How Do You Calculate Force in Biomechanical Movements?

[chris.hess]

Homework Statement

How much force is exerted when an individual leg presses 68 kg twelve times in 24 seconds?

The leg press machine is a horizontal press machine.

Distance pressed forward 14 inches

2 seconds forward followed by 2 seconds reversed for each rep was held constant through all twelve repetitions.

Second question:

If a 100 kg individual with a 28 inch stride length and runs with a hip flexion of 45 degrees and these values are held constant, runs .11 miles in a straight line in 26 seconds, how much force was exerted by this individual?

Are the two forces the same? Could the running individual use the force from running and convert it to a value that could be used to weight train (force in pounds or kg?)?

If not what are the appropriate equations to find out what I am trying to find? Is there variables that I am missing?

Thanks for any help.

Chris H.

 

[Basic_Physics]

http://www.pro-fitness.com/horizontal_leg_press.htm

 

[chris.hess]

Sure we can use that machine as the example

 

[Basic_Physics]

I was hoping someone else will pick it up, but here is my take on the first problem -
It seems that the force from the leg is used just to raise and lower the mass pieces.
The force supplied by the leg is then just the weight of the mass pieces - W = mxg
where g = 9.8 m/s². That is if we assume no acceleration during the motion, which strickly is not true, but this approximation should be close enough. One could come to a better approximation by assuming an acceleration of the mass pieces over the 14 inch distance over a time period of 2 seconds.

 

[chris.hess]

ok...I can work with that. So the equation I will use for the first part is Work = mass * x? * gravity?

I am still stumped on what to use for the straight shot run estimate...will the top equation be used as well since I assume you use the mass of the person and again gravity is acting down upon the person? I figure that you would want to know time, total distance, estimates of velocity from 0-20%, 20-40%, 40-60%, 60-100% of the total run, an average angle for hip flexion, average stride length...so many variables to think of but if I get the total amount of estimated force exerted or work done for that matter, can the value be converted into pounds of force? I would then like to take the final number and break it down by percentages to build a training program and see if using such an equation will contribute to increased performance. Luckly I have a labview program in play but will need to rework various equations to make sure my math is correct to the given situation.

 

[haruspex]

There's an inconsistency in the data. 2 seconds each way makes a 4 second cycle, 48 seconds for 12 repeats. Since the info on cycle time and distance is given, I would assume you are expected to take the acceleration needed into account, though it only adds a few percent.
For the second question, from a crude perspective, ignoring all losses, a runner needs no force to keep going at constant speed. The minimum forward force to complete the run is therefore that needed to provide a constant acceleration to cover the given distance in the given time. This will be tiny and wholly unrealistic.
However, the other info given allows you to draw a free body diagram of the runner (except - they don't give you the hip width or leg length, so I'm not sure you have all the necessary data). Based on the posture, gravitational force, and the assumption of constant speed you might be able to work out the compression in the leg.

 

[Basic_Physics]

The problem I foresee with this appoach is that we can evaluate the forces acting at the centre of gravity (green dot) of the runner. Light blue his weight and orange the force providing the acceleration, but due to the complicated lever system connecting this to the ground one cannot easily evaluate the force at the foot. Maybe measure it with accelerometers?
http://technabob.com/blog/2012/02/23/nike-plus-pressure-sensors-accelerometers/

 

[chris.hess]

So as I put this together:

#1 an individual uses a horizontal leg press to perform 12 repetitions of 68kg over a total of 24 seconds. Distance covered forward 0.3556 meters and the same back to the starting position. It takes the individual 1 second to press the mass forward and 1 second to return to the starting position. Thus the velocity to press out the weight over the distance is 0.3556 m/s? Thus the acceleration would be 0.3556 m/s2? Using the equation listed above W=mxg 68 kg * 0.3556 m/s2 * 9.8 m/s2 = 236.9 N? Do I then multiply the N times the total amount of time (5687.3N) or the total number of repetitions of 12 (2842.8 N)?

Funny thing is I work in a lab that has a system that has a inertial sensors system that we will eventually use to track people outside of our motion capture lab. But what I am proposing isn't easy to get in since I have other things to work on during the day.

I know some of the things I am proposing are crude in nature but even if I can calculate a rough estimate and one day use the lab equipment to verify this idea, maybe it could be used to help those who can't afford $100+ pair of shoes.

By the way the individual in question, his hip width is 0.4064 meters and leg length is 1.0414 meters. 0.6604 meters for stride length. Let's says 6 degrees of trunk flexion for the torso and 25 degrees of flexion at the ankle, what other variables will I need?

I appreciate your help.

#2 An individual who runs 177 m over a period of 26 seconds has a velocity of 6.8 m/s? Thus their acceleration is 0.26 m/s2? Do I use the same W=mxg as above (100 kg*0.26 m/s2 * 9.8 m/s2 = 254.8 N)?

 

[haruspex]

So as I put this together:

#1 an individual uses a horizontal leg press to perform 12 repetitions of 68kg over a total of 24 seconds. Distance covered forward 0.3556 meters and the same back to the starting position. It takes the individual 1 second to press the mass forward and 1 second to return to the starting position. Thus the velocity to press out the weight over the distance is 0.3556 m/s? Thus the acceleration would be 0.3556 m/s2?

No. If the force is constant over the second then so is the acceleration. That means the velocity increases steadily from 0, and the average velocity is only half the max. Use s = at²/2.

By the way the individual in question, his hip width is 0.4064 meters and leg length is 1.0414 meters. 0.6604 meters for stride length. Let's says 6 degrees of trunk flexion for the torso and 25 degrees of flexion at the ankle, what other variables will I need?

So this is a practical problem, not a set/theotretical one. With a deal of effort and modelling, you might be able to come up with answer, but I'm not sure I'd trust it. At least, not without consulting a sports scientist.

 

[Basic_Physics]

You can measure the exerted force with a bathroom scale on the horizontal press machine. Put it between his foot and the machine. We can discuss it more if you want to calculate an estimate for it. Basically the force that the leg exerts when it extends on the machine, T, will be such that
T - W = ma
where W is the raised weight and m its mass, a the acceleration of the weight.

For the running problem I would think that a real person would quickly accelerate up to their max speed (within about 3 strides?) and then one would get a slight drop in speed as the advantage of the lever system decreases as the leg pivots backwards and repeat with the other leg, so it would be a sort of sawtooth function with time. The force from the leg is thus used to accelerate the person from a lower to a higher speed with each stride. This suggest that one should rather use an average speed for the whole of the event.

 

[chris.hess]

BP,

I have a couple of biomechanics books. Any specific chapters or other suggested reading material to wrap my brain around as I put together an algorithm that could perform a rough estimate for the run?

Again I appreciate your help. I don't mind doing the research and I appreciate when you tell me what everything stands for... can you elaborate on s = at2/2. s= speed? s= a(acceleration)t^2(time{seconds})/2