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Find Magnitude & Direction for acceleration

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    Block B has acceleration of 4 m/s2... Relative acceleration of block A w/ respect to B is 4 m/s2. Find magnitude & direction of accel for A?

    2. Relevant equations
    a_A = a_B + a_A/B

    x_A = x_B + x_A/B

    y_A = y_B + y_A/B

    3. The attempt at a solution

    x & y components:
    -4cos(20) = -3.76
    -4sin(20) = -1.37

    -3.76 = 4 + x_A/B

    x_A/B = -7.76

    y_A/B = -1.37

    |a| = sqrt( -7.76^2 + -1.37^2 ) = 7.88 m/s2

    direction: tan -1 * (-1.37/-7.88) = 10.01 degrees
     

    Attached Files:

    Last edited by a moderator: Feb 2, 2015
  2. jcsd
  3. Feb 1, 2015 #2

    NascentOxygen

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    Hi ladkee. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    Check above.
     
    Last edited by a moderator: May 7, 2017
  4. Feb 2, 2015 #3
    what is the next step to find the magnitude of acceleration for a_A, are you saying use (-3.76, -1.37) as the x&y components?
     
  5. Feb 2, 2015 #4

    NascentOxygen

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    I just added an extra line to my post. Can you fix your line I've marked with a cross?
     
  6. Feb 2, 2015 #5
     
  7. Feb 2, 2015 #6

    I'm not sure what you mean by that? .....so are you saying for the x & y components I need to use are (-3.76, -1.37). How do you find the magnitude & direction for acceleration for this particular problem? please help, I need a clear response.
     
    Last edited by a moderator: May 7, 2017
  8. Feb 2, 2015 #7

    NascentOxygen

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    Those are the x and y components of the smaller block's relative acceleration, yes. You worked them out.

    I pointed out that you incorrectly applied the formula (which I highlighted by writing it in red). Can you apply that formula correctly to determine the x component of a's acceleration? So I'm saying there is something wrong in how you substituted values into that formula aa = ab + ...
     
  9. Feb 2, 2015 #8


    so should it be something like .... a_A = (4*i + (- 3.76*j) ) + ( 0*i + (-1.37*j) ) = 0.24*i - 1.37*j
     
  10. Feb 2, 2015 #9

    NascentOxygen

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    You can do it that way if you wish. Next, you're asked to express as a magnitude and direction ....
     
  11. Feb 3, 2015 #10


    Okay, for magnitude I did: |a| = sqrt ( (0.24^2) + (-1.37^2) = 1.39 m/s2

    and direction I did inverse tangent of (y/x) which is ( -1.37 / 0.24 ) = - 80.06 deg ....which I'm not sure is right
     
  12. Feb 3, 2015 #11

    NascentOxygen

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    Those look to be around what I was expecting.
     
  13. Feb 6, 2015 #12

    NascentOxygen

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    You should be able to go back to your first post in this thread, and without doing any further calculations, sketch the triangle showing this vector relation: a_A = a_B + a_A/B

    Hint: The first line in your first post tells you that the triangle will be isoscles.
     
  14. Feb 6, 2015 #13
    I got the right answers, 1.39 & - 80.06, thanks for all ur help, highly appreciated
     
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