# How to determine the optimal length away from the pivot point and angle to the radius to use the least amount of force to rotate the lever?

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. http://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.

• You can eliminate $\theta$ since $\sin\theta = {H\over\sqrt{H^2+R^2}}$. – amd Jul 11 '16 at 18:09
• Write a function for the value you are looking to maximise or minimise in terms of $\theta$, so if you want to maximise or minimise $y$, write $y=f(\theta)$. Then differentiate with respect to $\theta$ to find the min/max of $y$. – user334732 Jul 12 '16 at 8:31
• You need to be clear exactly what you want to maximise. It looks like you want to maximise the torque at the hand per unit of force in the muscle. Would this be correct? – user334732 Jul 12 '16 at 8:34

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.

• The only flaw in this answer is that R is not fixed. R, the distance of F from the pivot point, is changing. The force is always pointing towards H, a fixed constant, so as R increases, the hypotenuse increases, and since sin(theta)=o/h, sine(theta) decreases. Torque is the product of F, sine(theta), and R, and I'm trying to prove mathematically what are the best values for sin(theta) and R since if one increases, the other decrease. If R was constant then, yes, your answer is correct. However, R is not constant in this particular problem. – Micah Woodard Jul 12 '16 at 21:37
• @MicahWoodard sorry I didn't look at the picture; I'll fix. – user334732 Jul 13 '16 at 6:39
• @MicahWoodard ok i've fixed but the solution appears illogical. I apologise but I can't see the error. – user334732 Jul 13 '16 at 8:09
• @MicahWoodard ok I fixed the last part; this appears to be correct now. – user334732 Jul 13 '16 at 8:47