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Calculating work done with angles

  1. Jan 7, 2012 #1
    Hi all.

    I am trying to learn more on physics, and am at the subject on calculating work done. The formula to calculate work done is Work done = Force * Dislacement * consine(theta).

    I am on a question which explains: a 100N force is applied at an angle of 30 degrees to the horizontal to move a 15kg object at a constant speed for a horizontal distance of 5 metres.

    The answer given says "W = (100 N) * (5 m) * cos(30 degrees) = 433 J".

    I would like to know how why you get 433 joules? You multiply 100 by 5 to get 500, but how does multiplying by 30 give you 433. I am up to speed on realising that the work done should be less that 500 joules because of the angle, but the formula states you multiply the angle (30), so how do you multiply the angle to get a lower value than 500?

    Does every one follow what i'm asking?

    Thankyou.
     
  2. jcsd
  3. Jan 7, 2012 #2
    [itex]\ W = F * s * cos(θ) [/itex]

    [itex]\ W = 100 * 5 * cos(30) [/itex]

    [itex]\ W = 100 * 5 * 0.866... [/itex]

    [itex]\ W = 433J [/itex]
     
    Last edited: Jan 7, 2012
  4. Jan 7, 2012 #3
    The cosine can never exceed 1, so you're right, the maximal work is 5m*100N if the force is applied colinear to the displacement. The cosine of any other angle than 90° is smaller than 1, thus the resultant work is smaller than 500 J.
     
  5. Jan 7, 2012 #4
    Thanks guys.

    But, rollcast, how did you find the number 0.866 from a 30 degree angle???
     
  6. Jan 7, 2012 #5
    Because cos(30) = 0.866
     
  7. Jan 7, 2012 #6
    Do you have a calculator? Locate the 'cos'-button on it. Type '30', then press the 'cos' button!
    If the result comes out to something else than 0.866 verify that your calculator is in degree-mode (e.g. "DEG" instead of "RAD").
     
  8. Jan 7, 2012 #7
    Oh thanks. I didn't realise it was something you couldn't work out in your head.
     
  9. Jan 7, 2012 #8
    ..well, you could work out by hand that [itex]\cos{(30^{\circ})} = \frac{\sqrt{3}}{2}[/itex]
    But that would require that you know what the cosine actually is.
    Anyway it always a good idea to know where the formulae you use actually come from when doing physics.
     
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