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Math Help - forces

  1. #1
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    forces

    1) Suppose F is acting vertically downward on an object sitting on a plane that is inclined at an angle of 45 degrees to the horizontal. Express this force as a sum acting parallel to the plane and one acting perpendicular to it.

    F = || F|| cos (?) + ||F|| sin (?)




    2) an object is moving in the direction i + j is being acted upon by the force vector 2i + j, express this force as the sum of a force in the direction of motion and a force perpendicular to the direction of motion.
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    Re: forces

    Quote Originally Posted by Jonroberts74 View Post
    1) Suppose F is acting vertically downward on an object sitting on a plane that is inclined at an angle of 45 degrees to the horizontal. Express this force as a sum acting parallel to the plane and one acting perpendicular to it.

    F = || F|| cos (?) + ||F|| sin (?)



    2) an object is moving in the direction i + j is being acted upon by the force vector 2i + j, express this force as the sum of a force in the direction of motion and a force perpendicular to the direction of motion.
    1) Took me forever but I finally have a diagram for this. The rest of the diagram isn't yours, but I want you to note the angles. Why is the angle between the weight and the normal equal to the angle of incline? (This is always true.)
    forces-inclined_plane_-_a_simple_problem.png

    2) What have you tried so far?

    -Dan
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    Re: forces

    I thought maybe the cosine law would work because the direction of movement given and the force vector given aren't parallel so the perpendicular force wouldn't be orthogonal to both.

    but I couldn't get that to work

    not sure if I am wording this right but the angle between the normal and weight is equal to the angle of the incline because it's basically a rotation of 90 degrees. I know thats not worded correctly, those angles can form a congruent triangle to the triangle formed by the incline

    i'll give you what the book has the answer for (1) \vec{F} = \frac{||F||\vec{v}+ ||F||\vec{h}}{\sqrt{2}}; \vec{v}= (i-j)/\sqrt{2}, \vec{h} =-(i+j)/\sqrt{2}
    Last edited by Jonroberts74; June 22nd 2014 at 11:14 AM.
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    Re: forces

    still trying to solve these two

    1) would tangent be used?

    2) I need the normal for the perpendicular?
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    Re: forces

    Quote Originally Posted by Jonroberts74 View Post
    1) Suppose F is acting vertically downward on an object sitting on a plane that is inclined at an angle of 45 degrees to the horizontal. Express this force as a sum acting parallel to the plane and one acting perpendicular to it.
    First sketch the components of the vector F. There are components F_x and F_y. The component F_x is pointing down the plane and the component F_y is pointing into the inclined plane. From the triangle we know that there is an angle of 45 degrees between F and F_x, and that F_x is the side opposite to the angle in this case, so we know that F_x/F = sin(45). Similarly F_y/F = cos(45).

    Make sure you know this analysis back and front because you will see it over and over.

    Quote Originally Posted by Jonroberts74 View Post
    2) an object is moving in the direction i + j is being acted upon by the force vector 2i + j, express this force as the sum of a force in the direction of motion and a force perpendicular to the direction of motion.
    You are looking to find the components of F in the direction i + j (this is in the direction of motion) and the component of F is the direction perpendicular to i + j. Use the dot product here. We know, for the force component parallel to the direction of motion,that (\hat{i} + \hat{j} ) \cdot (2 \hat{i} + \hat{j}) = | \hat{i} + \hat{j} | \times | 2 \hat{i} + \hat{j} | \cdot cos ( \theta )
    where theta is the angle between the force and the direction of motion. So find alpha from this.

    For the perpendicular part use the dot product again. This time, though, you need to use the dot product to find a vector at right angles to the direction of motion. So
    ( \hat{i} + \hat{j} ) \cdot (a \hat{i} + b \hat{j} ) = 0 (because cos(90) = 0.)

    This will give you the value of b in terms of a, which you can arbitrarily choose to be 1. Now do the same as above with your new vector perpendicular to the direction of the motion.

    -Dan
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    Re: forces

    1) \vec{F} = <1,1> - <1,-1>

    2) \vec{F} = <1,-1> + <\frac{6}{\sqrt{10}}, \frac{3}{\sqrt{10}}>

    ??
    Last edited by Jonroberts74; June 28th 2014 at 09:44 PM.
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    Re: forces

    Quote Originally Posted by Jonroberts74 View Post
    1) \vec{F} = <1,1> - <1,-1>

    2) \vec{F} = <1,-1> + <\frac{6}{\sqrt{10}}, \frac{3}{\sqrt{10}}>

    ??
    when you say components, are you referring to this formula comp_{\vec{b}} \vec{a} = \vec{a} \frac{\vec{b}}{||\vec{b}||}

    I also watched a video about resolving the forces but that didn't seem to fit the question. I don't know much about physics
    Last edited by Jonroberts74; June 28th 2014 at 10:58 PM.
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    Re: forces

    the books gives the answer to number 1 as \vec{F}=\frac{||F||\vec{v}+||F||\vec{h}}{\sqrt{2}} where v= (i-j)/sqrt{2} and h=-(i+j)/\sqrt{2}


    so \vec{F} = \frac{||F||<1,-1>}{\sqrt{2}}+\frac{||F||<-1,-1>}{\sqrt{2}}
    Last edited by Jonroberts74; June 29th 2014 at 10:20 AM.
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    Re: forces

    1)  \frac{F}{\sqrt{2}}\hat{n} - \frac{F}{\sqrt{2}}\hat{p}

    where \hat{n} = \hat{i} - \hat{j} and [TEX] \hat{p} = -( \hat{i} + \hat{j})


    2) \vec{F} = (\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j})  + ( \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j})
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