Homework 7 Problem 3

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Monday, April 11th, 2016

3a)      Modeling the jumper’s center of mass as a particle, determine the minimum initial velocity required for the center of mass to reach the same height as the bar.  The jumper’s center of mass has an initial height of 1 meter.

3b)      Calculate the force required to accelerate the particle (jumper’s center of mass) from a velocity of 0 m/s to the initial velocity calculated (a).

3c)       Draw a 2D illustration of the jumper in the sagittal plane, taking off on one leg for the high jump.

3d)      Use the force calculated in (3b) and the anthropometric data to calculate the minimum net force at the knee if the planted foot is flat on the ground and the shin is at a 60 degree angle with the horizontal.


340 thoughts on “Homework 7 Problem 3”

  1. Nathan Shaw says:

    In 3a, we either need a distance from the bar or an angle for us all to get consistence results. If we want to truly minimize the initial velocity than 90 deg would do it. Since you would only jump straight up and down (no movement), should we find the delta x that would barely clear the width of the bar and back calculate the initial velocity from it? Any thoughts?

    In part 3b, do we need the jumper’s mass in order to calculate the force? I can’t see how to find the force without it. Do we simply look it up (who made the jump) or use a generic mass?

    1. Myra Fabro says:

      I was wondering what others did for this problem as well. I was assuming all vertical and a 90 degree angle.

      For 3b, I used the total mass (100 kg) from the table provided in 3d to keep things consistent since the force from 3b is to be used for 3d.

      1. Nathan Shaw says:

        I agree that we should use the table’s total mass of 100kg as the jumpers mass. The first time I read the problem it seemed like we were analyzing two jumpers, not one jumper just clearing the bar at each record. I am still hung up on the takeoff angle. If a jumper minimized his initial velocity and jumped at 90 deg he would not move horizontally and would have no chance of clearing the bar. I also agree with Erik that the shin angle is not the take off angle. I can jump at several take off angles while having my shin remain at a 60 deg angle.

        1. Benjamin Franklin Lee says:

          I treated it as if the jumper was jumping off 90 deg from the ground, such that it would be the “minimum” velocity because it would be infinitely small angle off of 90 degrees such that it could be approximated as 90 degrees.

          1. Nathan Shaw says:

            The problem statement asks for the minimum velocity needed for the CM to reach the same height as the bar, not clear the bar. I guess 90 deg would be that. It doesn’t seem to go with the problem context though. I would think we would want to look at the velocity needed to achieve a similar jump as the record holder (take off around 60-70 deg). Oh well.

          2. Myra Fabro says:

            That’s what I was thinking Nathan. Problem states jumper would just need to reach the same height as the bar and not clear it so based on how it’s worded, we can probably neglect some x distance.

        2. Erik Hansen says:

          90 degree provides the minimum velocity. The center of gravity for a high jumper is actually below the bar when they jump over it due to the arching of the body, so calculating it to the height of the bar should provide sufficient height to clear the bar.

    2. Katie Roth says:

      This paper seems to address common angle of attack (takeoff angle) for high jump. Looks like ~65 degrees would be a good average?

        1. David John Bonitsky says:

          Katie, that article is actually long jump and not high jump. I found a good picture by picture walkthrough of the high jump takeoff. It seems like the takeoff is essentially 85 to 90 degrees and their momentum carries them forward as well as shifting their COM in the air.
          http://www.coachr.org/rotation_files/rotati7.jpg

          1. Katie Roth says:

            Hi David,

            Thanks for pointing that out. I thought the angle seemed rather low — all makes sense now! 🙂

  2. Herberto Dutra says:

    I used the info from one if the subsequent items that stated :”the foot if planted flat on the ground and the chin is at a 60o angle”

    1. Erik Hansen says:

      The shin being at 60 degrees does not mean that the takeoff angle will be 60 degrees. It is just giving the position for the force calculation on the knee.

      1. Chris Valentino says:

        Erik, did you assume dorsiflexion or plantar flexion? I originally assumed dorsiflexion but am reconsidering after watching some high jump videos.

        1. Erik Hansen says:

          It will be plantar flexion since the ground reaction force is at the toes, so I think we may need to separate it into two FBD’s. One for the foot and one for the lower leg in order to connect the force at the toes to the knee. Thoughts?

          1. Chris Valentino says:

            But we are to also assume that the foot is flat on the ground so I think leaving the foot and leg connected is ok to find the reaction at the knee.

          2. Erik Hansen says:

            How are you going to apply the moment from the force at the toes to the lower leg?

          3. Erik Hansen says:

            Chris, I agree with you since the moment at the knee is not requested.

          4. Nathan Shaw says:

            I did it like Hwk 4, Prob 4a. TJR in one leg with the ground reaction force (previously calculate for the two heights) at the toes. The only difference is you now have the weigh of the foot to deal with.

  3. Herberto Dutra says:

    I got this text from a track and field blog that goes over high jumping techniques : “To achieve a maximal take-off, everything must be right. The take-off foot must be extended well forward of the COM (center of mass); the line of the body must angle backward at about 45 degrees (assuming a fast run); the take-off leg must be strongly braced momentarily so as to block all forward progress. With such a take-off, the jumper is forced upward, even without jumping. To understand this better, one might relate the action to the last two strides in the javelin throw.”
    Any thoughts ?

  4. Benjamin Franklin Lee says:

    This is probably a silly question, but I’m a bit confused as to what minimum “net” force is defined as for 3D.

    Is net force supposed to be only the X and Y forces or should that also include a moment?

    1. MacKenzie J Tweardy says:

      Benjamin, I am not sure if you will need a moment since knee extensions are typically modeled as single force, no moment. (Force in patellar tendon plus joint reactions)Because we are modeling this as a static problem, then the “jump” is just like an extension. Therefore, I am thinking that there should not be a moment.

      1. Erik Hansen says:

        I think the minimum net force at the knee is just the resultant forces in the E1 and E2 directions, and whether you choose to use total joint reaction or SFNM doesn’t matter because neither the moment or the patellar force are included in the net force.

        1. Herberto Dutra says:

          That’s what I found as well. You can plug the force components from the body pieces’ weights in the 60o direction and the ground floor calculated in the previous item. The forces in the E1 and E2 direction give two equations with two variables

    2. Emily Alexa Seyforth says:

      I am also using SFNM with the patellar tendon. If you use total joint reaction, the angle of the leg doesn’t make a difference for your resultant knee forces in the E1 and E2 direction, but I think Fpt will change based on the leg angle which will then affect your resultant forces.

      1. Emily Seyforth says:

        I am questioning this now though because if there is rotation about the knee then we would have to determine alpha for the sum of the moments equation in order to solve for Fpt

  5. Samantha Munoz says:

    What was everyone’s approach for part b? I was thinking of looking at the change in distance between bent leg and fully extended leg and trying to back out acceleration or time needed from that, then getting the force?

  6. Emily Elise Gill says:

    Samantha, I used the change in distance between the bent leg and fully extended leg with the work-energy equation to find the force, since we know the initial velocity and mass, and hence the kinetic energy at take-off.

    1. Samantha Munoz says:

      Thanks so much for your help Emily!

  7. Eryn Horton says:

    For part 3a don’t we need the height of the bar…? Or are we doing two calculations for both male and female assuming the bar is at 2.45 m and 2.09 m, respectively?

    1. Katie Roth says:

      I chose one and said I’d be assuming that height for the remainder of the calculations since the questions didn’t specifically call out a calculation for both heights.

    2. Nathan Shaw says:

      I calculated both. I used one jumper to see what it takes to make each record.

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