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Two ropes in a vertical plane... sin(theta) or cos(theta)?

  1. Aug 25, 2015 #1
    1. The problem statement, all variables and given/known data:
    Two ropes in a vertical plane exert equal-magnitude forces on a hanging weight but pull with an angle of 86 degrees between them. What pull does each one exert if their resultant pull is 372N directly upward?


    2. Relevant equations:
    x-component= magnitude*cos(theta)
    y-component= magnitude*sin(theta)
    Cy=Ay+By

    3. The attempt at a solution:
    Vectors A, B, and resulting vector C are all in the same plane. If they pull equally, we see the angle between A or B and C is 43 degrees. The x-direction forces cancel, so we are left with the y-direction. If we let A=B, we can simplify. If we add vectors A and B together in the x-direction, we get C=Ay+By, and we can equate it with C=372N. So I have 372=2*magnitude*sin(theta). However, the equation the book uses is 372=2*magnitude*cos(theta). Why? Shouldn't it be sin because it's in the y-direction?
     
  2. jcsd
  3. Aug 25, 2015 #2

    PeroK

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    You need a diagram showing where you have put ##\theta## in relation to the ropes.

    You're equations above in section 2 are only valid where ##\theta## is the angle with the horizontal.
     
  4. Aug 25, 2015 #3
    Ah, so I rotate the x,y axis by 90 degrees and I'm left with cos(theta).

    If I wanted to keep the x,y axis in the traditional location, is there an easy way to modify the equation?
     
  5. Aug 25, 2015 #4

    PeroK

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    You'd be better off remembering that sin = opposite/hypotenuse and cos = adjacent/hypotentuse. A triangle, in general, could have any orientation, so its sides will not always be aligned with the x and y axes.
     
  6. Aug 25, 2015 #5
    Okay.

    Is there a way to make it work using x-component=magnitude*cos(theta) and y-component=magnitude*sin(theta) if I use two different angles (47 and 133)? I had A use an angle of 133 and B use 47. Then since they cancel each other out, I did magnitude(A)cos(theta)= -magnitude(B)cos(theta) and got 2magnitude(A)=magnitude(B). Then I had magnitude(A)sin(47)+magnitude(B)sin(133), subbed using 2magnitude(A)=magnitude(B) and ended up with magnitude(A)=169N. A would then have a magnitude of 338 and B would have a magnitude of 169. Obviously that's not correct, but I can't tell if it was an assumption or it's just not a viable method.

    Edit: This is because it assumes the angle is always the angle from the object it's working on and not just from an axis, right?
     
  7. Aug 25, 2015 #6
    You need to draw a picture

    You will find the picture consists of two-right triangles, of exact same dimensions. You know length of the base, you know the angles, calculate the hypotenuese.
     
  8. Aug 25, 2015 #7

    BvU

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    yes
    wrong twice.

    ##|\vec A| \cos\theta_A = -|\vec B|\cos\theta_B\Rightarrow |\vec A| = |\vec B|## because ## \cos\theta_A = -\cos\theta_B##, and where the 2 comes from I don't know.

    Edit: This is because it assumes the angle is always the angle from the object it's working on and not just from an axis, right?[/QUOTE]

    Try again. And: You did make a drawing I hope ? Drawings are very, very useful.


    AddVecs.jpg
     
  9. Aug 25, 2015 #8
    I must've mistyped on my calculator because I did cos(47)/-cos(133) and got 2. Calculator memory is cleared so I'm not sure what I really typed in. So using the correct substitution, that the magnitudes are equal, I got the right answer. Thank you so much everyone, you've been really helpful!

    We don't need the angle from the combined vector because it's a combination of two vectors that we already know the positions of and not a separate and unrelated vector, right?
     
  10. Aug 25, 2015 #9

    BvU

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    In the problem statement you are given that the resultant pull is directly upward. So there is the angle of the sum vector. You use this when you write ##
    |\vec A| \cos\theta_A + |\vec B|\cos\theta_B = 0 ##: the 0 is the horizontal component of the resultant.
     
  11. Aug 25, 2015 #10
    Okay, thank you.

    One more question, in this problem's case, the expression for the x-component would be magnitude*tan(theta), right?
     
  12. Aug 26, 2015 #11

    BvU

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    No. The force points upwards, the ##\theta## you have chosen is wrt the X+ axis ("to the right") so it is ##\pi/2##. Just like with the others, the x component is ##R\cos\theta##, in this case 0.

    Try it out with a resultant at e.g. ##\pi/4##
     
  13. Aug 26, 2015 #12

    BvU

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    Coming back to this one. You are supposed to know and use the fact that ##\cos \phi = - \cos (\pi - \phi)##. No calculator should be needed - no typos on that thing can mess it up for you either.

    In case you do have to resort to using a calculator: never ever trust the result. So estimate the result any which way you can before you start hammering on the thing.

    And the most-often incurred mistake is typing degrees when the setting is for radians (and sometimes vice versa)
     
  14. Sep 2, 2015 #13
    Yes, thank you.
     
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