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A trough has a cross-section in the form of a trapezium. Its base has a length of $1m$, and the sides slope out at $45^o$. to the horizontal. The trough is filled with feed to a depth of $xm$. Find the value of $x$ given that the centre of mass of the contents of the trough is $0.5m$ above the base.

Working's out so far:

Centroid of a trapezium is $\frac{h}{3}\frac{(b+2a)}{(b+a)}$, where $h$ is the height and $b$ is the longest (base) and $a$ is the shortest base (in this problem, $a$ is on the horizontal).

The height of the trough is $sin(45)$, if we take the length of the sloped edge to be $1m$, as well. Although according to the answer it cannot be this low. How long are the edges of the trough? The sloping edge of the feed can be worked out using trig to get $x√2$.

The answer should be $0.866$

Answer:

The base is $b=1$ and if we call length of the top of the feed $a$ and so $a-2x$ is the length of the top directly above $b$. The height of the feed is $h=x$.

If we say that height of the centroid of the feed (inverted trapezium) is the height above the base then we can use $\frac{x}{3}\frac{(b+2a)}{(b+a)}$, with $a=2x+1$,

So we have $\frac{1}{2}=\frac{x}{3}\frac{1+2(2x+1)}{1+ (2x+1)}$ which gives $\frac{1}{2}=\frac{x}{3}\frac{(4x+3)}{(2x+2)}$

So $\frac{3}{2x}=\frac{(4x+3)}{(2x+2)}$, which gives $x=\frac{√3}{2}$= $0.866$.

J132
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  • The height that must be taken into account is that of of the feed, so $h=x$. Be careful that your formula for the centroid, as written, is wrong and that it represents the distance of centroid from the longest base: the distance from the shortest base is ${2h\over3}{b+2a\over b+a}$. – Intelligenti pauca Sep 13 '15 at 15:24
  • If we say that height of the centroid of the feed (inverted trapezium) is the height above the base then we can use $\frac{h}{3}\frac{(b+2a)}{(b+a)}$, with $a=2h+1$. The formula that you have given does not give $x=0.866$. So we can say that the centroid height is above the base regardless of whether it is the longest or shortest base, in terms of its length, and so use $\frac{h}{3}\frac{(b+2a)}{(b+a)}$, – J132 Sep 17 '15 at 09:16
  • You wrote at the beginning that in $\frac{h}{3}\frac{(b+2a)}{(b+a)}$, $b$ is the longest base and $a$ is the shortest base, but then you substituted $b=1$ and $a=2x+1$, so that bases got exchanged: that explains why you get the right result. If you had substituted $a=1$ and $b=2x+1$ you would have needed that factor of $2$ appearing in my formula. – Intelligenti pauca Sep 17 '15 at 09:40
  • But if we use your formula with $a=1$ and $b=2x+1$, we get $\frac{1}{2}=\frac{2x}{3}\frac{(2x+1)+2}{(2x+1)+1}$ which gives $\frac{1}{2}=\frac{2x}{3}\frac{2x+3}{2x+2}$, which rearranges to give $8x^2+6x-6=0$. – J132 Sep 17 '15 at 18:28
  • You are right: the formula should have been ${h\over3}{2b+a\over b+a}$, just the one you actually used. This is nothing but $h-{h\over3}{b+2a\over b+a}$. – Intelligenti pauca Sep 17 '15 at 20:42

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