I'm writing an essay for school in physics and my topic is torque. My topic deals with the elbow joint and the tendon that attaches the radial bone to the biceps, the force that rotates the forearm upward in supentination. The tendon attaches at a certain angle, which causes the force to be at an angle. My research goal is to find the optimal length away from the elbow joint in order to create a joint that requires the least amount of force to rotate it.
My set up is that of a third class lever with a weight at the end of the would be radial bone. The force pulls upwards at an angle and attaches at the humerus at a fixed height. H is constant as well and r and mg (the gravitational force on the weight). I know that the torque equation is $\text{torque}=RF\sin(\theta)$. To produce a certain amount of torque with the least amount of force, both $\theta$ and $R$ have to be at their highest values. However, as $R$ increases, $\theta$ decreases and vice versa. I'm assuming that both will be be the highest they can be in the middle but I don't know how to prove this. Any help at all in how to figure out mathematically which values are best would be greatly appreciated. The link of the picture is below. https://i.stack.imgur.com/38Cfk.jpg
Also, my teacher gave me the equation $\text{sum of torques}= 0 = RF\sin(\theta) - mgr$ (this is assuming that the set up is not rotating or has a constant acceleration). This turns into $RF\sin(\theta)=mgr$. I'm going to have to graph this equation with the data I collect but I'm not sure how because there are three variables: $\theta$, $R$, and $F$. My teacher vaguely referenced the lever equation and that this cancels out one variable. However, I found the lever equation to be exactly the same as the torque equation. Any help here would be greatly appreciated as well.
Let efficiency of the joint be $E = \frac{T}{F}$. We want to find $\theta$ such that $E$ is maximised.
$$T=RF\sin{\theta}$$ $$E=\frac{T}{F}=R\sin{\theta}$$
Now substitute in $$R=\frac{H}{\tan{\theta}}$$ to get $$E=H\frac{\sin{\theta}}{\tan{\theta}}=H\cos{\theta}$$
$H$ is fixed so we simply maximise $\cos{\theta}$ $$\frac{d\cos{\theta}}{r\theta}=-\sin{\theta}$$
To maximise, find where the derivative$=0:$
$$-\sin{\theta}=0\implies\theta=0+n\pi$$
Looking at a plot of $\cos{\theta}$ we can see $\theta=0$ is the maximum and it decreases approaching $\theta=\frac{\pi}{2}$ so the answer is to choose the most acute $\theta$ possible, and that means the greatest distance $R$ possible.