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Math Help - Vectors

  1. #1
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    Vectors

    < is angle sign in this case.

    C is the resultant of the forces, 2A and 3B, that is C=2A+3B.
    If A=300N<43 deg. and C =100N<97 deg. , use the component method to determine the magnitude and direction of B.

    I am not sure if this problem is done correctly. Here is my attempt:

    Ax = 600 cos 43 = 438N
    Ay = 600 sin 43 = 409 N

    Bx = ?
    By = ?

    Cx=100 sin 43 = 68 N
    Cy=100 cos 43 = 73 N

    438 + Bx= 68
    409 + By = 73

    Therefore
    Bx= -370N
    By= -336N

    B = sqrt( (-370)^2 + (-336)^2) So B = 499N

    Is this correct? What about the 3B?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    your method is basically correct, however it would be the result

    If A + B = C

    You have 2A + 3B = C

    so rework 2Ax + 3Bx = Cx

    etc
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  3. #3
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    Quote Originally Posted by Calculus26 View Post
    your method is basically correct, however it would be the result

    If A + B = C

    You have 2A + 3B = C

    so rework 2Ax + 3Bx = Cx

    etc
    Ok so I did notice one error in my attempt above..This is my second attempt. Does it look correct? Can someone please work it out and let me know

    Ax = 600 cos 43 = 438N
    Ay= 600 sin 43 = 409 N

    Bx=?
    By=?

    Cx=100 sin 7 = 12N
    Cy=100 cos 7 = 99N

    438 + 2Bx = 12
    409 + 2By= 99

    Bx=-213
    By=-155

    B = 263 N

    Theta = 216 deg. from due east(+)
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Ok

    Note Cx should be negative (draw it out)

    You should go with Cx = 100cos(97) = -12

    Cy = 100 sin(97) = 99

    438 + 2B = -12

    B= -225

    which makes theta 211
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