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More Forces with some trig; I don't know where the mistake is

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Mass m sits on a frictionless, horizontal table. Assume:
    - All vertical forces acting on m (such as gravity and the normal force) sum to zero.
    - All applied forces act parallel to the table
    When a force of magnitude F is applied to mass m, it accelerate at magnitude a = 4.49 m/s2.

    NOTE: The axes lie along the table.

    Suppose force F1 has magnitude 2F and points at 45 degrees to the x-axis. If forces F1 and F2, with a magnitude F, act on mass m at the same time, what will the magnitude of its acceleration be now?


    2. Relevant equations
    Components


    3. The attempt at a solution
    I attempted to solve this by drawing a diagram, and F1 is pointing out at 45 degrees relative to the x-axis. I found its components, with the x-direction vector being 8.98cos(45) = 6.3498. F2 is already in that direction as well, so I just added the two vectors to get a sum of 10.8398. When I entered this answer as 10.84 into Webassign (online homework - I don't know who is familiar with it), it said that the answer is incorrect. Is it that I didn't put enough digits, or did I solve the problem incorrectly?
     
  2. jcsd
  3. Feb 20, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hi iJamJL! welcome to pf! :smile:
    but that's only the x-component of the total force …

    what about the y-component? :wink:
     
  4. Feb 20, 2012 #3

    jedishrfu

    Staff: Mentor

    your description is a bit confusing you probably need to draw a force diagram to show what you mean. You mention one force F that accelerates the m horizontally and then shift
    into F1 and F2 which I thought were components of a second force but then I got lost when you said F1 has a magnitude of 2F and that F1 and F2 have a magnitude of F?
     
  5. Feb 20, 2012 #4
    Re: welcome to pf!

    Thanks for the welcome!

    Hm, I see what you mean, tiny-tim. Let me see if I can get this right.

    You're saying that I've got the x-component down for the F1 where it is 8.98 as the resultant. The x-component becomes 6.3498, and the y-component becomes 6.348. I can then carry these both on with the 4.49 from F2, making the vectors like this:

    -> 10.8398
    ^ 6.348

    Then the resultant vectors, with the Pythagorean theorem, results to 12.5618. Is this correct?
     
  6. Feb 20, 2012 #5
    They do not give any values for the forces. That is how the problem is given.
     
  7. Feb 20, 2012 #6

    tiny-tim

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    yes, that's correct :smile:

    (though it would be better, and easier, if you added the forces

    2F/√2 + F and 2F/√2 :wink:)​
     
  8. Feb 20, 2012 #7
    You lost me there, tiny-tim. LOL but I understand it the other way, so I guess we could let it rest. Thank you!
     
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