Why when you run then kick a wall you feel an upwards push?

[Thoams Jerome]

Sounds wird but I tried it and people use this it help clime walls. How does this happen?

 

[sophiecentaur]

I think it works because the pressure against the wall, due to your rapid horizontal deceleration allows you to provide a lot of downwards friction force (without hitting the wall hard, you feet would just slip downwards). There is another factor, too. By virtue of the fact that your foot doesn't slip down the wall you can transfer some of your Kinetic Energy to Gravitational Potential Energy (taking you higher) in the same way that a pole vaulter uses the pole or as you can hit a steep ramp on a bike. That will 'feel' like an upwards push due to the free energy.

 

[Thoams Jerome]

I think it works because the pressure against the wall, due to your rapid horizontal deceleration allows you to provide a lot of downwards friction force (without hitting the wall hard, you feet would just slip downwards). There is another factor, too. By virtue of the fact that your foot doesn't slip down the wall you can transfer some of your Kinetic Energy to Gravitational Potential Energy (taking you higher) in the same way that a pole vaulter uses the pole or as you can hit a steep ramp on a bike. That will 'feel' like an upwards push due to the free energy.

Do you know were I find the equation for the kinetic to gravitational energy? I saw some one doing this, So I'm trying to find a way to take two steps up a wall instead of one. I needed to know if it was friction or something else at play thank you.

 

[sophiecentaur]

Do you know were I find the equation for the kinetic to gravitational energy

Starting with initial KE = final GPE (and that is the maximum, not allowing for loss)
KE = mv²/2
PE = mgh
So you can equate those two and m cancels out, giving you
h = v²/2g
But that value of h will be pretty optimistic.
Without the friction, you couldn't push down.

 

[Mister T]

Do you know were I find the equation for the kinetic to gravitational energy? I saw some one doing this, So I'm trying to find a way to take two steps up a wall instead of one. I needed to know if it was friction or something else at play thank you.

It's entirely friction. Looking at the energy instead of the force doesn't change this, it's simply another way of looking at the situation.

The friction force is an upward force on your body, which causes an upward acceleration. This allows you to convert the kinetic energy you gained during the acceleration to potential energy.

 

[sophiecentaur]

It's entirely friction. Looking at the energy instead of the force doesn't change this, it's simply another way of looking at the situation.

The friction force is an upward force on your body, which causes an upward acceleration. This allows you to convert the kinetic energy you gained during the acceleration to potential energy.

That's not consistent with rest of your post. The friction force is very much dependent on the retarding force, of course, but the KE is very relevant and can be used if you 'get the angles right'.
If you had magnetic boots, you could climb the wall without starting with any KE but the Power you can deliver is very limited (around 1kW even over a very short period). Using your KE will take you faster to a given height. Once your forward speed has been lost, and the normal force on the wall has dropped to zero, you have no traction but KE can take you higher than that point.
There is a fair degree of skill in doing this trick properly. Unless you get it right, your body never gets any lift and you fall back on the ground. You have to attempt to do the same thing with your legs as a curved ramp would achieve for a wheeled vehicle. It looks annoyingly easy when an expert does it.

 

[Mister T]

Once your forward speed has been lost, and the normal force on the wall has dropped to zero,[...]

By that time you have used the friction force to accelerate yourself upward, so you now have some upward speed.

[...]you have no traction but KE can take you higher than that point.

Yes, because you have some upward speed.

 

[sophiecentaur]

By that time you have used the friction force to accelerate yourself upward, so you now have some upward speed.
Yes, because you have some upward speed.

I think we are actually singing from the same hymn sheet and that actually both friction and energy need to be considered in the context of the 'trick'.

 

[Khashishi]

Friction is approximately proportional to the normal force and coefficient of friction.
$F_f = \mu N$
By kicking a wall, you provide a normal force and a downward force on the wall. If the downward force is lower than the static friction, then you will grip to the wall, and you can push yourself up. Otherwise, you will slip, but still accelerate up a little bit due to dynamic friction, which is lower than static friction.

 

[sophiecentaur]

By kicking a wall, you provide a normal force and a downward force on the wall.

There is more to it than that. By running at the wall, you have Momentum and time for which you can exert a force can be much greater than with a standing start. The Change of Momentum is equal to the Impulse (=Force X time). Initially, your leg is bending, reducing your momentum and applying the necessary force to provide friction. If the contact point with the wall is below the level of your Centre of Mass there will be a vertical component of force which will push you upwards even if you don't attempt to climb.

 

[Mister T]

I think we are actually singing from the same hymn sheet and that actually both friction and energy need to be considered in the context of the 'trick'.

They are two equivalent ways of explaining the phenomenon.

 

[Mister T]

By kicking a wall, you provide a normal force and a downward force on the wall. If the downward force is lower than the static friction, then you will grip to the wall, and you can push yourself up. Otherwise, you will slip, but still accelerate up a little bit due to dynamic friction, which is lower than static friction.

The downward force exerted on the wall is the friction force. If its magnitude is smaller than the maximum value of static friction then you don't slip, if not you do slip. But I agree that either way, it's the upward friction force exerted on the person by the wall that causes the upward acceleration of the person.

 

[jbriggs444]

Starting with initial KE = final GPE (and that is the maximum, not allowing for loss)
KE = mv2/2
PE = mgh
So you can equate those two and m cancels out, giving you
h = v2/2g

The justification for equating PE with KE seems to be that one can strike the wall with the feet in an elastic collision without slipping. Both suppositions are questionable, but reasonably close.

If one could strike the wall and bounce elastically as if it were a 45 degree slope and if the coefficient of friction were at least equal to 1 then this could be achievable with no additional energy input. [Imagine a stiff but perfectly elastic pogo stick hitting the wall at a 45 degree angle without slipping]

With my legs and age, the rebound would be considerably less energetic than the impact. With a ski jumper's strength and skill, the rebound could plausibly be more energetic than the impact. If this were the case, one would have to increase the coefficient of friction. Or go with a non-vertical rebound angle, sacrificing some vertical rebound velocity for horizontal rebound velocity and the corresponding increase in horizontal impulse.

A conservation of momentum argument indicates that vertical impulse is limited to $\mu$ times horizontal impulse where $\mu$ is the coefficient of static friction.

 

[CWatters]

+1 to most of the above posts.

The reaction force where a foot meets the wall will contain components due to friction and the normal force. Another way to think about it is that the resultant provides a centripetal force acting on your body. The resultant is also unlikely to act through your centre of mass so it might also produce a torque that rotates your body.

I've shown a more or less constant radius but that's unlikely to be true in most cases. Might depend on your objective (simple jump up or somersault etc).

 

[Mister T]

The justification for equating PE with KE seems to be that one can strike the wall with the feet in an elastic collision without slipping. Both suppositions are questionable, but reasonably close.

If you model the person as a particle, yes.

[Imagine a stiff but perfectly elastic pogo stick hitting the wall at a 45 degree angle without slipping]

With the pogo stick spring originally compressed somewhat so that some of the spring potential energy present prior to the collision gets converted to kinetic energy of the jumper. This is the conversion that a skilled jumper does, with his body supplying the energy rather than the spring of a pogo stick. This is the thing that the youngster can do better than the old dude.

 

[Mister T]

The reaction force where a foot meets the wall will contain components due to friction and the normal force.

True. They indeed are perpendicular components of that force, even though it's conventional to refer to them as separate forces.

But it's the vertical component (friction) that's responsible for the vertical component of the acceleration.

 

[jbriggs444]

If you model the person as a particle, yes.

Well, I was imagining the person as being passive so that "elastic" is the best collision you can get. I think that is what you were getting at with the "particle" characterization.

Edit: I also was not thinking of something with rotation such as a Superball striking the wall with lots of top-spin.

With the pogo stick spring originally compressed somewhat

Yes, I agree that one can do better than elastic with an appropriate active mechanism such as a jumper or a pre-stressed pogo stick.

 

[Thoams Jerome]

Starting with initial KE = final GPE (and that is the maximum, not allowing for loss)
KE = mv²/2
PE = mgh
So you can equate those two and m cancels out, giving you
h = v²/2g
But that value of h will be pretty optimistic.
Without the friction, you couldn't push down.[/QUO
thank you

 

[Mister T]

Well, I was imagining the person as being passive so that "elastic" is the best collision you can get. I think that is what you were getting at with the "particle" characterization.

Yes. Having no internal structure, a particle is not capable of storing (or releasing) internal energy.

Skilled athletic moves, such as the one we're discussing here or perhaps a skateboarder, who rolls down a ramp that's one side of a "U"-shaped track, then up the other side, gaining kinetic energy in the process even though his initial and final potential energies are the same (that is, he started and ended at the same height) is another good one.

 

[jbriggs444]

Yes. Having no internal structure, a particle is not capable of storing (or releasing) internal energy.

And yet we are willing to contemplate elastic collisions between pointlike particles. There is something not completely realistic in that idealization.

 

[sophiecentaur]

With my legs and age,

Haha. Mine's older than yours.

A conservation of momentum argument indicates that vertical impulse is limited to μμ\mu times horizontal impulse where μμ\mu is the coefficient of static friction.

That sounds reasonable. it would rely on the performer using just the right amount of normal force by controlling the leg - but it would also rely on energy conservation, which doesn't apply to muscles. OTOH, the performer can put some energy in during the manoeuvre. It gets too hard to make a good prediction.

If you model the person as a particle, yes.

The particle model can't deal with the necessary geometry for achieving the upwards result. You have to use a model with a finite size, I think. It would be one step too far in idealising the situation.