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Homework Help: Finding the horizontal accerelation using forces

  1. May 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Interactive Solution 4.11 offers help in modeling this problem.

    Two forces, vector F 1 and vector F 2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 30.5 N and F2 = 47.5 N. What is the horizontal acceleration (magnitude and direction) of the block?

    link to picture (let me know if it works): http://www.webassign.net/CJ/04_11.gif

    2. Relevant equations
    SOHCAHTOA equations
    Newton's 2nd law: F=ma

    3. The attempt at a solution
    Because F1 is at an angle, I broke it up into its y and x components:

    sin(-65)*30.5 = -27.6 N
    cos(-65)*30.5=-12.89 N

    Since the question only asks for the accerelation in the x direction, I got the following equation:
    (-47.5 N+-12.89N)=m*a=5*a

    Then I solved for a and got 12.078 m/s^2 in the x direction. But the website of my homework is telling me this is wrong. help?
    Last edited: May 29, 2009
  2. jcsd
  3. May 29, 2009 #2


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    Homework Helper

    Except your drawing shows the direction of the force F1 to be positive x.

    This means that the horizontal component of F1 is + and that opposes the direction of F2.

    Apparently you have taken the sum where the difference would seem to be the correct approach.
  4. May 29, 2009 #3
    i did that too but the computer is telling me thats the wrong answer as well. is it possible that 65 is the wrong angle to use ?
  5. May 29, 2009 #4
    Your question is asking for acceleration. You are giving force as an answer.
  6. May 29, 2009 #5
    The horizontal component of the force F1 and the force F2 are in opposite directions, therefore you have to substract 12.89 N from 47.5 N, and divide that net force by the mass to get the magnitude of the acceleration. The direction is obviosly to the left since F2's magnitude is larger than that of F1s horizontal component.

    This is my first post, I hope I helped, great forum btw.
    Last edited: May 29, 2009
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