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M2 Statics of rigid bodies

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    ABCD is a uniform square lamina of side 4m and mass 8Kg. It is hinged at A so that it is free to move in a vertical plane. It is maintained in equilibrium with B vertically below A by a horizontal force acting at C and a vertical force acting at D, each of magnitude F newtons. Find:

    (a) the value of F
    (b) the magnitude and direction of the force exerted by the hinge on the lamina.

    I could do b if it weren't for a.

    Any help?? - I have tried resolving around A to remove the hinges forces and I get \displaystyle 2\sqrt{2} \times{8g} - 4F - 4\sqrt{2} F=0

    I may post some more up later.

    Thanks

    -Sohan-
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    You've done it correctly, but you've forgotten to take into account the direction in which the forces are acting. The first and last terms in your moment-taking above should be resolved, if you know what I mean. If you don't understand, I'll try and post some working.
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    (Original post by tommmmmmmmmm)
    You've done it correctly, but you've forgotten to take into account the direction in which the forces are acting. The first and last terms in your moment-taking above should be resolved, if you know what I mean. If you don't understand, I'll try and post some working.
    I'm confused. I missed the week we did this in class and I totally have no idea what to do. Some working would be great. Thanks.
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    Let me draw a diagram on paint, I'll be a few minutes
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    ReputationRep:
    Did you mean statics?
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    (Original post by Mush)
    Did you mean statics?
    yes I did
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    There you go.

    Do you see why the sin(theta)s are needed?
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    Am i correct in saying, everything has to be at right angles to the line connecting it with the moment point??

    And why can it not be parrallel?
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    You need to use the component of the force resolved perpendicular to the line connecting it with the moment point.

    If it was parallel, there would be no moment. Imagine if you had a sheet of square paper which was attached to the table at one corner with a drawing pin, if you pulled on the opposite corner in the direction of the diagonal from the pin to your hand, then there would be no moment clockwise or anti-clockwise. If you changed the direction you were pulling in, how would this affect the moment of your hand's force on the paper?
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    (Original post by tommmmmmmmmm)
    You need to use the component of the force resolved perpendicular to the line connecting it with the moment point.

    If it was parallel, there would be no moment. Imagine if you had a sheet of square paper which was attached to the table at one corner with a drawing pin, if you pulled on the opposite corner in the direction of the diagonal from the pin to your hand, then there would be no moment clockwise or anti-clockwise. If you changed the direction you were pulling in, how would this affect the moment of your hand's force on the paper?
    I'd rep but I already did 2 days ago. Maybe I'll do what I have with Bruceleej and make you wait for your next one

    Thanks so much.

    -Sohan-
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    No problem Sohan
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    Another one: I get so many variables here though.

    A uniform ladder rests with one end on rough horizontal ground and the other end against a rough vertical wall. The coefficient of friction between the ground and the ladder is \frac{3}{5} and the coefficient of friction between the wall and the ladder is \frac{1}{3}. The ladder is on the point of slipping when it makes an angle \alpha with the horizontal. Find \tan \alpha.
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    At 2 in the morning? :p:

    Anyway, you should get five forces on your diagram:
    - the normal reaction from the floor
    - the normal reaction from the wall
    - the frictional force exerted by the floor (make sure it's in the right direction)
    - frictional force exerted by wall (ditto)
    - weight of the ladder

    Let the length of the ladder = 2a or something similar.
    Take moments (remember to use the sine of the acute angle that the force forms with the line of action of the moment), and resolve vertically and horizontally. You should get a nice set of simultaneous equations to solve
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    is this correct for a diagram?
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    (Original post by tommmmmmmmmm)
    At 2 in the morning? :p:
    Oh and what the hell are you doing at 2 in the morining here? :p:

    I :sugar: you really. If you weren't here then I'd have to wait a long long time lol.

    Thanks sooooo much. Bruceleej's Rep timetable applies to you too now. I shall rep you every 28 days for a long long time.

    And you shall go on my Sig.

    -Sohan-
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    Nearly, you've got the direction of the frictional force at the top of the ladder wrong, though. Think about which way the ladder would move if it slipped; the frictional force must be opposed to it. It helps to picture the situation in your head.
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    (Original post by tommmmmmmmmm)
    Nearly, you've got the direction of the frictional force at the top of the ladder wrong, though. Think about which way the ladder would move if it slipped; the frictional force must be opposed to it. It helps to picture the situation in your head.
    I still cannot solve xmg \cos \alpha - 2xF_2 \cos \alpha = 2xX \sin \alpha

    with F_2=\frac{1}{3}X , X=F_1 , Y=mg-F_2 , F_1=\frac{3}{5}Y
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    Erm, no idea why I'm here at this time :p:

    I don't mind doing any of this, keep the questions coming because I've got to prepare for M2 as well (5 weeks and 3 days, scary thought, eh?)

    Anyway, it's bedtime for me, much man-love
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    I'll leave you with a hint: divide through by cos(alpha)
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    love you in a very non homosexual manner

    :love:

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