Calculating Force to Propel 160-lb Body Up 35 Inches

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Discussion Overview

The discussion revolves around calculating the force required to propel a 160-pound body upward by 35 inches, with considerations of elasticity and the complexities of human biomechanics. Participants explore theoretical approaches, practical implications, and the challenges of measurement in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks to understand the force needed to jump a specific height and questions the validity of simplifying their body to a point mass.
  • Another participant notes that the force required depends on the time interval during which the force acts, emphasizing that it is an initial force causing acceleration.
  • A suggestion is made to use conservation of energy principles to relate kinetic and potential energy to find the initial velocity needed for the jump.
  • Discussion includes the need to consider the duration of force application and how it relates to the work done and kinetic energy gained.
  • One participant proposes measuring jump time using slow-motion video as a more accurate method than a stopwatch.
  • Another points out that the velocity needed to reach the height is independent of mass, but mass must be considered when calculating work done to achieve kinetic energy.
  • Concerns are raised about the impact of elasticity, with questions about its significance in the force calculation and the complexities of the human body as a non-point mass system.
  • One participant argues that the force exerted during a jump is not constant and is significantly higher than the weight used on an exercise machine.
  • Another participant suggests that the problem is fundamentally about energy and that any force must exceed the weight to achieve lift.
  • A more practical approach is proposed, involving measuring maximum force exerted during a jump to inform training regimens.

Areas of Agreement / Disagreement

Participants express various viewpoints on the complexities of calculating the required force, with no consensus on a definitive method or answer. There is acknowledgment of the limitations and assumptions involved in the discussion.

Contextual Notes

Participants note the challenges of measuring force accurately, the dependence on various factors such as time and elasticity, and the simplifications made in modeling the human body. The discussion highlights the need for further empirical investigation and consideration of individual biomechanics.

Who May Find This Useful

This discussion may be of interest to athletes, coaches, biomechanics researchers, and individuals exploring the physics of human movement and force application in sports.

royzizzle
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I am a basketball player. how would I solve for the force needed to propel my 160 pound body(simplifying my body into a dot) up a vertical distance of 35 inches.
Also would elasticity, etc have a large impact. Is it a big error to simplify the situation in this manner?
I want to know this so I can put the weight on the squat machine equal to this force

thanks
 
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The force needed depends on the time interval (actually distance) during which that force acts.
 
Matterwave said:
The force needed depends on the time interval (actually distance) during which that force acts.

the force which propels my body comes from the leg muscles. so it is an initial force that causes my body to experience acceleration in the upward direction. That is, the force is not acting the entire distance of 35 inches.
So I guess my question becomes how fast must my initial velocity be if I hope to go upwards 35 inches. and how do I calculate the force needed for this velocity?

thanks
 
Hi, the first one of your questions has an easy answer. If you're familiar with conservation of energy you can work it out knowing that at the top of the trajectory, your kinetic energy is zero, while at the start, your potential energy is zero. Since the energy is the sum of these two terms and it remains constant, then you can put your K.E ((mv^2)/2) at the beginning equals to the P.E at the end (mgh), equate this two, solve for v, and you get your initial velocity as a function of h, the upward distance you want to travel (make sure you get the right units!).
For the second one, you need to take into account what matterwave said. The force in your muscles is not an instantaneous one, and it acts on your body for a certain period of time. The work done, is equal to the change in kinetic energy (as long as your potential remains constant), and if you are familiar with integrals, the work in terms of the force is just the integral of the force with respect to time. So if this work done is the change in your kinetic energy, and you started out with a K.E. of zero (at rest in the vertical direction), then the work would equal to mv^2/2, where v is the velocity you would've gained.

I would suggest that you pay close attention to the way you propel upwards when you jump, the way your legs push downwards to the ground as you do it, and then try to measure the length of time you do this push (you could do it holding a stopwatch while you jump).
I believe the way you would apply the force would look like a 'spike' function, like a bell; my first guess would be to take your force function as F_max * Exp((-x^2)/k), where k is a factor you'd need to tune, so as to have most of the work done (say, 99%) at the time interval you measured and then you could find F_max as to make this work equal to the kinetic energy you need to gain. Ask someone to help you solve for these parameters.
If my guess is correct, what would happen is that you start doing a bit of a force, then build up this one until you reach the maximum, and then start to reduce the force as you raise. In this case you'd need to work out using the F_max, or a bit more maybe (again, i might be wrong, let's see if anyone corrects me, or take this argument and discuss it with someone you know).
If anything here sounded awfully complicated, ask and i'll try to put it in a simpler form. Good luck :)
 
Last edited:
i think that measuring the time it takes you to jump would be complicated with just a stopwatch. maybe you can film yourself and then play it in slowmotion to see how much time it took you to make the jump.
 
one more thing. when you solve for the velocity needed to reach the height, realize that it's independent of your mass. But you have to take into account your mass when you calculate the work you need to do to gain the needed kinetic energy (that is, the work = mv^2/2, where v is the velocity you calculated before, the one you need to reach the height) and in this case, the mass is not equal to your weight (just in case, sorry if I'm saying things you already know)
 
thanks!

one final point. will the effects of elasticity have a significant impact on my answer? if so, about how much weight should I add on the machine? thanks
 
Elasticity of what? Of your tendons?

I can't see any elasticities coming into the force equation; however, do realize that the body is an extended and highly complicated object with many different connections and levers. To approximate your body with a point is pretty ridiculous. To build the strength necessary isn't just to be able to apply the same force over and over. The way you apply the force also matters, especially in high resistance cases where you don't want to hurt anything. To get any sort of indication on what you may want to do you should see a strength trainer or your coach.
 
To be honest, you're not going to be able to compute the force. The force is not constant, but even if it were, it'd depend on how long that force was applied or over what distance. The problem is the time is literally how long your feet are pushing off the ground which is something you can't measure without special sensors or possibly estimating with high-speed video.

Also, the force is going to be extremely high compared to the amount of weight you'd want to use on an exercise machine. The weight on a machine is a steady force while a jump is a quick, large force.
 
  • #10
As everybody tells you there is no answer just constrains.
One is that the force has to be bigger than your weight so F>m*g
and from than on any answer is correct, given a certain distance upon that force pulls.
It is an energy problem m*g*h(eight)=F*x(distance)
 
  • #11
well, yeah. even with all the approximations i gave up there, I'm pretty sure your body is a lot more complex and what i gave was an overly simplified (possibly wrong) way to see the problem. I was thinking that maybe you can get a good estimate by jumping over a big balance (like those you find in the airports, maybe bigger) and check what's the maximum force you do to get as high as you can. Then train your muscles to be able to make this force easily, measure again, train again, and repeat until your desired height is achieved. That's what I would do, but my career does not depend upon it, if you know what i mean ;)
 

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