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    A hoop of weight 20 newtons can rotate freely about a pin fixed to a wall. A string has one end attached to the pin, runs round the circumference of the hoop to its lowest point, and is then held horizontally at it's other end. A gradually increasing horizontal force is now applied to the string, so that the hoop begins to rotate about the pin. Find the tension in the string when the hoop has rotated through 40 degrees.

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    (Original post by joyoustele)
    ...
    Attachment not found.
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    (Original post by RDKGames)
    Attachment not found.
    It works now. But how can i make it so the image stays on the page, instead of just providing a link?
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    (Original post by joyoustele)
    It works now. But how can i make it so the image stays on the page, instead of just providing a link?
    Haven't encountered this type of problem before, but just to check before I guide you to the answer, what's the answer to the question?
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    (Original post by RDKGames)
    Haven't encountered this type of problem before, but just to check before I guide you to the answer, what's the answer to the question?
    The Answer: Tension =7.28N

    I've spent so long on this
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    (Original post by joyoustele)
    The Answer: Tension =7.28N

    I've spent so long on this
    Oh fair enough, OK then.

    So, on that diagram, the CoM of the hoop in the diagram is dead in the middle of it. Mark that point and draw a force of 20N going down (for obvious reasons). Then, you have F(t) going horizontally from the bottom of the hoop in the diagram. (I chose F(t) since F is a gradually increasing force as a formality, but since you are looking at an instant of time, just say that force is F). Then you want to take moments about the pin, and considering those two forces, proceed to solve for F. Call the radius r, you'll quick notice that it cancels out so it isn't really a big thing here as to what it is.
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    (Original post by RDKGames)
    Oh fair enough, OK then.

    So, on that diagram, the CoM of the hoop in the diagram is dead in the middle of it. Mark that point and draw a force of 20N going down (for obvious reasons). Then, you have F(t) going horizontally from the bottom of the hoop in the diagram. (I chose F(t) since F is a gradually increasing force as a formality, but since you are looking at an instant of time, just say that force is F). Then you want to take moments about the pin, and considering those two forces, proceed to solve for F. Call the radius r, you'll quick notice that it cancels out so it isn't really a big thing here as to what it is.
    where did i go wrong?

    F(2r)=20r ?
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    (Original post by joyoustele)
    where did i go wrong?

    F(2r)=20r ?
    Well I dunno how you ended up with that equation so I dunno where you went wrong.

    The perp distance to F from the pin is not 2r since the pin isn't all the way at the top of the circle when at the angle.
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    (Original post by RDKGames)
    Well I dunno how you ended up with that equation so I dunno where you went wrong.

    The perp distance to F from the pin is not 2r since the pin isn't all the way at the top of the circle when at the angle.
    Oops.....Rookie mistake...
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    (Original post by RDKGames)
    Well I dunno how you ended up with that equation so I dunno where you went wrong.

    The perp distance to F from the pin is not 2r since the pin isn't all the way at the top of the circle when at the angle.
    I cant find the Distance. Thats the main bit i have struggled with
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    (Original post by joyoustele)
    I cant find the Distance. Thats the main bit i have struggled with
    Perp distance to F is the black line + the red dashed line. What's the length of the red dashed line...?

    [NOTE: This is a skill you'll need to use if you'll study vertical circular motion - to find this sort of length in a circle]

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    (Original post by RDKGames)
    Perp distance to F is the black line + the red dashed line. What's the length of the red dashed line...?

    Note: This is a skill you'll need to use if you'll study vertical circular motion.

    rcos40?
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    (Original post by joyoustele)
    rcos40?
    Should be good to finish it off now then.
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    (Original post by RDKGames)
    Should be good to finish it off now then.
    F(r+rcos(40)=20 \sqrt (r^2-r^2cos40)

    F=\dfrac{ 20\sqrt (r^2-r^2cos40) }{r+rcos40}

    .....have i gone wrong? ..
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    (Original post by joyoustele)
    F(r+rcos(40)=20g(\sqrt (r^2-r^2cos40)

    F=\dfrac{ 20g\sqrt (r^2-r^2cos40) }{r+rcos40}

    .....have i gone wrong? ..
    Yeah.

    First line:
    1. LHS is fine though missing a bracket at the end
    2. Where did g come from...?
    3. The perp. distance to the weight is just, yknow, r\sin(40) maybe...??
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    (Original post by RDKGames)
    Yeah.

    First line:
    1. LHS is fine though missing a bracket at the end
    2. Where did g come from...?
    3. The perp. distance to the weight is just, yknow, r\sin(40) maybe...??
    Nm my last post. I figured everything out.

    I got T=7.26, I'm about to check why i'm 0.02N off. But this is the correct method.

    Thanks RDK. Im going to be one of the only people in my class, to have figured this Out. Lol
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    (Original post by joyoustele)
    Nm my last post. I figured everything out.

    I got T=7.26, I'm about to check why i'm 0.02N off. But this is the correct method.

    Thanks RDK. Im going to be one of the only people in my class, to have figured this Out. Lol
    Not sure why you're offset by that.

    Without rounding, \displaystyle F(r+r\cos(40))=20r\sin(40) \Leftrightarrow F=\frac{20\sin(40)}{1+\cos(40)} \approx 7.28
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    (Original post by RDKGames)
    Not sure why you're offset by that.

    Without rounding, \displaystyle F(r+r\cos(40))=20r\sin(40) \Leftrightarrow F=\frac{20\sin(40)}{1+\cos(40)} \approx 7.28
    Got 7.28,

    Damn. This question seems a lot easier than i thought it was....
    Thank you very much.
 
 
 
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