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Simple Q about moments

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Have you had any embarrassing moments? Share them here.. 27-02-2015
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    The first question in the attachment.

    Say I take the angle with horizontal to be theta, and the length of rod be 2L.

    To find the moment about A,

    Why can't I take the vertical component of the force of 4 N, i.e. 4 cos theta.

    And then multiply it with the perpendicular distance from A, i.e.

    2L sin (90-theta) x (4cos theta)

    And equate it to 16 x L sin (90- theta)

    to get the answer?

    Click image for larger version. 

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    I keep getting cos theta = 2
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    bump
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    If you split 4N into components then you have to take into account the fact that there will be two components that provide moments.

    It is much simpler to multiply 4 by its perpendicular distance to A, 2l. Then you are dealing with two moments acting in different directions.
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    (Original post by Killjoy-)
    If you split 4N into components then you have to take into account the fact that there will be two components that provide moments.
    thank you!!!!
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    (Original post by Killjoy-)
    If you split 4N into components then you have to take into account the fact that there will be two components that provide moments.

    It is much simpler to multiply 4 by its perpendicular distance to A, 2l. Then you are dealing with two moments acting in different directions.
    Yup, was just wondering why that didn't work, whether it was because of a gap in understanding of the concept
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    (Original post by bmqib)
    The first question in the attachment.

    Say I take the angle with horizontal to be theta, and the length of rod be 2L.

    To find the moment about A,

    Why can't I take the vertical component of the force of 4 N, i.e. 4 cos theta.

    And then multiply it with the perpendicular distance from A, i.e.

    2L sin (90-theta) x (4cos theta)

    And equate it to 16 x L sin (90- theta)

    to get the answer?

    Click image for larger version. 

Name:	Untitled.png 
Views:	35 
Size:	41.1 KB 
ID:	141823
    Hi

    Part i) is only a two marker, so I don't even think you need to involve the length.
    We know that for a system in equilibrium, the total upwards force = total downwards force.
    So if you view the rod horizontally,
    total upwards force (4N) = total downwards force (the vertical component of the weight, 16sin alpha N)
    Alpha corresponds to the angle the rod makes with the vertical. So once you have that angle, it should be straightforward to work out the value for theta. When I did it, I got the value of theta to be 75.5 degrees

    ignore what's above :facepalm:
    Take moments about A.
    have the length of the rod be 2L.
    Total Clockwise moments = Total Anti-Clockwise Moments
    16cos theta L = 4 x 2L
    Cancel the Ls to get
    16cos theta = 8
    so cos theta = 1/2
    theta = 60
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    (Original post by magdaplaysbass)
    Hi

    Part i) is only a two marker, so I don't even think you need to involve the length.
    We know that for a system in equilibrium, the total upwards force = total downwards force.
    So if you view the rod horizontally,
    total upwards force (4N) = total downwards force (the vertical component of the weight, 16sin alpha N)
    Alpha corresponds to the angle the rod makes with the vertical. So once you have that angle, it should be straightforward to work out the value for theta. When I did it, I got the value of theta to be 75.5 degrees
    hi! i think you made a mistake somewhere, the angle of the rod with the horizontal is 60 degrees

    This is a moments question so you have to consider the lengths the forces are perpendicualrly from the pivot!
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    Or at least consider the forces at A
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    :facepalm: just realised where i went wrong. do you still need help with it?

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