1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

W = F d cos θ Is Algebra Okay for Work?

  1. Aug 21, 2010 #1
    Greetings all :blushing:

    1. The problem statement, all variables and given/known data
    Greetings all this is the basic type of problem that I am having. W = F d cos θ. This is not for homework this is for my preparation for an exam.


    2. Relevant equations
    it has to do with the Work:: W = F d cos θ



    3. The attempt at a solution
    I have the answer which is given as:

    W = F d cos θ
    = 76.0 N (12.7 m) cos30.0°
    W = 836 J

    The problem is that I do not know how to multiply cos30.0 to know how you get 836 J

    I have tried to turn it into .5 so that it would be = 76.0 N(12.7m) (.5) but when I multiply it all together I get 482 and not 836 J

    Can you please help me figure this out is it the way I am doing the algebra that is not correct or my conversion of the cos 30? Thank you
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 21, 2010 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF.

    The cosine of 30 degrees is NOT 1/2. That's why you didn't get the right answer. Generally speaking, All you're expected to do is to use a calculator to figure out what cos(30o) is.
     
  4. Aug 21, 2010 #3
    Yes but for this exam that I will take I am not allowed to use a calculator. And I really would like to know how to use cosine and multiply with it not only for this but for other things as well. Do you know how to get to get to836 J without a calculator, or is it too hard to compute without a calculator please? Thank you. The examples seem to feel that we can do it on a scratch piece of paper. Thank you for your response
     
  5. Aug 21, 2010 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The sine and cosine of 30° (π/6 radians) and 60° (π/3 radians) are supposed to be easy to remember or derive because in this situation you have a special right triangle known as a 30°-60°-90° right triangle:

    http://en.wikipedia.org/wiki/Special_right_triangles#30-60-90_triangle

    By the way, your algebra and method is fine. You just had the wrong numbers.
     
  6. Aug 21, 2010 #5
    Thank you for the information and the welcome I am still confused as to how to solve the equation. I will work on it some more though. God bless
     
  7. Aug 23, 2010 #6

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Then how did you multiply (76.0 N) * (12.7 m)? I guess you could have done it by hand, but I don't think a physics exam that doesn't allow calculators would ask you to do that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: W = F d cos θ Is Algebra Okay for Work?
Loading...