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    Attachment 727460
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    need help pin all parts
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    "Attachment not found". What year and question number was this?
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    (Original post by mupsman2312)
    "Attachment not found". What year and question number was this?

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    (Original post by Asad234)
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    Well, you'd have to use moments in two dimensions. It looks like a pretty tough question, though!
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    (Original post by mupsman2312)
    Well, you'd have to use moments in two dimensions. It looks like a pretty tough question, though!
    I know I have to use moments. How do I do it though?
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    (Original post by Asad234)
    I know I have to use moments. How do I do it though?
    Erm...

    I should know this - I did M2 last year!
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    (Original post by mupsman2312)
    Erm...

    I should know this - I did M2 last year!
    not helpful.
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    (Original post by Asad234)
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    You've got two forces acting on the handbrake : F and T and any reaction forces from the pivot. Try taking moments about the pivot so you only need to consider F and T.

    Post all your working if you get stuck.
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    (Original post by Notnek)
    You've got two forces acting on the handbrake : F and T and any reaction forces from the pivot. Try taking moments about the pivot so you only need to consider F and T.

    Post all your working if you get stuck.
    I know you take moments from the pivot, but I don't know the angle for T

    F x 350 x cos40= T x 60 x sin?

    So just help me there.

    If you know how to do it, please post the working instead, I'll be able to figure out my mistake myself. Thank you
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    (Original post by Asad234)
    I know you take moments from the pivot, but I don't know the angle for T

    F x 350 x cos40= T x 60 x sin?

    So just help me there.

    If you know how to do it, please post the working instead, I'll be able to figure out my mistake myself. Thank you


    The perpendicular distance from the tension force to the pivot is the distance BP. You have angle CBP = 35 and BC = 60mm so you can find BP.

    Then multiply the magnitude of T by BP to get the moment.

    By the way you should really convert mm to m although it doesn't make a difference for this question.
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    (Original post by Notnek)


    The perpendicular distance from the tension force to the pivot is the distance BP. You have angle CBP = 35 and BC = 60mm so you can find BP.

    Then multiply the magnitude of T by BP to get the moment.

    By the way you should really convert mm to m although it doesn't make a difference for this question.
    By your logic that would be wrong.

    The answer to part two is 162N

    I get this

    F x cos 40 x 350= 1000 x 60 x cos 35

    F = 183N

    If you think I did it wrong, I am more than happy for you to correct me.
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    (Original post by Asad234)
    By your logic that would be wrong.

    The answer to part two is 162N

    I get this

    F x cos 40 x 350= 1000 x 60 x cos 35

    F = 183N

    If you think I did it wrong, I am more than happy for you to correct me.
    cos 40 is incorrect. I'll give an explanation - give me a few mins.
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    (Original post by Asad234)
    By your logic that would be wrong.

    The answer to part two is 162N

    I get this

    F x cos 40 x 350= 1000 x 60 x cos 35

    F = 183N

    If you think I did it wrong, I am more than happy for you to correct me.
    Notice that AB isn't horizontal so you can't just use cos(40).



    Does this diagram help? If not, please explain how you calculate moments because I don't know which method you use.
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    r
    (Original post by Notnek)
    Notice that AB isn't horizontal so you can't just use cos(40).



    Does this diagram help? If not, please explain how you calculate moments because I don't know which method you use.
    How do you know that angle of F is 60?
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    (Original post by Asad234)
    r

    How do you know that angle of F is 60?
    Wait, Now I know why.


    Thanks, your method helped me.
    I got 162
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    (Original post by Asad234)
    r

    How do you know that angle of F is 60?
    The angle between BA and the hirizontal is 10 (alternate angles) so I marked 10 on the diagram. Then the angle between my red line and the vertical is a right-angle so you get 90 - 40 - 50 and I marked that on.
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    (Original post by Asad234)
    Wait, Now I know why.

    Thanks, your method helped me.
    I got 162
    Cool - that's what I got too. Sorry I couldn't help you out with this question - I think that I was just having a little bit of an off-day yesterday, and got scared by the look of the question...

    Spoiler:
    Show

    Oh, well - I'm more than happy to help you out with any stats problems (mechanics is probably my type of weakest applied maths, but I can give it a go for you anyway if you need any more help in the future).
 
 
 

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