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I am really struggling with the last part of this question and would really appreciate some help. The question, regarding centres of gravity, is from an old O level text and reads as follows:

$ABCD$ is a uniform square plate, of side $2a$, $M$ and $N$ are the midpoints of $AB$, $CD$. If $E$ is a point on $MN$ such that $NE = $$\frac{1}{2}a$ and the triangle AEB is cut away, find the distance from $N$ of the centre of gravity, $G$, of the remainder.

Find the length of $NE\:$ if $G$ is coincident with $E$.

Solutions give $\frac{7}{10}a$ for part one and $(\sqrt3 -1)a$ for part two.

I have been successful in gaining the correct solution for part one but am not at all sure about part two. I am not even sure how to set up the question to obtain the correct answer. My only method so far has been to consider that an equilateral triangle is removed. The reason for this choice of triangle was the $\sqrt3$ in the answer and the altitude of an equilateral triangle of base $2a$ is $a\sqrt3$. However, try as I might I am unable to obtain correct answer. Also, if an equilateral triangle is the right choice, then I don't know why and so can only attempt this second part by checking solution and by using trial and error - clearly an unsatisfactory approach.

I should add that I have used the $\bar{x}$$=\frac{\Sigma wx}{\Sigma w}$ and $\bar{y}$$=\frac{\Sigma wy}{\Sigma w}$ approach with origin at $D$.$\:$ Also, as weight is proportional to area, I have let unit weight equal unit area.

Many thanks in advance for any advice offered as I have been stuck on this for some time.

GR L
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    The descriptions of $M,$ $N,$ and $E$ are explicit and imply that $G$ and $E$ are distinct points, so what can be meant by "the length of $NE$ if $G$ is coincident with $E$"? Some part(s) of the previous description of the figure has to be amended to make sense of this. I think the intention was simply to delete the statement that $NE=\frac12a.$ Was that mentioned in problem statement as originally presented to you? – David K Sep 02 '21 at 12:32
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    I think for the second part we're supposed to understand: "Now, suppose $E$ is instead an arbitrary point on $MN$. Find the length... " etc – AakashM Sep 02 '21 at 12:34
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    The solution for the second part should arise from a generalization of the first part. Can you show your work for the first part? That would likely make it much easier to supply a suitable hint. – David K Sep 02 '21 at 12:36
  • @DavidK Thank you for your replies. I'm sorry if I've not made myself very well understood. My working is a mess, probably the reason for the need to amend, and I'm afraid that my skills in writing are rather limited. I have tried to visualise this and at one point thought $NE$ had zero length but am unsure. Your question from your first comment regarding 'what can be meant' sums up my difficulty. The first answer gives the c.g. outside the plate I'm not sure if this is relevant. I'm really not sure where I'm going wrong, I'm going to have to think about this some more. – GR L Sep 02 '21 at 12:53
  • @DavidK From your original message - Apologies if I'm unclear. I should have mentioned in original post, the question given is complete and exactly the same as that given in the text. I'm not sure whether you meant that this question or my further description below needed amending. Thank you once again. – GR L Sep 02 '21 at 13:07
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    I was wondering if you had forgotten to mention something from the original problem statement for part two, but I believe you when you say you did not. In that case you should not feel too badly about being confused, because the problem statement actually is broken. But I agree with the comment above that for part two you suppose $E$ can be any point on $MN.$ Depending on where we put $E$ the c.g. might be on the remaining plate or in the part that was cut out. But we need equations, numbers, and variables to work with, even if they look messy at first. – David K Sep 02 '21 at 13:09
  • @DavidK I think I understand now. I will have another go and try to obtain two equations. I will also study the answer given. Thank you for all your help. – GR L Sep 02 '21 at 13:15

2 Answers2

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Let $A = (0, 0), B=(2a, 0), C(2a, 2a), D= (0, 2a)$

Then $M = (a, 0 ), N = (a, 2a ) $

Point $E$ is on $MN$, if the distance $NE = y$ then $E = (a, 2a - y) $

Hence, the centroid of $\triangle AEB = ( a , \dfrac{y}{3} )$

And therefore, after removing $\triangle AEB$, the centroid of the remainder is given by

$G = \dfrac { a^2 (a, a ) - a y (a, \dfrac{y}{3} ) } { 4a^2 - a y } $

Simplifying,

$G = (a, \dfrac{ 12 a^2 - y^2 }{12 a - 3 y} )$

If the y-coordinate of $G$ is equal to that of $E$ then we have

$ y = \dfrac{ 12 a^2 - y^2 }{12 a - 3 y}$

Therefore,

$ y (12 a -3 y) = 12 a^2 - y^2 $

Re-arranging,

$ 2 y^2 - 12 a y + 12 a^2 = 0 $

Hence, $ y = \dfrac{1}{4} ( 12 a - \sqrt{ 144 a^2 - 96 a^2 } ) =\dfrac{1}{4} ( 12 a - 4 \sqrt{3} a ) = (3 - \sqrt{3} )a $

From which $NE = 2a - y = 2a - (3 - \sqrt{3}) a = a (\sqrt{3} - 1) $

Hosam Hajeer
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enter image description here

In figure $a_2$ (i.e A'B'DC )is equivalent to $a_1$ and $b_2$(i.e MA'CDB') is equivalent to $b_1$.

For figure a we have:

$CB'^2=(\frac{3a}2)^2+(2a)^2=\frac{25a^2}4\rightarrow CF=\frac52 a$

$CG=\frac12 CF=\frac 54 a$

$NG=\sqrt{(\frac54a)^2-a^2}=\frac 34 a$

For figure b we have $DB'=x$:

$G_1$from A'B'=$\frac13(2a-x)$

Area of MA'B'=$2a(2a-x)\frac12$

Area of A'B'DC=$\frac 12x(2a\times x)$

so we must have:

$[\frac13(2a-x)][2a(2a-x)\frac12]=\frac 12x(2a\times x)$

which gives:

$x^2+2ax-2a^2=0\rightarrow x=a(\sqrt3-1)$

sirous
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