How do I find torque on an axis when no weight or force is given?

Question

I need some help with the following problem:

Find the torque around the x axis of the triangle with vertices (0, 0), (1, 4) and (1, 0). Assume the density equals one. — Life of Fred Calculus, Stanley F. Schmidt

To be perfectly clear, this is the entire problem: there is no weight or force given!

Here are my thoughts on the problem / what I have already tried.

1. The triangle is essentially the graph $f(x) = 4x$ on $[0, 1]$

2. Elsewhere in the book, the author makes the assumption that weight is equal to $density \times x$, so $\tau = weight \times arm = 1 \times x \times arm$

3. It seems to me that the y-values of the points on the hypotenuse of the triangle (the values of $f(x)$) represent the arm length of torque. Thus,$$\int_0^1 4x\, dx = 2x^2 \Big]_0^1 = 2$$

$2$, therefore, is what I believe the answer to the problem should be. The answer book says that the answer is $\frac{8}{3}$. I have checked the author's errata page for the book, and this problem isn't listed. How do I solve this problem?

Answer 1

For any point at $r = \langle x, y\rangle$ the differential torque produced by gravity $g$ and the point's density $\rho$ is:

$$\vec \tau_{x,y} = \rho (\vec r \times \vec g)$$

Then the magnitude of this is : $$\begin{align} \vec \tau_{x,y} & = \rho (\vec r \times \vec g) \\[2ex] \parallel \tau_{x,y}\parallel & = \rho \, g\,\parallel r\parallel\sin \theta \\ & = \rho\, g \,y & \text{NB: for the first quadrant} \end{align}$$

Hence the magnitude of the torque over the entire triangle is:

$$ \parallel\tau\parallel \; =\; \int_0^1\int_0^{4x} \rho\, g\, y\; \operatorname d y\; \operatorname d x $$

Which shall resolve to $\frac 8 3$, counter-clockwise if we assume $g = 1$ in whatever units are being used.

Answer 2

It seems to me that the y-values of the points on the hypotenuse of the triangle (the values of f(x) represent the arm length of torque.

This would be a good approach if the entire mass of the triangle lay along the hypotenuse. It does not. The mass assumed to be uniformly distributed over the entire area of the triangle.

You can do a double integral in which you account for the moment arm of every point within the triangle, or you can take an integral over $y$ of the mass at each distance from the $x$-axis. (This mass decreases linearly to zero as you go from $y = 0$ to $y = 4$.)

An easier way is to find the centroid (center of mass) of the triangle, concentrate all the mass at that point, and use that mass and the distance from the centroid to the axis to find the torque. The centroid has $y$ coordinate $\frac 43$ and the mass of the triangle is $2$, so the torque is $2 \cdot \frac43 = \frac 83$. But that requires you to know already some non-trivial facts about how this notion of torque works.