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    •A uniform square lamina ABCD has side oflength 2 m. E is a point on AD such that ED = 1.5 m. The triangular portion EDCis removed, resulting in a centre of gravity that is 0.7 m from AB. If thelamina (which now has weight W) is in the vertical plane with side AE on arough horizontal surface, find the smallest magnitude force that can be appliedto point C to stop the lamina from toppling over.
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    Tips to help you. (1) This is a moments question. (2) Make a scale copy of the situation to help you visualise. (3) Draw and label a diagram with the letters and where it is sensible to take moments. (4) If you are still stuck, report (1), (2) & (3) back here for further advice.
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    (Original post by nerak99)
    Tips to help you. (1) This is a moments question. (2) Make a scale copy of the situation to help you visualise. (3) Draw and label a diagram with the letters and where it is sensible to take moments. (4) If you are still stuck, report (1), (2) & (3) back here for further advice.
    http://m.imgur.com/enP8nTO
    Im not really sure where to take moments from. Is it even possible to take moments as i dont know where the normal force or frictional force are acting in terms of measurements.
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    (Original post by ErniePicks)
    http://m.imgur.com/enP8nTO
    Im not really sure where to take moments from. Is it even possible to take moments as i dont know where the normal force or frictional force are acting in terms of measurements.
    If the lamina is going to topple over, which point is it going to rotate about? And that's the place to take your moments from.

    PS: Good diagram.
    Edit: Apart from the friction - see later.
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    (Original post by ghostwalker)
    If the lamina is going to topple over, which point is it going to rotate about? And that's the place to take your moments from.

    PS: Good diagram.
    Apparently my diagram is wrong, the frictional force is supposed to be acting in the other direction, why si that the case? i figured if it toppled over to the right the frictional force would have to act towards the left. OH WAIT IS IT BECAUSE THE BOTTOM WILL ROTATE IN CLOCKWISE DIRECTION SO IT WILL ACTUALLY PIVOT TO THE LEFT SO THE FRICTION FORCE WILL ACT ON THE RIGHT OHHHHHHHHHHHHHHHHHHHHHHHHHH AAAAAGGGGHHHHHHHHHHHHHHHHHHHHHHH
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    (Original post by ErniePicks)
    Apparently my diagram is wrong, the frictional force is supposed to be acting in the other direction, why si that the case? i figured if it toppled over to the right the frictional force would have to act towards the left. OH WAIT IS IT BECAUSE THE BOTTOM WILL ROTATE IN CLOCKWISE DIRECTION SO IT WILL ACTUALLY PIVOT TO THE LEFT SO THE FRICTION FORCE WILL ACT ON THE RIGHT OHHHHHHHHHHHHHHHHHHHHHHHHHH AAAAAGGGGHHHHHHHHHHHHHHHHHHHHHHH

    The centre of gravity of the lamina is to the right of the base (AE), so the mass will topple to the right, pivoting about the point E.

    Friction is irrelevant (but as it topples to the right, the base will tend to move to the left, so friction acts to the right to stop that). Whichever direction friction acts in, it's line of action will go through the pivot E. So it's moment about E will be zero.
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    (Original post by ghostwalker)
    The centre of gravity of the lamina is to the right of the base (AE), so the mass will topple to the right, pivoting about the point E.

    Friction is irrelevant (but as it topples to the right, the base will tend to move to the left, so friction acts to the right to stop that). Whichever direction friction acts in, it's line of action will go through the pivot E. So it's moment about E will be zero.
    would that be the case of the normal force as well? yes it would. But i am confused on how to find the minimum force required. taking moments the perpendicular distance to e from the line of action would have to be at max, and apparently that is when it is perpendicular to ec. Why is that the case? Couldn't we make the angle even smaller for even greater distance?
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    (Original post by ErniePicks)
    would that be the case of the normal force as well? yes it would.
    Yes. When it's on the point of toppling, all the reaction will be through E, and hence no moment about E.

    But i am confused on how to find the minimum force required. taking moments the perpendicular distance to e from the line of action would have to be at max, and apparently that is when it is perpendicular to ec. Why is that the case? Couldn't we make the angle even smaller for even greater distance?
    Making the angle smaller would make the distance smaller. The maximum distance for the line of action through C, is when that line of action is perpendicular to the line connecting C and E.

    Try drawing a few cases, or do it algebraically if you wish expressing the perpendicular distance in terms of CE and the angle
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    (Original post by ghostwalker)
    Yes. When it's on the point of toppling, all the reaction will be through E, and hence no moment about E.



    Making the angle smaller would make the distance smaller. The maximum distance for the line of action through C, is when that line of action is perpendicular to the line connecting C and E.

    Try drawing a few cases, or do it algebraically if you wish expressing the perpendicular distance in terms of CE and the angle
    You are a legend and i will remember you on my deathbed
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    (Original post by ErniePicks)
    You are a legend
    :blush:

    and i will remember you on my deathbed
    I would hope you had more meaningful things to remember by then, but thanks for the thought.
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    (Original post by ghostwalker)
    :blush:



    I would hope you had more meaningful things to remember by then, but thanks for the thought.
    sorry to bother but got into bit of a pickle. This is the question

    "
    •Whatis the moment of inertia about the centre of mass of a rod of length 1.5 m,with bodies of mass 2 kg and 3 kg respectively, on either end of the rod(ignore the mass of the rod itself)? What is the kinetic energy of thisassembly if it rotates about its centre of mass at 30 revolutions per minute(RPM)?"

    http://imgur.com/rz0y4af

    Well this was my attempt and apparently i drew drew it wrong. C of G is apparently closer to 2kg and i got Xg and 1.5-Xg the wrong way around. I am not sure why though, shouldn't C of G be closer to 3kg as it's the heavier weight?
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    (Original post by ErniePicks)
    sorry to bother but got into bit of a pickle. This is the question

    "
    •Whatis the moment of inertia about the centre of mass of a rod of length 1.5 m,with bodies of mass 2 kg and 3 kg respectively, on either end of the rod(ignore the mass of the rod itself)? What is the kinetic energy of thisassembly if it rotates about its centre of mass at 30 revolutions per minute(RPM)?"

    http://imgur.com/rz0y4af

    Well this was my attempt and apparently i drew drew it wrong. C of G is apparently closer to 2kg and i got Xg and 1.5-Xg the wrong way around. I am not sure why though, shouldn't C of G be closer to 3kg as it's the heavier weight?
    You have the centre of mass correct. Why do you think it is wrong?
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    (Original post by ghostwalker)
    You have the centre of mass correct. Why do you think it is wrong?
    Because this is the correct solutionName:  moments.jpg
Views: 64
Size:  34.2 KB

    So I'm just wondering why is the C of G closer to 2kg instead of 3, and how am i supposed to know which distance is Xg in a question like this.
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    (Original post by ErniePicks)
    Because this is the correct solutionName:  moments.jpg
Views: 64
Size:  34.2 KB

    So I'm just wondering why is the C of G closer to 2kg instead of 3, and how am i supposed to know which distance is Xg in a question like this.
    x_g is whatever you define it to be - there is no one correct method.

    They defined it relative to the 2kg mass, you defined it relative to the 3kg mass. Both answers are correct, just refer to different distances.

    0.6m from 3kg mass is the same position as 0.9m from 2kg mass.
 
 
 
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