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Math Help - Law of cosines/vectors confusing

  1. #1
    Senior Member Paze's Avatar
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    Law of cosines/vectors confusing

    Hi MHF.

    I'm trying to calculate angle theta using law of cosines but I am getting varying results of which none make sense to me.

    The problem:

    A log is 10 meters long and weighs 1 ton. A 16 meter long rope is bound to each end of the log and a hook is subsequently hooked to the middle of that rope with the intention of lifting the log, so that each 'side' of the rope becomes 8 meters. The forces y and x (Don't know how to make vector signs) are equally powerful and their sum is 1 ton. The rope also has to endure the pull of x and y regardless of the direction of the pull. Find out how much weight the rope has to support.



    First, let's have a look at my hand-drawn attempt to find angle theta with the law of cosines. I intend on finding angle theta so I can subsequently use that for law of sines in order to find 'side c' (pictured as 10m) using my other unit (ton). I have 0.5 ton on each side and I want to find out how many tonnes side/vector c is (the sum of vectors y and x).



    I calculate angle theta to be 77 but it seems by all accounts that the 'long angle' should be more than 90. Also ignore the sin on my picture, I meant cos and realized later that I was doing sin, so my answer is wrong because of that, I later corrected it to 77..

    Can someone please point me to my error in calculating angle theta (or if I am wrong by assuming that it should be larger than 90) and also tell me if I am doing this problem right at all? I'm not very confident with my approach and it is likely wrong.

    Thanks.
    Last edited by Paze; April 10th 2013 at 10:50 AM.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Law of cosines/vectors confusing

    Your first equation: 10^2 = 8^2 + 8^2 -2(8)(8)\cos \theta is incorrect. The hypotenuse is 16, one side is 10, and the unknown side is  2 \times 8 \cos \theta. Thus from Pythagoras:

     16^2 = 10^2 + (2 \times 8 \cos \thetas)^2

    Rearrange to get:

     \cos \theta = \sqrt {\frac {16^2-10^2} {16^2} } = 0.781, so  \theta = \cos^{-1} 0.781 = 38.7^{\small o}

    Or (I think) an easier way is to consider the attached fig - from it you can see that  \sin \theta = 5/8.
    Attached Thumbnails Attached Thumbnails Law of cosines/vectors confusing-crane.jpg  
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    Senior Member Paze's Avatar
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    Re: Law of cosines/vectors confusing

    Quote Originally Posted by ebaines View Post
    Your first equation: 10^2 = 8^2 + 8^2 -2(8)(8)\cos \theta is incorrect. The hypotenuse is 16, one side is 10, and the unknown side is  2 \times 8 \cos \theta. Thus from Pythagoras:

     16^2 = 10^2 + (2 \times 8 \cos \thetas)^2

    Rearrange to get:

     \cos \theta = \sqrt {\frac {16^2-10^2} {16^2} } = 0.781, so  \theta = \cos^{-1} 0.781 = 38.7^{\small o}

    Or (I think) an easier way is to consider the attached fig - from it you can see that  \sin \theta = 5/8.
    Your answer is consistent with my calculation of angle theta (2*(arcsin(5/8))=77).

    My question is more vector-oriented I suppose. I need to understand if I am calculating the weight-support of the rope correctly using the methods that I have displayed...They seem pretty far-out from the text-book's descriptions..
    Last edited by Paze; April 10th 2013 at 11:56 AM.
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Law of cosines/vectors confusing

    Each rope supports 1/2 ton. That does not mean that the tension in each rope is 1/2 ton; what it means is that the vertical component of the tension in each rope is 1/2 ton. The total tension is the vector sum of the vertical and horizontal components of tension. Given the vertical component of the tension force = 1/2 ton, and the angle theta is 38.7 degrees, the tension in the rope is = (1/2)ton/cos(38.7 degrees) = 0.64 ton.

    And sorry, but the angle I calculate for theta is different that yours. You define theta in your drawing as the half-angle between the ropes, which is how I've used it as well, not the full angle between the ropes.
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    Senior Member Paze's Avatar
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    Re: Law of cosines/vectors confusing

    Quote Originally Posted by ebaines View Post
    Each rope supports 1/2 ton. That does not mean that the tension in each rope is 1/2 ton; what it means is that the vertical component of the tension in each rope is 1/2 ton. The total tension is the vector sum of the vertical and horizontal components of tension. Given the vertical component of the tension force = 1/2 ton, and the angle theta is 38.7 degrees, the tension in the rope is = (1/2)ton/cos(38.7 degrees) = 0.64 ton.

    And sorry, but the angle I calculate for theta is different that yours. You define theta in your drawing as the half-angle between the ropes, which is how I've used it as well, not the full angle between the ropes.
    I see how people might misinterpret my drawing. My angle theta is actually the full angle. The line that appears to be dividing the triangles is actually a part of the head of the 'arrow' on that vector, lol.

    But thank you for your answer. It seems consistent with my book's answer. Can you tell me why you divide (1/2)ton/cos(38.7degrees)? What does that do exactly?
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    MHF Contributor ebaines's Avatar
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    Re: Law of cosines/vectors confusing

    If you consider the tension as being the vector sum of the horizontal and vertical components, so that  \vec T = \vec T_x + \vec T_y, then from the attached drawing you can see that  \cos \theta = \frac {T_y} T. Rearrange to get  T = \frac {T_y}{\cos \theta}
    Attached Thumbnails Attached Thumbnails Law of cosines/vectors confusing-crane2.jpg  
    Last edited by ebaines; April 10th 2013 at 01:39 PM.
    Thanks from Paze
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    Senior Member Paze's Avatar
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    Re: Law of cosines/vectors confusing

    Quote Originally Posted by ebaines View Post
    If you consider the tension as being the vector sum of the horizontal and vertical components, so that  \vec T = \vec T_x + \vec T_y, then from the attached drawing you can see that  \cos \theta = \frac {T_y} T. Raarrange to get  T = \frac {T_y}{\cos \theta}
    Thank you!
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