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Resultant Vector - Chin Brace

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Q: Make a scale drawing showing both the forces produced by the cables and the resultant force. Estimate the angle carefully or measure it with a protractor.

    Supplement:
    A person with an injured jaw has a brace below his chin. The brace is held in place by two cables directed at 65 above the horizontal. (See the figure.) The cables produce forces of equal magnitude having a vertical resultant of 2.30 upward.


    I don't have a protractor and would like to go about the estimate mathematically. I am having trouble finding the right and left lengths. I am guessing the resultant vector is length 2.30N and 90 degrees? If someone could walk me through it for understanding that would be great.
     

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    Last edited: Jan 22, 2012
  2. jcsd
  3. Jan 23, 2012 #2
    or if someone could point me in the right direction i have no idea where to begin
     
  4. Jan 27, 2012 #3

    berkeman

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    Staff: Mentor

    Why do you need to estimate some angles if the angles are labeled already for you? I'm missing something there.

    And just draw the components of the two angled forces. The two vertical components need to sum to the total weight of the head, right?
     
  5. Jan 27, 2012 #4
    I need to find the magnitude of the left pulling force and right pulling force. The upward resultant is 2.30N and 90 degrees? The angle of the left is 115 degrees and angle of the right is 65 degrees not sure N force though.
     
  6. Jan 27, 2012 #5

    berkeman

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    So just draw a free body diagram (FBD) for the bottom of the chin brace, showing the resultant force, and the components of the two strings. The vertical components of the string forces must sum to what? And the horizontal components must sum to what?

    Use the angles and basic trig to calculate the magnitudes of the vertical and horizontal string force components. Please show us your work...
     
  7. Jan 27, 2012 #6
    I did the vertical force at 90 degrees by 2.3N. The left diagonal force I did 115 degrees and the right diagonal force I did 65 degrees. I am having trouble finding the magnitude of the left and right forces.

    I tried (2.3)^2 = sqrt(x^2 + x^2) I did x^2 twice because of two equal forces pulling in opposite directions to find each magnitude. I also tried 4.6 for each magnitude. I feel like I don't have enough info or I am missing something.

    Check my image.
     

    Attached Files:

  8. Jan 27, 2012 #7

    berkeman

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    The diagram looks reasonable, except you need to also show the force down due to gravity. What is the magnitude of the downforce due to gravity?

    Write an equation for the sum of the horizontal forces in the horizontal direction. What is the sum equal to?

    Write an equation for the sum of the vertical forces in the vertical direction. What is the sum equal to?
     
  9. Jan 27, 2012 #8
    Good place to start, I am guessing 1N is the downforce due to gravity and that will help me find the horizontal and vertical forceS?
     
  10. Jan 27, 2012 #9

    berkeman

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    First it is important to understand what the sum of the forces in each of the two dimensions must equal.

    Since F=ma, what do the vertical forces need to all add up to? And the horizontal forces?
     
  11. Jan 28, 2012 #10
    We don't have the mass of the head. I set Yleft and Yright to 2.30 and solved for Xleft and Xright in which I got 1.075 and magnitude of 2.54 for each direction (after left = sqrt(x^2 + y^2). That answer was wrong. But the resultant magnitude is 2.30N and the resultant angle is 90 degrees. Are we assuming m = 1 and having F = 9.8? or are we setting 2.30N = m * 9.8? I can find resultant forces when I have the magnitude and angles of each direction already but I am having difficulty without the magnitude.
     
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