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Math Help - some more mechanics help please, (moments)

  1. #1
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    some more mechanics help please, (moments)

    Can anyone have a look at question 5 on this paper I am really stuck.

    i can do 5a but i am stuck on 5b



    thanks a lot,

    btw i have tried but don't seem to get the right answer,
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  2. #2
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    There are 3 forces acting on the beam.

    Weight, tension at A, tension at C. You should have gotten

    C = 140
    A = 70

    The beam is in equilibrium, which means no movement. No moments about any point.

    No moments about point B!

    Let distance CB = x

    Moment about B, from force at C:

     M_c = 140(x)

    Moment about B from force at A:

     M_a = 70(x+90)

    Moment from weight(which acts at the centre):

     M_w = 210(\frac{x+90}{2})

    The sum of these should be zero, and then you have an equation in x to solve!

    Remember that anticlockwise moments are positive, and clockwise moments are negative!
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  3. #3
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    Quote Originally Posted by Mush View Post
    There are 3 forces acting on the beam.

    Weight, tension at A, tension at C. You should have gotten

    C = 140
    A = 70

    The beam is in equilibrium, which means no movement. No moments about any point.

    No moments about point B!

    Let distance CB = x

    Moment about B, from force at C:

     M_c = 140(x)

    Moment about B from force at A:

     M_a = 70(x+90)

    Moment from weight(which acts at the centre):

     M_w = 210(\frac{x+90}{2})

    The sum of these should be zero, and then you have an equation in x to solve!

    Remember that anticlockwise moments are positive, and clockwise moments are negative!
    Hi,

    I dont understand why its  70(x+90) should it not be  70 \times xm ? because your taking moments about B, and the distance from B to A is 'x meters?

    thanks for your help
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  4. #4
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    Oh sorry you edited your post, now it makes sense lol

    I typed before you edited it

    thanks a lot!
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  5. #5
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    Quote Originally Posted by Tweety View Post
    Oh sorry you edited your post, now it makes sense lol

    I typed before you edited it

    thanks a lot!
    Sorry about the confusion. Slip of the fingers .
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  6. #6
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    Am I doing this correctly?

    here is my working for solving x

     140x + 70(x+90)= 210( \frac{x+90}{2})

     210x + 6300= 210(x+45)

     210x +6300 = (-210x) + (-9450) ????

    this can't be right? what am i doing wrong?
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  7. #7
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    Quote Originally Posted by Tweety View Post
    Am I doing this correctly?

    here is my working for solving x

     140x + 70(x+90)= 210( \frac{x+90}{2})

     210x + 6300= 210(x+45)

     210x +6300 = (-210x) + (-9450) ????

    this can't be right? what am i doing wrong?
    my answer book says the equation should be like this taking moments about A   140 \times 90 = 210 \times d

    d=60
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  8. #8
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    The moments from the forces at A and C cause CLOCKWISE motion, which is taken as negative. The weight causes an ANTICLOCKWISE moment, which is positive.

    Also you made an error when you went:

     210(\frac{90+x}{2}) = 210(45+x)

    Fraction do not work like that, it should be:

    210(\frac{90+x}{2}) = 210(45+\frac{x}{2})

    How's this look:

    -140(x)-70(x+90)+210(\frac{x+90}{2}) = 0

    -140(x)-70x-70(90)+210(\frac{x}{2})+210(\frac{90}{2}) = 0

    -140(x)-70x-70(90)+105x+105(90) = 0

    -105x = 70(90)-105(90)

    105x = 3150

    x = 30cm
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