# How does one exert greater force on the ground by jumping?

When one jumps, how does he/she manage to exert greater force on their ground than their weight? Also, what is normal force and the reaction force (are they the same thing?) and by newton's third law, shouldn't the reaction(weight) when we are standing on the ground that the ground exerts on us send us flying above the ground- why doesn't the law apply here? Finally, when we drop a hard stone on the ground why doesn't it bounce? Plus, why is the force exerted by the stone on the ground greater than its weight?

In a standing state, your legs exert a force on the ground that correlates to your weight. Any motion or acceleration caused by the movement of your legs is added to this.

Think of jumping as generating upward momentum in your body. Your legs are causing you to accelerate upwards and gravity is causing you to accelerate downward. If you legs don't generate more acceleration than gravity, you will never leave the ground. Otherwise, your body builds up momentum until the acceleration caused by you legs stops. At this point, you leave the ground and travel upwards until gravity has reduced you upward momentum to 0. Then you start to fall.

Don't forget that momentum is stored energy and must be accounted for.

Sometimes a stone will bounce. This has to do with elasticity.

• I would not describe momentum as stored energy, but rather as inertia in motion. A battery stores energy but has no momentum. – KvdLingen Mar 14 '14 at 17:00
• Newton’s Second Law can be rewritten: force = rate of change of momentum. galileoandeinstein.physics.virginia.edu/lectures/momentum.html – Hoytman Mar 14 '14 at 17:09
• Yes, but force still is not energy. Force changes momentum, work changes energy. – KvdLingen Mar 14 '14 at 17:10

There's still something missing from all the answers so far. When you drop something on the ground, say, a rock of mass $m$, by the time it makes contact with the ground it's traveling at a velocity $v$ and thus has momentum $p = mv$. To be stopped completely, its momentum has to equal $0$ at the end. So you have a total change in momentum of $\Delta p$. According to (the most literal, I think) Newton's 2nd law, you have $\Delta p = F \Delta t$, where $F$ is the force slowing down the object over the timespan $\Delta t$ (in reality time is continuous and $F$ is probably changing continuously, but this is enough to illustrate the point).

So, if the $m = 1\ kg$ rock goes from falling at $v = 10\ m/s$ to $0$ in a millisecond or so, you might have $F = \Delta p/\Delta t = 10\ kg\ m/s /(.001s)=10000\ N$, which is obviously much bigger than just the gravitational force of $F_g \approx 1\ kg \times 10\ m/s^2 = 10\ N$.

• Actually I put it in, but your description is better. I edited mine a bit to better it – KvdLingen Mar 15 '14 at 8:29

When you jump, your muscles are using their chemical energy to contract or extend appropriately to create an additional force on the ground; by Newton's third law, the ground exerts additional force on you, overcoming your weight and pushing you off the ground.

In this case, the normal force is indeed the reaction force, as it is equal and opposite to (the perpendicular component of) the force you exert on the ground. There is indeed a normal force even when you don't jump, but this is balanced by the force of gravity pulling you towards the Earth. When you jump, these forces are still present, but the force of your jump increases the normal force such that it is greater than your weight (the force of gravity) and there is a net upward force, resulting in a jump.

The reason any object bounces is because it deforms very slightly when it hits the ground, but its natural elasticity sort of "pops out" the deformation (think of a basketball), pushing on the ground in a similar manner to a jump. Stones, however, are terribly rigid and inelastic, so by the time the ground applies enough force to deform it, it tends to simply chip or shatter rather than deform elastically.

I don't know why you say the force of the rock on the ground is greater than its weight, so I'm not sure how to answer that bit.

I think one reason it seems odd (at least for me) that we can jump, swim or walk for that matter, is because organic or inorganic motorized systems have the ability to make their center of mass move without the cause being an external force. I insist on the term "cause" here because the force from the legs extending is first transmitted to the ground (that is the cause) that then pushes up in return (that is external force).

The point is then that if we imagine our own body as a dead body (like a rock), there is no way we can understand how we jump in the first place; after all a rock doesn't jump!

Also, saying that there is a "reaction" from the ground doesn't solve the problem but merely avoids it and gives too much credit to the ground in my opinion.

I think the simplest way is to model it as "relativistic" collision which is just a simplification of (for people who haven't done QFT yet) or a refinement of (for those who have) Newton's 3rd law.

Let us describe the state before the jump in the frame of reference of the center of mass of the jumper and the Earth:

• The jumper has a mass $M_j$ and a momentum $p_j = 0$
• The Earth has a mass $M_E$ and a momentum $p_E = 0$
• The total momentum of the system is $0$ (because of the frame of reference) and the total energy is $(M_j+M_E)c^2$

For simplicity, let us now describe the state of the system after the jump once the jumper has reached the maximal distance and is about to fall bac on Earth:

• The jumper has a mass $M'_j$ and a momentum $p'_j = 0$
• The Earth has a mass $M'_E$ and a momentum $p'_E = 0$
• Taking as a reference potential the initial condition, we need to consider the distance $d$ separating the jumper from the Earth
• The total momentum of the system is still $0$ and the total energy is now $(M'_j+M'_E)c^2 + M'_jgd$

As far as I am concerned, nothing much has happened to the Earth and therefore $M'_E = M_E$. Applying conservation of energy we thus find that:

$M_j-M'_j = M'_jgd/c^2$

Hence the only way to jump is to loose mass! But obviously here the explanation is not that because you lost mass the gravitational pull is smaller and hence you jump, the change in mass is probably of the order of the mass of cell in your body and plus the potential energy minimum is always the ground regardless of your mass.

No, the explanation lies in the energy interpretation of mass. If I consider an 80 kg person, rounding the gravitational pull to $10\:m.s^{-2}$ and $d=1m$, we get that the mass-energy difference is $800$ Joules.

Again, it has been provided by a mass loss that originates in fact from exothermic bio-chemical reactions (mostly coming from ATP I presume) used to make molecular motors, responsible for muscle contraction, work. A rough estimate from the data I could gather would be to say that for this jump, about 0.1 moles of ATP have been used to process it. Trying to go further is a bit delicate but that could correspond to making 50% of the ATP in your legs work 10 times in a row (not necessarily the same molecule) over a fraction of second.

In summary, from a physical point of view, it is as if the jumper was in an excited state before jumping and then went down in internal energy by emitting this energy which is then converted into internal kinetic energy of the legs parts finally translated into motion of the center of mass of the jumper after "collision" with the ground.

Sorry for the novel ;).

We can also use the idea of work and energy here. When we jump, the chemical energy in our muscles get converted to kinetic energy that exerts a force on the ground (that using Newton's 3rd law), will trigger a reaction that will 'push' us up- and thus a jump. This resultant normal force = Mass x Distance of jump. When we are jumping, we accelerate initially by F=ma, but then by gaining GPE and thus, a fall in kinetic energy, the speed reduces to 0, in the apex of our jump. Then, we fall- accelerated by gravity and while falling, we have a velocity and thus, a kinetic energy and when we strike the ground, we do work on it, transferring energy into it and thus, force (as the ground has too much inertia to move a lot- but enough kinetic energy can cause this- craters etc.). This force exerted + the weight is the total force we exert once we fall on the ground after jumping. Newton's third law is when A exerts a force on B and B re-exerts the force on A as a simultaneous reaction. When we are standing on the ground, we are 'stretched' by two forces- normal and the weight that cancel out and leave us at equilibrium. That stone will not bounce simply because it is inelastic and incompressible- the ground would rather crack than let the more incompressible rock bounce. However, dropping the stone on a hard ground- like steel or any incompressible surface will result in the rock shattering due to the collision and momentum or bouncing.

A net force is required on a object to give it an acceleration. In this case the net force on the body is given by $F_{net} = F_{grav} - F_{normal}$

To jump up, the normal force has to be greater than the force of gravity. As the normal forces forms a reason pair with the force the body exerts on the ground (usually called weight) increasing the force of your body on the ground by stretching your legs, increases the normal force.

The same applies to the falling rock. Because of the momentum of the rock has to be changed, the force of the rock on the ground is bigger than the force of gravity, therefore the normal force on the rock is also bigger than the force of gravity, generating a net force on the rock which opposes the direction the rock is traveling in. So the velocity of the rock decreases.

Jumping of the rock does not depend solely on the rock. If the ground has some elasticity, kinetic energy is transferred back to the rock causing it to jump. You can observe this by letting a rock fall on a trampoline.

## protected by ACuriousMind♦Dec 27 '17 at 13:54

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