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Math Help - resultant force problem

  1. #1
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    resultant force problem

    I know I am going to have a problem like the following on an exam tomorrow. I have a few ideas but I am not getting far enough along with it. So, if someone could help me out it would be much appreciated. Here's the problem:

    Two forces, one of 50lbs and the other of 30 lbs act on the same object. The magnitude of the resultant force is 70 lbs. Find the angle between the two forces.

    Here are my thoughts. I know that three force vectors have the following relationship: F1 + F2 = F3. I know that the formula for deriving the cosine of the angle between two vectors is their dot product divided by the product of their magnitudes. And, I know that a vector can be described in terms of its magnitude and direction by v = ||v|| (cos@ + sin@) where @ = the angle to the horizontal axis. But I cannot figure out how to put it all together in order to solve the equation.
    Thank you.
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  2. #2
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    Quote Originally Posted by pireskwared View Post
    I know I am going to have a problem like the following on an exam tomorrow. I have a few ideas but I am not getting far enough along with it. So, if someone could help me out it would be much appreciated. Here's the problem:

    Two forces, one of 50lbs and the other of 30 lbs act on the same object. The magnitude of the resultant force is 70 lbs. Find the angle between the two forces.

    Here are my thoughts. I know that three force vectors have the following relationship: F1 + F2 = F3. I know that the formula for deriving the cosine of the angle between two vectors is their dot product divided by the product of their magnitudes. And, I know that a vector can be described in terms of its magnitude and direction by v = ||v|| (cos@ + sin@) where @ = the angle to the horizontal axis. But I cannot figure out how to put it all together in order to solve the equation.
    Thank you.
    There is another formula that you will find to be useful. The three vectors, when sketched (using the head-tail method or parallelogram method of vector addition) will form a triangle. Then you can apply the Law of Cosines.

    -Dan
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  3. #3
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    Yes, that works.

    Quote Originally Posted by topsquark View Post
    There is another formula that you will find to be useful. The three vectors, when sketched (using the head-tail method or parallelogram method of vector addition) will form a triangle. Then you can apply the Law of Cosines.

    -Dan


    Thanks Dan. Yes, that works. I get approximately 88.7 degrees for the answer.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by pireskwared View Post
    Thanks Dan. Yes, that works. I get approximately 88.7 degrees for the answer.
    I keep getting 120 degrees.

    70^2 = 30^2 + 50^2 - 2 \cdot 30 \cdot 50 \cdot cos(\theta)

    cos(\theta) = \frac{70^2 - 30^2 - 50^2}{-2 \cdot 30 \cdot 50}

    cos(\theta) = -\frac{1}{2}

    \theta = 120^o

    -Dan
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  5. #5
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    Quote Originally Posted by pireskwared View Post
    Thanks Dan. Yes, that works. I get approximately 88.7 degrees for the answer.
    I don't know what went wrong but the included angle should be exactly 60.

    By the Law of Cosine you can calculate the angle \alpha

    \cos(\alpha)=\frac{70^2-50^2-30^2}{-2 \cdot 50 \cdot 30} = -\frac12

    The angle in question is: 180^\circ - \alpha
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  6. #6
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    How I got my answer

    I think there are two angles whose sum is the angle between the vectors. The parallelogram that makes up the force vectors involves two triangles. The angle that forms one corner of the parallelogram is what is required, I think. So law of cosines must be applied twice with the two triangles:

    50sq = 70sq + 30sq - 2*4200 cos @1
    @1 = 66.8676036

    and

    30sq = 50sq + 70sq - 2*3500 cos @2
    @2 = 21.7867893

    @1 + @2 = approx. 88.65
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  7. #7
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    Quote Originally Posted by pireskwared View Post
    I think there are two angles whose sum is the angle between the vectors. The parallelogram that makes up the force vectors involves two triangles. The angle that forms one corner of the parallelogram is what is required, I think. So law of cosines must be applied twice with the two triangles:

    50sq = 70sq + 30sq - 2*2100 cos @1
    @1 = 66.8676036

    and

    30sq = 50sq + 70sq - 2*3500 cos @2
    @2 = 21.7867893

    @1 + @2 = approx. 88.65
    ...
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  8. #8
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    Yes, I re-checked and made that error

    Thanks erboth -- When the law of cosines math is done correctly it yields to angles who sum is 60 degrees.
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