5

Using Figure 6 below, the two arms incermsop lte the motor unit recruitment pattern ofthe muscle fibers of fiber types, number and 'strength'. pts) 1000 9...

Question

Using Figure 6 below, the two arms incermsop lte the motor unit recruitment pattern ofthe muscle fibers of fiber types, number and 'strength'. pts) 1000 900 Figure 6 800 1 700 Dominant Arm 1 600 Non-Dominant Arm 500 3 400 300 J 200 100Force Generated (kg)

Using Figure 6 below, the two arms incermsop lte the motor unit recruitment pattern ofthe muscle fibers of fiber types, number and 'strength'. pts) 1000 900 Figure 6 800 1 700 Dominant Arm 1 600 Non-Dominant Arm 500 3 400 300 J 200 100 Force Generated (kg)



Answers

(I) Suppose the point of insertion of the biceps muscle into the lower arm shown in Fig. 9-13a (Example 9-8) is 6.0 cm instead of 5.0 cm; how much mass could the person hold with a muscle exertion of 450 N?

Young smarter. This is given by stress over strain. F over a over tells the l over l equals F l over a delta l equals F s over a delta l equals If the length is 0.2 meter divided by area ease 50 centimeter square. So 50 times 10 to the negative four made their squared And delta l is three centimeter or three times 10 to the negative two meter now for the two different forces. So you're going to find the young moderators for 25 Newton Force and 500 Newton Force If we put the values are force over here, we find that young spotted us to be 3.33 I understand to the fore Pasko's and for 500 new down 86.67 Time Stand to the five basket

So Party asks us to draw free by the diagram. And this would be the free by diagram for this system and for part B, they want us to find the magnitude of the tension force. So in order to do this, we need to use us the sum of the torque. This is in rotational equilibrium, in part be so we can say that the sum of the torque is going to be equal to zero Newton meters and we can say that this is going to be equal. Teo T. CO sign of theta times 5.5 meters the distance away from the axis of rotation. And then it will be minus the weight of the arm. Times Co sign fate again, times again, the distance from the axis of rotation. My apologies. This is actually centimetres. My apologies. And this would be 16 centimeters, and then it would be minus the weight of again 38 centimeters times coast on a fada. So this would be the weight that is in the hand. And essentially you can see that all the co signs they're gonna cancel out this equal zero, by the way, so we can say that T the tension force is simply going to be equal to 16 times the weight of the arm, plus 38 times the weight in the hand divided by 5.5 5.5 of the distance from the tension force to the axis of rotation. And we find that this is going to be equal to begin. Say T is going to be equal to 16 times 9.8. Yeah, gravity times 2.16 That would be the mass of the arm plus 38 times gravity times 7.5 kilograms of the mass of the weight. And then this would be divided by, of course, 5.5. We don't have to convert two meters in this case because the centimeter unit is of course, going to cancel out. And we'll only have an acceleration times a mass s o. This was going to be equal to 569 Newtons. So this would be your answer for a party for part. See, they simply want us to find the reaction force is so essentially east of H and e Savi. And so we can say that some of forces in the ex direction equals zero Newtons and the only force in the extraction is actually east of age. So east of H words, of course, equals their own humans. And we can say that the sum of forces in the wider action however, this is going to equal attention force minus e sabi minus the weight of the arm minus the weight in the hand So essentially east of V and rather, this would equal zero Nunes because it's ah in vertical. Oh, gather equilibrium in the white direction Translational equilibrium in the white direction. Rather and this would be equal Tio east of the would be equal to the tension minus the weight of the arm minus the weight in your hand. And this is giving us 569 Nunes minus 2.1 60 mass of your arm, plus the mass of the weight times the acceleration due to gravity. And we find that Isa V is going to be able to 474 noon. So this would be your answer for the y, the Y component, and then this would be the ex component for part D. They're asking us. Ah, here. How is the tension force changing as she raises her arm. Well, attention force We found to be independent of the angle. So here the tension force will be constant as she raises on. And we can say t independent of Fada. The angle measure that is the end of the solution. Thank you for watching.

So we need to find the torque needed and want to throw the ball and then the force exerted by the triceps. Since we know that the mass of the ball or are the last of the arm is three point six kilograms, we know the linear acceleration is seven point zero meters per second squared and the radius is going to be point three one meters. So we know that for part, in order to find the torque unnecessary, we find the moments of inertial times, the angular acceleration there's going to be m r squared times, the angular rather times the linear acceleration divided by our. So this will simply be a R. And this is giving us three point six times seven times point three one and this is equaling seven point eight one new in meters needed now. Given this, what's the force needed? So the force Ah rather we know Tower would equal the forced times the force perpendicular times a radius, which means that the force perpendicular is going to be equal to the Tao or the torque divided by the radius, and in this case it's going to be seven point eight one two divided by our so point three one. And we're getting that the force needed to force needed exerted by the triceps is going to be three point one two five. Noone's again exerted. Yeah, my apologies exerted. Bye, triceps! That is the end of the solution. Thank you for watching.

Once again. Welcome to new problem. This time we have one arm, Uh, that's holding on to a short boot. So these are these are, like fingers right here. Um, you know, just having an unlike that. And it's it's holding onto a short boot. Um, the arm itself, the short put itself is right here. So the mass, uh, of the moss off the short route happens to be given us, um, 77.3 kilograms. So that's that's the, um, the mask off the short for 7.3 kilograms. And then there are the piece of information that we're given is that he's a specific distance, This one right here. The distance between the elbow on the extensive in that distance are gonna call it, um d equals two small, the equals to two point five centimeters. Also, we're supposed to find, So this is the information was given. Uh, we're also given that the entire distance from this point up until the center off the short poot is early centimeters, the distances in centimeters Onda. We do see that the law um, this law I'm right. D'oh! I'm gonna call it M l Mass Law uh, has, um, us off 2.3 kilograms with the center of gravity. So this distance right here, that's the center of gravity. And up until there, that's 12 centimeters. This 12 is 12 centimeters, because that's the information that's given. Um, that's from the elbow. So was saying from the elbow, right here it's 12 centimeters. That's the distance. Our purpose is to find this force f m that holds on to the ball itself. Such that the ball doesn't get doesn't fall off. It's it's holding not the ball, but the short route. That's that's what's going on with this problem eso was supposed to find we want to find but force Mmm f m. You know, that's that's what we want to find. So that's the information we were given. Well, gonna use equilibrium and according to our equilibrium will get to see that this sum of all the forces on it's gonna be equals to zero. And also the some off all the talk about the elbow eyes also gonna be equivalent to zero. So gonna use talk and we'll interested in the talk around the elbow festival. Ah, we take the the mass the mass off the law firm. Uh, times 12 centimeters, actually. The weight, the weight off the law arm, remember, talk talk is defined us the perpendicular distance, times the force. This is the force. So we have to change all of these two weights a male G. So this would be m l G times 12 centimeters plus the must off the short poop, which is, um M s G ah, this time it's not 12 centimeters, but it's gonna be 30 centimeters, 30 centimeters, because you can see from the short boot up until the elbow right there. That whole distance is his 30 centimeters. And then, ah, all of that is you know, all this one is pushing down, and the lower arm is also pushing down. Uh, so this is clockwise is positive, But this force the one we're looking for, uh, counterclockwise is gonna be negative. And so in this problem, we have to subtract the talk you to that. So f m times the small distance D that sam is gonna be equivalent to zero. Okay, It's gonna be equivalent to zero. So in the next page next page, we simplify the problem. We have if enemy close to go back M l g 12 centimeters m l g tall centimeters, then, uh, plus em, um, m s plus m s g 30 centimeters. But remember, ml is the muss off the law, Um, and M iss is the must Oh, the short put. And in this case we're using, we're using talk to solve this problem, and then we want to divide by, you know, the small Dee Dee Dee's this distance right here, the distance between the elbow under under force itself. You know, this is the gap. So we wanna have this distance right here and fucked. Uh, assuming was fuming this elbow, it's gonna be right here. This is this is no assumption, right? Here is our elbow. But then there's a gap right there in that distance. Gap is 2.5 centimeters, and so the next step would be to plug in the numbers responsible for this problem. So 2.3 kilograms times nine points, eight me. This second squared times 12 centimeters plus. Remember? You know, you can change the centimeters two majors by doing that one meter, 100 centimeters. Uh, which gives you zero point 12 meters. Or you can leave it like that because of centimeters they're gonna cancel out. So mass off, uh, s is a mass of the shot 47.3 kilograms and multiply that by nine point h meet his second squared. And then you multiply that but 30 centimeters and divide everything by, uh, 2.5, but centimeters, So this sentimental will cancel that one and that one. So we just left with, uh, the product of a kilogram and meet it the second squared. And remember, one newton is equivalent to one kilogram up. Made a meet up one kilogram. He does, uh, one kilogram. Um, meat is per second squared. That's one mutant. So the problem is being transformed right away into Newton's, and that becomes approximately 900 70 mutants. So I hope you enjoy the problem again. It was, uh, on, um, at a law firm and an upper arm holding a short boat. And we used to talk because of some of the talk around the elbow right here is gonna be zero. So we define the talk as a result off the the law, um, and then sum it up with the talk off the shot put. Both of these are positive. And the reason why the positive is because it's choice clockwise would be positive. And then the force that we're looking for would have Ah, uh uh counterclockwise. And so it's negative. And then we end up simplifying the problem. Using our bro, we end up having 970 mutants, so I hope you enjoy the problem and have a wonderful day. Okay, thanks. Bye. You can send any questions or comments walking by.


Similar Solved Questions

5 answers
9 2 1)2 + 46y 3)?
9 2 1)2 + 46y 3)?...
4 answers
Point)Math 216 Homework webHW1O, Problem 3Find the solution to the linearization around zero of the systemx =-X-y_3xt,y 4x - y+ 3xy4with initial conditions x(0) =0. and y(0)0.8Complete the following two statements: The critica point (0,0} isunstable B. stable asymptotically stableand is a(n)A. spiral point improper node saddle point proper node center
point) Math 216 Homework webHW1O, Problem 3 Find the solution to the linearization around zero of the system x =-X-y_3xt,y 4x - y+ 3xy4 with initial conditions x(0) =0. and y(0) 0.8 Complete the following two statements: The critica point (0,0} is unstable B. stable asymptotically stable and is a(n)...
5 answers
Simple pendulum of mass Zmn and length is atrached t0 IMas 3m that _ constrained slide along frictionless horizontal (rack: Find the nortal mnde [requencies and corresponding normal modes of oscillation 3m
simple pendulum of mass Zmn and length is atrached t0 IMas 3m that _ constrained slide along frictionless horizontal (rack: Find the nortal mnde [requencies and corresponding normal modes of oscillation 3m...
5 answers
Poini)Find the position vector R(t) given the velocity v(t) = = 9e"ti + 8cos(4t) j + 12t2k and the initial position vector R(O) 2i + 2j +kR(t)
poini) Find the position vector R(t) given the velocity v(t) = = 9e"ti + 8cos(4t) j + 12t2k and the initial position vector R(O) 2i + 2j +k R(t)...
5 answers
025 3 PointsConsider the electron potential maps for the molecules shown below Which of the molecules is the most polar?Hoc poteme (o +0 A
025 3 Points Consider the electron potential maps for the molecules shown below Which of the molecules is the most polar? Hoc poteme (o + 0 A...
5 answers
Is / = 4 an eigenvalue ofIf so, find one corresponding eigenvector:Select Ihe correct choice below and, if necessary; fill in the answer box within your choiceYes; 2 =4 Is an eigenvalue ofOne corresponding eigenvector is(Type vector or Iist of vectors. Type an Integer Or simplified (raction for each matrix element )No; 1 =4 is not an eigenvalue ol
Is / = 4 an eigenvalue of If so, find one corresponding eigenvector: Select Ihe correct choice below and, if necessary; fill in the answer box within your choice Yes; 2 =4 Is an eigenvalue of One corresponding eigenvector is (Type vector or Iist of vectors. Type an Integer Or simplified (raction for...
5 answers
Assignment Gradebook ORIONEvaluate the integral 25 10xY x dx J16using Part 1 of the Fundamental Theorem of Calculus_click here to enter or edit Your answer1OxVxdxClick if you would like to Show Work for this question: Open Show Work Show HINTLiNK To TexTIn order to contin close the following SafariB0'0o86 'Qj
assignment Gradebook ORION Evaluate the integral 25 10xY x dx J16 using Part 1 of the Fundamental Theorem of Calculus_ click here to enter or edit Your answer 1OxVxdx Click if you would like to Show Work for this question: Open Show Work Show HINT LiNK To TexT In order to contin close the following ...
5 answers
5.6possible equation of state for & gas takes the form PV = RT exp VRT in which & and R are constants. Calculate expressions for 3) and show that their product is ~1, aS stated in section 5.4
5.6 possible equation of state for & gas takes the form PV = RT exp VRT in which & and R are constants. Calculate expressions for 3) and show that their product is ~1, aS stated in section 5.4...
5 answers
Find the particular solution to the differential equation$$y^{prime}=4 x+3$$passing through the point $(1,7)$, given that $y=2 x^{2}+3 x+C$ is a general solution to the differential equation.
Find the particular solution to the differential equation $$ y^{prime}=4 x+3 $$ passing through the point $(1,7)$, given that $y=2 x^{2}+3 x+C$ is a general solution to the differential equation....
5 answers
Find F(s)_ H{(1Se ~4f)cos(6t)}F(s)
Find F(s)_ H{(1 Se ~4f)cos(6t)} F(s)...
5 answers
Factor the following expressions completely _ 16.2162 + 7b17.x? _8x+1518.x2 _ 4
Factor the following expressions completely _ 16.2162 + 7b 17.x? _8x+15 18.x2 _ 4...
1 answers
Determine the minimum applied force $\mathbf{P}$ required to move wedge $A$ to the right. The spring is compressed a distance of $175 \mathrm{mm}$. Neglect the weight of $A$ and $B$. The coefficient of static friction for all contacting surfaces is $\mu_{s}=0.35 .$ Neglect friction at the rollers.
Determine the minimum applied force $\mathbf{P}$ required to move wedge $A$ to the right. The spring is compressed a distance of $175 \mathrm{mm}$. Neglect the weight of $A$ and $B$. The coefficient of static friction for all contacting surfaces is $\mu_{s}=0.35 .$ Neglect friction at the rollers....
5 answers
Assume that we are using three-digit decimal arithmetic. For € 0.0001 _ compute02Q0 +for @0 equal to each of 1, 2,and 3. Comment
Assume that we are using three-digit decimal arithmetic. For € 0.0001 _ compute 02 Q0 + for @0 equal to each of 1, 2,and 3. Comment...
5 answers
Question 2 (45 points) Show YOur steps clearly -2 3 Let A = -1 4 52-4 7(a) (10 points) Find the cofactors: Am, A21 and A3q (b) (15 points) Find det(A) , and det(2A). (c) (20 points) Find det(A) , by reducing A into a triangular matrix:
Question 2 (45 points) Show YOur steps clearly -2 3 Let A = -1 4 5 2 -4 7 (a) (10 points) Find the cofactors: Am, A21 and A3q (b) (15 points) Find det(A) , and det(2A). (c) (20 points) Find det(A) , by reducing A into a triangular matrix:...
5 answers
Lml Com Qaciso TestJekanine Leah ( CorJt 6r_ digdal3
Lml Com Qaciso Test Jekanine Leah ( CorJt 6r_ digd al 3...
1 answers
Use the Gauss-Jordan method to find $\mathrm{A}^{-1}$, if it exists. Check your answers by finding $\mathbf{A}^{-1} \mathbf{A}$ and $\mathbf{A} \mathbf{A}^{-1}$ $$\mathbf{A}=\left[\begin{array}{rrr} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]$$
Use the Gauss-Jordan method to find $\mathrm{A}^{-1}$, if it exists. Check your answers by finding $\mathbf{A}^{-1} \mathbf{A}$ and $\mathbf{A} \mathbf{A}^{-1}$ $$\mathbf{A}=\left[\begin{array}{rrr} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]$$...
5 answers
Scenario 3 Suppose we have a random sample Yi,Yz. Ya be idependent and identicully distributed gamma random variables (assume n is large) , with 7 but p unknown Derive large sample 95% Confidence interval for p.
Scenario 3 Suppose we have a random sample Yi,Yz. Ya be idependent and identicully distributed gamma random variables (assume n is large) , with 7 but p unknown Derive large sample 95% Confidence interval for p....
5 answers
The periodic time of the physical pendulum is Znk(gl)-42 where k Is the radius of gyration.
The periodic time of the physical pendulum is Znk(gl)-42 where k Is the radius of gyration....

-- 0.058000--