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When performing calf raise 66 kg man experiences a darsiflexion torque of 64.6 Nm created by the ground reaction force under histoes If his Achilles tendon has mome...

Question

When performing calf raise 66 kg man experiences a darsiflexion torque of 64.6 Nm created by the ground reaction force under histoes If his Achilles tendon has momert armof 2,7 cm, how much force must he generate with the calf muscle (Fal t0 resist this torque?#Rourid LIAI GSNET(Wo declmalplaces You do not need Iriclude the Uallts In YOUF answer but you should Indicatc if it is positive orncgalne

When performing calf raise 66 kg man experiences a darsiflexion torque of 64.6 Nm created by the ground reaction force under histoes If his Achilles tendon has momert armof 2,7 cm, how much force must he generate with the calf muscle (Fal t0 resist this torque? #Rourid LIAI GSNET (Wo declmalplaces You do not need Iriclude the Uallts In YOUF answer but you should Indicatc if it is positive or ncgalne



Answers

A 75-kg man stands on his toes by exerting an upward force through the Achilles tendon, as in Figure 9.43. (a) What is the force in the Achilles tendon if he stands on one foot? (b) Calculate the force at the pivot of the simplified lever system shown—that force is representative of forces in the ankle joint.

So let us draw this in horizontal. We have got normal force over here if b over here and if a over here, these distances for centimeter and that distance is to us sandy meter. So this that's born a and this point be And that's point C Taking dark about a to B zero, we find, if b times for equals and chance to have us for so f B comes out to be 12 those four equals 16. Divide very four times the normal forces mg hooches 75 kilogram times 9.8 meters per second squared and that comes out to the 29 40 Newton. Now we can't drive that if a bus and it was f B. So if a comes out to be f B, which we just calculated to be 29 40 new done minus end blinky 75 times, 9.8 New done, and that comes out to be two. So 05 millions

So we have here is a question about a person who's hanging their lower leg. We want to know what force their femur is exerting on their lower leg. So we noticed that the leg is angled down at 30° and Patella Tendon is angled another 20° from there. So what we need to know is the angle of the skeleton you compared to horizontal. So if you know that the leg is angled down at 30° we also know that that angle between the horizontal And our 20° line Is going to be 30 plus 20. So the angle of our patella tendon Is going to be 50° compared to the horizontal. So what we need to know is that what we need to do first is draw a free body diagram on our level. Yeah. Honor where your leg, we got the force of the femur perfectly horizontal. The force of gravity materia, plus the force of gravity on the ankle rate. And then we have he forces the patella At an angle of 50°. Mhm. That's what we can do here is we got to break up our force of patella into two directions. So I'm going to get rid of my force of tele line and change this into The horizontal component which is going to be f. c. f. p. co sign 50° and vertically. Our forces patella times aside, 50°. Just figuring out the components of my force, We know everything has to be balanced because the leg is just sitting there, not accelerating. So we can write the following two equations. The force of the patella times a sign of 50° has to equal the force of gravity acting on the tibia plus the force of gravity acting on the ankle weight. And we also know the force of the patella outcomes to cosign about 50 degrees has the equal force of the femur. So we're gonna use this top equation first. So what we do know is that each of the forces of gravity is mass times gravity. So we can sell for our forces the patella here. So that's the teller and the sign about 50 degrees has to equal the massive a tibia Which is five kg Times 9.8 m/s squared experience plus the massively ankle weight. Mhm. Times the acceleration of gravity. Yeah. So when we solved here, so 57 20 plus three times in 20 divided by sine 50. You get the force of the patella as 102 .349 meets right. And now we can go ahead and use our second equation. Our force of patella terms of co sign 50 degrees equals force of femur. So potential force Come to co sign at 50°. And this gives us a force in the femur mm. Yeah 65 points 79 newtons to the wrecked. Yeah. Mhm

All right, This problem, you're asked to consider first free body diagram. Excuse me? Have a person in a barbeau and was to force exerted by the ground onto the athlete. So I've drawn the weight lifter and a drawn the barb. Ellis is a brick because the barbells just like an object he moves that 0.6 meters, 1.6 seconds. That's a velocity of 0.375 meters per second. So we have 21st. I want to draw two for by diagrams. Don't be the person one. And on the person one, we have their weight, which we could calculate. It's just, uh, 90 times before everybody ground onto the athlete. Okay, so we have, um, their weight, which we calculate to start, and that is going to be 90 times 9.8. So but my calculator, the Indy Times 9.8 you don't need to There's with Okay. Yes, we have 882 Nunes. Now it's going up on the person is the normal force because even equal ops reaction. So the normal force from the ground was athlete. And that's what I need to solve before we consult that we need to consider the free by diagram of the weight. All right, well, the wait or I should say Bar Bell saying, Wait twice in this kind of confusing. So the bar? Well, yes. Better. Okay, Barbeau. It has a downward wait of 400 Newtons that's given to us. All right. Now we need to think about the Upward Force and the Net force on the objects that goes up Sunnis. Be some acceleration. So we know using her equations. You know that the final velocity squared west initial squared is equal to two times acceleration on the object times, its distance minus original position. Its final velocity was 0.375 start at zero. Own explorations in the X directions would say extra time with the vertical execute. These wise would be more appropriate. Yeah, they were appropriately is wise. They were using vertical direction. We should be more careful and try to use labels that represent the axes were working with. All right, so we're trying to solve for the solution, so that's still unknown. It moved a total of 0.6 meters and we can say that started out. Uh, zero has been from 0.6 meters, where zero is a starting position. All right, is he will 1.2 times acceleration. The direction that is be equal to whatever 0.375 square it is. I think I typed in the wrong one, second 0.14 So the exploration of why is able to 0.14 divided by 1.2 0.117 or 0.1 to a mystic using two decimal places. Well, I could say used. Yeah, they used three desperately is 8.375 So I'm just gonna use three here. 30.117 is with that rounds to is get around. Um, since acceleration, the Barbeau has we know that it's 490 Newtons will be divide that by 9.8 we can get its mass is equal to so for 90 divided by 9.8 is 50 kilograms nice and even. All right, so that means that the we know that the force needed to be If we take 50 and we multiply that times 0.117 get 5 25 So the net upward force needed to be 5.85 Newtons, but remember that it has a downer force of 4 90 So if our net upward force was 5 25 Newtons, we needed to overcome the weight of the barbell which was working a downward direction with a force great enough that I when I overcome with that, I get this net amount of 5.5 and so that's pretty straightforward. You could calculate it. We could no, to Ah X, minus 4 90 needs to equal 5.85 And then we would sat foot on their side and we get that The upward force had to be 495 point 85 Nunes, which would give us the desired net force. And we did all this work because now we know they are free. Body diagram is almost complete, but now we can finish the person free by diagram. It's the person. If I push upward with the force of four or 95.85 Newtons, that means there has to be a force in the opposite direction and equal opposite force, which would hit which would push on the person and therefore pushed the person into the ground with that same force. So 4 95.8 five Newtons. So the person is at rest. They're not falling through the ground when this forces exerted on them. Therefore, the normal force from the ground onto the person has to equal this some. So we know that normal force is equal to do A two plus 495.85 13 soups So many commas 1377 0.85 Nunes So what is the told forces exerted by the ground on Fadhli? Um what if circled here and white or, in other words, this number here and then Our free by diagrams are as do are as shown here in the red and the blue and we are done.

We have got F Juan at an angular off 20 degree. So that means if one can be decomposed and we have if one co sign 20 degree and in this direction we have F one sign ready daily, no contentedly in the other direction. We have got our force off. If do so, the horizontal competent because if to signed 20 degree and the Vatican's competent becomes if to go sign 20 dd so we could just write them like this and we can see that if X it was, if one signed 20 minus F two sign 20 it was, If one minds F two signed 20 degree now F one equals F two. It was 200 new done. So that is 200 minus 200 times. Signed 20 degree equals zero. And if why eat was, if one does have to course, signed 20 daily, it was $200. 200. We'll sign 20 degree equals 376 New term in the aboard deduction


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