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    ARGH! Somebody help please!

    I cant do these questions to save my life..

    i just cant see how to do moments! my god its so hard! any ideas? good notes? tips, tricks? if you need a demonstration question...

    A uniform straight rod AB has mass M and length 2a. The end A is smoothly hinged at a fixed point so that the rod can turn freely in a vertical plane. Horizontal and vertical forces of magnitudes Mg and 0.3Mg respectively are applied to the end B. These forces and the weight of the rod are shown in the diagram. The rod rests in equilibrium at an angle θ to the horizontal.
    a) By taking moments about A, find the value of tan θ.
    b) Calculate the magnitude, in terms of M and g, and the direction, to the nearest degree, of the force exerted by the hinge on rod AB
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    (Original post by kikzen)
    ARGH! Somebody help please!

    I cant do these questions to save my life..

    i just cant see how to do moments! my god its so hard! any ideas? good notes? tips, tricks? if you need a demonstration question...

    A uniform straight rod AB has mass M and length 2a. The end A is smoothly hinged at a fixed point so that the rod can turn freely in a vertical plane. Horizontal and vertical forces of magnitudes Mg and 0.3Mg respectively are applied to the end B. These forces and the weight of the rod are shown in the diagram. The rod rests in equilibrium at an angle θ to the horizontal.
    a) By taking moments about A, find the value of tan θ.
    b) Calculate the magnitude, in terms of M and g, and the direction, to the nearest degree, of the force exerted by the hinge on rod AB
    Moment = force x perpendicular distance to line of action of force

    Moments about A:

    Weight acting at the centre, the distance you want is along the horizontal, so the moment = Mg.acos t

    Same thing for the 3Mg/10 force: (3Mg/10).2acos t

    For the last force, the distance is along the vertical: Mg.2asin t

    Clockwise moments = anti-clockwise moments in equilibrium

    => Mgacos t = 2Mgasin t + 0.3Mgacos t

    Obviously you can get tan t from there.
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    (Original post by kikzen)
    ARGH! Somebody help please!

    I cant do these questions to save my life..

    i just cant see how to do moments! my god its so hard! any ideas? good notes? tips, tricks? if you need a demonstration question...

    A uniform straight rod AB has mass M and length 2a. The end A is smoothly hinged at a fixed point so that the rod can turn freely in a vertical plane. Horizontal and vertical forces of magnitudes Mg and 0.3Mg respectively are applied to the end B. These forces and the weight of the rod are shown in the diagram. The rod rests in equilibrium at an angle ? to the horizontal.
    a) By taking moments about A, find the value of tan ?.
    b) Calculate the magnitude, in terms of M and g, and the direction, to the nearest degree, of the force exerted by the hinge on rod AB

    Here goes, Taking clocckwise moments Mgcosx*a
    anticlockwise gives Mgsinex*2a + 0.3mgcosx*2a

    Mgsinx*2a + 0.3mgcosx*2a = Mgcosx*a

    Ms and a cancel giving

    2sinx + 0.6cosx = cosx
    2sinx = 0.4cosx

    then solve.

    Hope that's correct.

    MB
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    (Original post by kikzen)
    ARGH! Somebody help please!

    I cant do these questions to save my life..

    i just cant see how to do moments! my god its so hard! any ideas? good notes? tips, tricks? if you need a demonstration question...

    A uniform straight rod AB has mass M and length 2a. The end A is smoothly hinged at a fixed point so that the rod can turn freely in a vertical plane. Horizontal and vertical forces of magnitudes Mg and 0.3Mg respectively are applied to the end B. These forces and the weight of the rod are shown in the diagram. The rod rests in equilibrium at an angle θ to the horizontal.
    a) By taking moments about A, find the value of tan θ.
    b) Calculate the magnitude, in terms of M and g, and the direction, to the nearest degree, of the force exerted by the hinge on rod AB
    you need to do 3 things - moments about a point - vertical forces and horizontal forces

    moments about A = 0, clockwise is postive
    0= Mg * a * Cos θ - Mg * 2a * Sin θ - 3/10 * Mg * 2a * Cos θ
    cancel Mg and a out
    0= cosθ - 2 sinθ -3/10*2*sinθ
    divide by cos θ
    0 = 1 -2 tanθ -6/10 tanθ
    rearrange
    26/10 tan θ = -1
    tanθ = -10/26
    cant do the rest as i have to go im afriad
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    (Original post by Nylex)
    Moment = force x perpendicular distance to line of action of force

    Moments about A:

    Weight acting at the centre, the distance you want is along the horizontal, so the moment = Mg.acos t

    Same thing for the 3Mg/10 force: (3Mg/10).2acos t

    For the last force, the distance is along the vertical: Mg.2asin t

    Clockwise moments = anti-clockwise moments in equilibrium

    => Mgacos t = 2Mgasin t + Mgacos t

    Obviously you can get tan t from there.
    what happenned to the 0.3???

    MB

    P.S can someone check my post - answer looks odd.
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    Blimey, we've all got different answers!

    I mostly agree with musicboy's solution, except that
    2sinx + 0.6cosx = cosx should lead to
    2sinx = 0.4cosx

    For the second part, resolve forces along the rod.

    In direction A is pointing minus direction B is pointing (sorry, poor mathematical language :rolleyes: )

    mgsinx + mgcosx - 0.3mgsinx

    Follow through to obtain an expression involving mg and x (I got 5.7mgsinx) and use your answer to part i to get one in terms of m and g.
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    (Original post by musicboy)
    what happenned to the 0.3???

    MB

    P.S can someone check my post - answer looks odd.
    Oops, sorry. Yeah, there should be a 0.3 in that equation. /me goes to edit post.
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    (Original post by el GaZZa)
    Blimey, we've all got different answers!

    I mostly agree with musicboy's solution, except that
    2sinx + 0.6cosx = cosx should lead to
    2sinx = 0.4cosx

    For the second part, resolve forces along the rod.

    In direction A is pointing minus direction B is pointing (sorry, poor mathematical language :rolleyes: )

    mgsinx + mgcosx - 0.3mgsinx

    Follow through to obtain an expression involving mg and x (I got 5.7mgsinx) and use your answer to part i to get one in terms of m and g.
    For b, you can just stick horizontal and vertical components at A and then resolve from there:

    Fy + 0.3Mg = Mg => Fy = Mg - 0.3Mg => Fy = 0.7Mg

    Fx = Mg

    => |F| = [(Fx)^2 + (Fy)^2]^1/2, etc

    and tan t = Fy/Fx and go from there.
 
 
 
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