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    Hi,

    I have a homework question and am not sure how to get to the answer. The question is:

    Four uniform rods AB, BC, CD, and DA are each 4 metres in length and have masses of 2kg, 3kg, 1kg and 4kg respectively. If they are joined together to form a square framework ABCD, find the position of its centre of gravity.

    The answer that the book wants is 1.8 m from AD and AB

    I am not sure how to get to that as that is not in the centre and I thought the centre of mass was at the intersection of the lines of symmetry


    Many thanks in advance for your help
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    (Original post by bendent1234)
    Hi,

    I have a homework question and am not sure how to get to the answer. The question is:

    Four uniform rods AB, BC, CD, and DA are each 4 metres in length and have masses of 2kg, 3kg, 1kg and 4kg respectively. If they are joined together to form a square framework ABCD, find the position of its centre of gravity.

    The answer that the book wants is 1.8 m from AD and AB

    I am not sure how to get to that as that is not in the centre and I thought the centre of mass was at the intersection of the lines of symmetry


    Many thanks in advance for your help
    it would be the intersection of the lines of symmetry if all the rods had the same mass, however since they dont so you cant use this method.

    firstly I would draw a diagram with A at (0,0) and make a coordinate system. with the other points being at (0,4) (4,4) and (4,0)

    then you have to use  \displaystyle \sum m_i x_i = M \bar{x} where  \displaystyle m_i is the mass of a rod and  \displaystyle x_i is the x value of the centre of gravity of the rod,  \displaystyle M is the whole mass of the system and  \displaystyle \bar{x} is the C.O.M of the whole system.

    Use this to find  \displaystyle \bar{x} . Then use a similar procedure to find  \displaystyle \bar{y}
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    (Original post by DylanJ42)
    it would be the intersection of the lines of symmetry if all the rods had the same mass, however since they dont so you cant use this method.

    firstly I would draw a diagram with A at (0,0) and make a coordinate system. with the other points being at (0,4) (4,4) and (4,0)

    then you have to use  \displaystyle \sum m_i x_i = M \bar{x} where  \displaystyle m_i is the mass of a rod and  \displaystyle x_i is the x value of the centre of gravity of the rod,  \displaystyle M is the whole mass of the system and  \displaystyle \bar{x} is the C.O.M of the whole system.

    Use this to find  \displaystyle \bar{x} . Then use a similar procedure to find  \displaystyle \bar{y}
    When I did this I got that  M \bar{x} = 20
    The whole mass would be 10 (2+3+1+4)
    So  \bar{x} would be 2, which is not right
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    (Original post by bendent1234)
    When I did this I got that  M \bar{x} = 20
    The whole mass would be 10 (2+3+1+4)
    So  \bar{x} would be 2, which is not right
    going on the coordinates A(0,0) B(0,4) C(4,4) and (4,0) like so; http://imgur.com/a/2H7rh

    x coord of com of AB is 0, mass is 2
    x coord of com of BC is 2, mass is 3
    x coord of com of CD is 4, mass is 1
    x coord of com of AD is 2, mass is 4

    do you agree?
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    (Original post by DylanJ42)
    going on the coordinates A(0,0) B(0,4) C(4,4) and (4,0) like so; http://imgur.com/a/2H7rh

    x coord of com of AB is 0, mass is 2
    x coord of com of BC is 2, mass is 3
    x coord of com of CD is 4, mass is 1
    x coord of com of AD is 2, mass is 4

    do you agree?
    Oh of course, silly me, that makes sense now. So just do the same for  \bar{y} ?
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    (Original post by bendent1234)
    Hi,

    I have a homework question and am not sure how to get to the answer. The question is:

    Four uniform rods AB, BC, CD, and DA are each 4 metres in length and have masses of 2kg, 3kg, 1kg and 4kg respectively. If they are joined together to form a square framework ABCD, find the position of its centre of gravity.

    The answer that the book wants is 1.8 m from AD and AB

    I am not sure how to get to that as that is not in the centre and I thought the centre of mass was at the intersection of the lines of symmetry


    Many thanks in advance for your help
    If you imagine one side of the framework had mass 50kg and the other side 5kg then it should be clear that you can't assume that the COM will lie in the centre of the square.

    For a lamina with uniform mass, the COM will lie on the lines of symmetry.
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    (Original post by bendent1234)
    Oh of course, silly me, that makes sense now. So just do the same for  \bar{y} ?
    yea thatll be you done

    Also a note about the form of your answer, always write "a distance 1.8 from AB" etc unless you are given a coordinate system, if you create a coordinate system to make it easier just remember not to quote your answer as coordinates (in case you get a strict marker, it would be harsh to lose marks but you could)
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    (Original post by DylanJ42)
    yea thatll be you done

    Also a note about the form of your answer, always write "a distance 1.8 from AB" etc unless you are given a coordinate system, if you create a coordinate system to make it easier just remember not to quote your answer as coordinates (in case you get a strict marker, it would be harsh to lose marks but you could)
    Actually I got the value of  \bar{y} wrong

    I did 10  \bar{y} = (2x4) + (3x2) + (1x0) + (4x2)

    So  \bar{y} came out to be 2.2, rather than the needed answer of 1.8.  \bar{x} is fine
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    letmethink
    Here's a problem for you!
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    (Original post by bendent1234)
    Actually I got the value of  \bar{y} wrong

    I did 10  \bar{y} = (2x4) + (3x2) + (1x0) + (4x2)

    So  \bar{y} came out to be 2.2, rather than the needed answer of 1.8.  \bar{x} is fine
    does your diagram look like this?

    Name:  gg.png
Views: 66
Size:  2.6 KB

    since 4-2.2= 1.8 I would guess that you are correct its just that your y is defined with C and D being at y= 0 and A and B being at y = 4. So y = 2.2 when referring to your diagram means "the COM is a distance 2.2 from DC" which is exactly the same thing as "the COM is 1.8m from AB".

    does that make sense?

    Note: this is a perfect example of why using coordinates as your answer is a bad idea, so its good this happened and can be cleared up
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    (Original post by DylanJ42)
    does your diagram look like this?

    Name:  gg.png
Views: 66
Size:  2.6 KB

    since 4-2.2= 1.8 I would guess that you are correct its just that your y is defined with C and D being at y= 0 and A and B being at y = 4. So y = 2.2 when referring to your diagram means "the COM is a distance 2.2 from DC" which is exactly the same thing as "the COM is 1.8m from AB".

    does that make sense?

    Note: this is a perfect example of why using coordinates as your answer is a bad idea, so its good this happened and can be cleared up
    Thank you very much for all your help with this question
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    (Original post by bendent1234)
    Thank you very much for all your help with this question
    my pleasure
 
 
 
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