacement vector.

    The rest is simple trigonometry. You want the component of displacement in the y-axis. However since gravity pushes down and the displacement is upward, the work should be negative. So the answer should be mgd sin (-45). Using some trig identities you could also write that as either -mgdsin (45) or as you said mgd cos (45 +90).

    Sorry I made a mistake earlier. It helps if you draw out the problem, you'll see the answer clear as day.

    You could also solve it using vector notation. Fg = -mgj, D = .71di + .71dj. Then the work is -.71mgd by taking the dot product. [.71 = sin/cos (45)]
     
  14. Aug 4, 2011 #13

    Doc Al

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    You need the angle between the force and the displacement vectors. The force is gravity, which acts down and thus 90 degrees below the x-axis. The displacement is at a 45 degree angle above the x-axis. What's the angle between those vectors? (It's not 45 degrees!)
     
  15. Aug 4, 2011 #14
    Chunkysalsa: Thank you very much. I was able to solve it. I kind of got to understand via using sin(-45). But i understood it completely using the vector notation.
     
  16. Aug 4, 2011 #15
    Doc Al: Thank you as well. You made me see the big and obvious picture!
     
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