- #1

- 50

- 0

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

- Thread starter royzizzle
- Start date

- #1

- 50

- 0

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

- #2

Matterwave

Science Advisor

Gold Member

- 3,965

- 326

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

- #3

- 50

- 0

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

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

- #4

- 45

- 0

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 beggining 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 wanna 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 :)

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:

- #5

- 45

- 0

- #6

- 45

- 0

- #7

- 50

- 0

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

- #8

Matterwave

Science Advisor

Gold Member

- 3,965

- 326

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.

- #9

Pengwuino

Gold Member

- 4,989

- 16

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

- 555

- 23

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

- 45

- 0