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Forces and Torques on a bicep

Why does the humerus, the elbow joint, exert a downward force in the following setup. I understand that the bicep must be involved in a force pair, but I identified the counterpart to the bicep's force (Fb) as the bone (basically, another body part) pulling the bicep downwards. The explanation I found is:

"Because muscles can contract, but not expand beyond their resting length, joints and muscles often exert forces that act in opposite directions and thus subtract."

Why does the muscle's inability to expand give rise to the humerus's downward force?

(http://cnx.org/contents/d703853c-6382-4035-8d6a-dcbca00a15ca@6/Forces_and_Torques_in_Muscles_)

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There are two force balancing considerations here.

On the one hand the torque about the joint must be zero, otherwise the arm would have some angular acceleration. Treating positive forces as directed in the directions they are drawn, this tells us $$ r_1 F_\mathrm{B} = r_2 w_a + r_3 w_b. $$ Given all the distances and weights, you can solve for the tension $F_\mathrm{B}$ in the biceps.

However, you also need the whole system to also not be linearly accelerating in the vertical direction: $$ -F_\mathrm{E} + F_\mathrm{B} - w_a - w_b = 0. $$ Solving for $F_\mathrm{B}$ in the first equation and plugging into the second, one finds $$ F_\mathrm{E} = \left(\frac{r_2}{r_1} - 1\right) w_a + \left(\frac{r_3}{r_1} - 1\right) w_b, $$ which is the same relation physicus derived by treating the torques about the biceps-forearm attachment point. Since $r_2,r_3 > r_1$, $F_\mathrm{E} > 0$ and must be directed downward.

The reason the text attributes this to the inability of muscles to extend is probably because one could imagine having $F_\mathrm{E} = 0$ but requiring the triceps to exert a force. Let's see what happens if there is an upward-directed force $F_\mathrm{T}$ a positive distance $r_4$ to the left of the joint. The two force balance equations are \begin{align} r_4 F_\mathrm{T} & = r_1 F_\mathrm{B} - r_2 w_a - r_3 w_b, \\ F_\mathrm{T} + F_\mathrm{B} & = w_a + w_b. \end{align} This system can be simultaneously solved for $F_\mathrm{B}$ and $F_\mathrm{T}$: \begin{align} F_\mathrm{B} & = \frac{r_2+r_4}{r_1+r_4} w_a + \frac{r_3+r_4}{r_1+r_4} w_b, \\ F_\mathrm{T} & = \frac{r_1-r_2}{r_1+r_4} w_a + \frac{r_1-r_3}{r_1+r_4} w_b. \end{align} Again, $F_\mathrm{B} > 0$ for any positive weights. However, $F_\mathrm{T} < 0$ no matter what $r_4$ is (so long as it's positive), since $r_2,r_3 > r_1$. That is, the triceps could not achieve force balance by contracting -- it would have to push on the back of the joint.

As a result, the bone undergoes a compression to balance things with $F_\mathrm{E}$. In fact much of what the skeletal structure does is provide pushing forces to balance the compressive load being put on it.

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I don't see how the inability to expand gives rise to this force. I would explain the forces like this:

In order for the situation to be stable, the total torque does not only have to be zero in the joint, but also at the point where the biceps makes contact with the bone, i.e. $$ r_1 F_E = (r_2-r_1) w_a+ (r_3-r_1) w_b\\ \Rightarrow F_E = (\frac{r_2}{r_1}-1) w_a+ (\frac{r_3}{r_1}-1) w_b $$ Plugging in the numbers from the image you provided one gets $F_E=407\,N$.

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Force $F_E$ is downward to prevent the hand from rotating. It is due to the balance of moments. Take $F_E$ away, and the arm is going to accelerate clockwise.

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