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 $x=\frac{√3}{2}$= $0.866$.