When the palmaris longus muscle in the forearm is flexed_ the wrist moves back and forth: If the muscle generates force of 43.5 N and it is acting with an effeclive...

Medical When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth (Figure 8-51). If the muscle generates a force of 45.0 $\mathrm{N}$ and it is acting with an effective lever arm of $22.0 \mathrm{~cm}$, what is the torque that the muscle produces on the wrist? Curiously, many people lack this muscle. Some studies correlate the absence of the muscle with carpal tunnel syndrome. Example 8-12

For the elbow for the elbow. Toby Pie voted. We must balance. We must balance talk, do toe palm and do toe triceps so we can write late F T is equal to triceps falls. Now I am drawing the diagram with the help off the data that are given in the question. So the diagram looks like this. Yeah. Now, with the help of this diagram, we will write some equation so we can write a equation. 90 into 30 into 10 to power minus two is equal to 2.4 into 10 per minus two into empty. So after we can be calculated from this. So on solving this we get F t is equal to 1125 Newton.

Over. Okay, so now we're going to be doing Chapter one problems. 65 2 This one is kind of complicated. We're talking about a patient in therapy, and they have a forearm that weighs 24 5 News, but and lit that can lift 112 thie. Only other significant forces on the harm come from the biceps muscle, which acts perpendicular, perpendicular early, 24 on and a force at the elbow. If the bicep spirits of pole of 232 7 magnitude of the bicep forces 232 years of that, when the forum is raised at 43 degrees, okay, above the horizontal, and then we want to find the magnitude direction of the mobile force. So we know that elbow plus bicep is going to equal this resulting. So now lets you do all this to figure out where we're talking about. So char X y plain our resultant force needs to be just straight up. We know that, and this is the bicycle force, and this is at a angle of 43 degrees. Okay, Cool. So we can rearrange this equation here toe right? In terms of you know this is our minus B. Okay, so in components we can write, he acts are exploiters the X, which equals zero minus 2 32 Time's a sign of for these three to get the X component. So it's 158 Newton's and the ex So sue the same with the why here This becomes 130 2.5 because that is the total weight from the forum, plus the wait. So this is the week er that resulted vector magnitude, and we can subtract B y. So this is 132.5 minus 2 32 Times Co signed a 43 so that comes out to be native. 37. Newton's Okay, clear. So now that we have that, we could find the magnitude of the elbow force. This is a square root of 37 squared plus 1 58 square. I could put that on a calculator. He's our elbow, has a magnitude force of 162 Newtons. So let's draw this and our ex wife trying to figure out what direction the elbow is, right? So if it's 158 Newton's not degrees on the wire, about that 158 newbies north her up in the X direction right such this way 1 50 feet in the negative, 37 down here. So the elbow wasn't doing this kind of force. So for wanting to find this angle here, we can take data equals arc Tangent of negative 37 over 1 58 and we see that this becomes 13 degrees south of horizontal, which is also the same as negative 15 degrees. Cool, then we have the magnitude and the direction of our resulted vector our elbow after not a result of awesome.

So we're given an arm to analyze and the forces in it When someone is holding their arm, I'm vertically or horizontally. I mean, so we know that we're told that the arm weighs 41.5 Newtons. And since I'm using gonna use positive being up, I'm going to say that that's a negative load because I'm using factors. I want to make sure I'm I'm using the correct signs for my coordinate system. We're also told that the muscle that keeps your that will keep your arms supported acts at an angle of 12 degrees and has a force of that. I'll note by F key. And so the X component is this. And then the Y component is here, the distance from the shoulder so where your arm is supported by your shoulder socket. Shoulder blades is 29 centimeters, and the distance from that point to where the muscle attach is is eight centimeters, and so were asked to determine the magnitude of the tension force in the muscle, and then the force exerted by the shoulder on the humerus, the upper arm bone. So again the shoulder passed through, uh, with forest passed through your shoulder come into the upper arm bone. So first we do is we take moments and I picked 0.0, which is at the point where you're showing your arm. Attach this to your shoulder. Here, take moments about that. We have the moment from the weight of your arm and then the moment from the muscle that will counteract this way to keep your arm supported and from rotating downwards. We also have so from this weekend just we can solve for f t. Why? Because that's the only thing unknown in this equation. We find that that's 150 Newtons. Once we know this, we know angle here. Sorry. Once we know, once we know this and we know this, we could find the tension in the force in the muscle. And that is 724 Newton. And once we know this, we confined the force in the X direction or the horizontal force. And that is minus 700 eight. So that means it is again pulling in the negative X direction on the arm. Judge, you expect? Um, no, we have the want to figure out what the forces here and the joint are. So we write a vector equation, of course balances. So the some of the forces must equal zero. And we know this factor. And we know this factor so we can find this factor. And it has components in the X direction of 708 Newtons and in the Y direction, um, minus 109 news. And so the magnitude of that force is 716.

For this question, we have the magnitude of the force that exerts has exerted by the bicep equal a 232 Newton's call that beak, and it's raised an angle of 43 degrees above horizontal. So if they did, B is equal to 43 degrees are here is the result of Vector that we're told the vector. Some are must be equal to 132.5 Newtons, which comes from the 20 point find Newton's from the weight and the 112 Newton's that is lifted there at the beginning. So then, to find the resulting force exerted by the elbow, we can consider the fact that our is going to be equal to the force of the bicep. Since everything is an equilibrium, are has people to force the bicep, plus the force of the elbow. Or, in other words, the elbow is equal to the uh resulting vector are minus the force of the bicep bi, so we can break this into his X and Y components. So we have B of X is equal to or vex minus B of X, but the result in vector are is all in the wind direction. So this is going to be equal to zero. So we have is negative b of X. So in the ex direction, uh, B is negative. So this is minus b times the co sign or excuse me? Uh, time to the sign of the angle data be. So this comes out to equal 158.2 Newton's and its sign here from our target a metric identity because of because sign is opposite over high pot news. Okay, so that's why we need to consider e of X to be the sign. Uh, because be here is tthe e high pot news. Andy of X is the opposite, uh, force there. This would be 158.2 moons e supply likely are likewise is equal to our supply minus b supply. So all the value of ours in the UAE direction So this is equal to the magnitude of our, which was 100 32.5 millions minus b survive, which is be times the coastline of the angles. They to be okay, so plugging those values and we find that this is equal to negative 37 point to Nunes. Okay, so now that we know you have X, any of why we confined the magnitude of the force exerted by the elbow in choosing Pythagorean theorem that's gonna be equal to the square root of you have X squared plus e of y squared. So plugging in the values we just found for e of X and free of why we find that the magnitude of V is equal to 162.5 Nunes 162.5. Now we want to find the angle that this makes with the horizontal. So the angle we can call that Alfa is gonna be equal to the inverse tangent via via ovary of ex since through our triggered a metric identity of lie over A of X is equal to the tangent of Alfa so we can solve for output by taking inverse tangent of you've I already of X playing in those values, we find that this is equal to 13 degrees of course, that is 13 degrees below the horizontal gun box and as their solution