Torque on a right angle triangle I am attempting a question regarding the torque on a right angle triangle. The triangle is pivoted along the adjacent axis, there is no force acting on the opposite side, however there is a 12N force acting on the 1m hypotenuse into the plane of the diagram (the opposite side is 0.6m and the adjacent is 0.8m).

How do I go about calculating the tourque using Tourque=Force x Distance because the distance of the hypotenuse from the fulcrum is changing as you move along the length of the hypotenuse. I was considering taking the mean distance (0.3m) and multiplying it by the total force however now I am wondering if it is only the furthest point (0.6m) which will affect the torque of the system, however not all the force acts at the furthest distance and hence I cannot see how that is a valid option. This left me considering if this is an integral problem something like $\tau=\int_0^R dF.dr$ however this will just give me $0.6∗12=7.2Nm$.
Many thanks if someone can clear this up for me!
 A: Your intuition is correct, and you are close to getting the right answer. The proper way to treat this is with an integral, although the "average distance the force acts is 0.3 m" argument would get you to the same answer.
But let me show you the integral way.
Since we are only interested in the vertical dimension, we can simplify the problem and look at the force per unit vertical displacement - $f=\frac{F}{L_o}$.  Then the force on a segment $dy$ is $f\cdot dy$ , the torque due to that segment is $f\cdot y \cdot dy$ and the total torque is given as
$$\Gamma = \int_0^{L_o} f \cdot y \cdot dy = \frac12 f L_o^2 = \frac12 F L_o$$
Which is the same result you would have obtained from the simple argument I gave in the first paragraph
More generally, you would do the integral per unit distance along the hypotenuse; in that case, there would be a factor $\sin\theta$ that appears twice - once in the numerator, where it converts you from "distance measured along the hypotenuse" to "distance perpendicular to the axis", and again in the denominator, because the length along the hypotenuse is $\frac{1}{\sin\theta}$ of the vertical size $L_o$ of the triangle. So it would be more complicated, but all cancel in the end.
