Why is there no work for someone walking on flat plane

[Jewish_Vulcan]

for example when you climb up an inclined plane why is there no work in the horizontal direction yet there is work in the vertical direction? I do not want examples I want a direct answer.

 

[rock.freak667]

In the definition of work, the work done is the product of your force vector and your displacement vector which works out as |F|*|d|cosθ where |F| is the magnitude of the force, |d| is the magnitude of the displacement and θ is the angle between the two.

What happens if θ = 90 ?

 

[Jewish_Vulcan]

cos90=0, I found that out a while ago but why is it cos and not sin?

In the definition of work, the work done is the product of your force vector and your displacement vector which works out as |F|*|d|cosθ where |F| is the magnitude of the force, |d| is the magnitude of the displacement and θ is the angle between the two.

What happens if θ = 90 ?

I know knew that all along but it seems like there are still forces acting in the horizontal direction like the legs being pushed forward maybe...

 

[Drakkith]

cos90=0, I found that out a while ago but why is it cos and not sin?

I know knew that all along but it seems like there are still forces acting in the horizontal direction like the legs being pushed forward maybe...

A person is a bad example to use. The human body is a very inefficient machine and it takes work just to keep walking forwards on a flat plane. Replace 'person' with a rolling ball or something and assume no friction.

 

[Jewish_Vulcan]

A person is a bad example to use. The human body is a very inefficient machine and it takes work just to keep walking forwards on a flat plane. Replace 'person' with a rolling ball or something and assume no friction.

There would be no horizontal force for a rolling boll on a flat surface ehh?

 

[Chestermiller]

Let's just analyze the case of a single horizontal step (no vertical component) on a horizontal surface. If we can do that, then we can extend it to walking. And we can extend it to an inclined plane.

When you take a single horizontal step, you convert some of the chemical energy stored in your body into mechanical energy. The mechanical energy takes the form of your foot pushing backwards against the ground, the ground pushing forward on your foot, the force on your foot being transmitted to your center of mass through your body parts, and your center of mass moving forward. So the force of the ground causes your center of mass to displace forward, and this translates into work: Fxd = kinetic energy of your body. Now your leading foot sets down on the ground, and your body stops moving. The kinetic energy you had developed is lost, and gets transformed into heat. So the overall effect is chemical energy converting to mechanical work, mechanical work converting to kinetic energy, and kinetic energy converting to heat. Was any work done? Yes, it was done along the way, but it got dissipated. Was there any change in potential energy? No. Was there any permanent change in kinetic energy? No. So, there was no conservative work (i.e., potential energy plus kinetic energy) done? No. But there was mechanical work done.

Does this make sense so far?

Chet

 

[collinsmark]

cos90=0, I found that out a while ago but why is it cos and not sin?

It comes down to what is meant by the vector dot product in, $W = \vec F \cdot \vec d = |F||d| \cos \theta.$

Work is not just the magnitudes of force and displacement multiplied together, but only those components that are parallel to each other. To find the work, you must find the projection of one of the vectors onto the other (it doesn't matter which is projected onto which; either way produces the same answer in the end), and the result of the projection is two parallel vector components. The work is the multiple of those.

This projection operation is where the cosine comes from, and why it is not the sine.

So now ask what is the result if the two, original vectors are completely perpendicular?

(And by the way, as Drakkith mentioned, the human body is a bad example for demonstrting this type of "work.")

 

[collinsmark]

Oh btw, there is also an alternative way to approach the vector dot product interpretation.

When we say that $\vec F$ and $\vec v$ are vectors, what we really mean is that they each have three components, one for each spacial direction.

$$\vec F = F_x \hat x + F_y \hat y + F_z \hat z$$
$$\vec d = d_x \hat x + d_y \hat y + d_z \hat z$$

And the vector dot product is

$$W = \vec F \cdot \vec d = F_x d_x + F_y d_y + F_z d_z$$

And $F_x d_x + F_y d_y + F_z d_z$ is mathematically equivalent to $|F||d| \cos \theta$. So that's another way to approach the problem.

 

[sophiecentaur]

This topic comes up very frequently and the only reason for the confusion is the conflict between a formal definition and practical experience. Work done to a Mass, when raising it is the same as the Potential Energy it gains - mgh in many cases. Any practical system for getting it to this height will involve more (perhaps a lot more) energy input. Animal bodies are particularly inefficient.
People seem to get very stroppy about this conflict when they forget the ideal situation that they are starting off their reasoning with. There is no problem when you apply the Laws and Formulae as they are supposed to be applied. Next step is to introduce more practicalities (more complicated model) to a greater and greater depth until everything has been included.

 

[Jewish_Vulcan]

the answer is: the force in the horizontal direction has a net of 0, when they say for a certain distance like 10m they make you "assume" that the person came to a stop after the 10 m, for him to come to a stop an equal but opposite force must act against the motion. because w=fd, w=0(10)=0, so please put away the calculus

 

[Drakkith]

the answer is: the force in the horizontal direction has a net of 0, when they say for a certain distance like 10m they make you "assume" that the person came to a stop after the 10 m, for him to come to a stop an equal but opposite force must act against the motion. because w=fd, w=0(10)=0, so please put away the calculus

That doesn't look correct to me. There's no reason to assume they come to a stop after 10 meters. What happened to the person before or after encountering the inclined plane doesn't matter when it comes to understanding the work done.

 

[Jewish_Vulcan]

That doesn't look correct to me. There's no reason to assume they come to a stop after 10 meters. What happened to the person before or after encountering the inclined plane doesn't matter when it comes to understanding the work done.

then describe how the net force is 0 in the horizontal direction please.

 

[Drakkith]

then describe how the net force is 0 in the horizontal direction please.

I don't see what the person is doing before or after the inclined plane has to do with the net force while moving up the plane.

 

[Jewish_Vulcan]

I don't see what the person is doing before or after the inclined plane has to do with the net force while moving up the plane.

I understand that you do not agree with me but I do and I gave a logical reason... I do not mean to be rude to a staff mentor but I may believe you if you give me a reason...

 

[Drakkith]

I understand that you do not agree with me but I do and I gave a logical reason... I do not mean to be rude to a staff mentor but I may believe you if you give me a reason...

What are you talking about? That was my reason.

 

[Jewish_Vulcan]

What are you talking about? That was my reason.

I did not see a scientific reason that stated why there is no work in the horizontal direction for say a person walking forward on a flat surface, please restate your reason of why there is no force in the horizontal direction. So far the only valuable information I have obtained from you is "What happened to the person before or after encountering the inclined plane doesn't matter when it comes to understanding the work done.", You could be correct but I would need other members to agree with you to convince me that you are correct. Furthermore that qoute does not explain why or how there is a net force of 0 in the horizontal direction.

 

[Drakkith]

I did not see a scientific reason that stated why there is no work in the horizontal direction for a person climbing up an inclined plane...

Because of the reasons stated earlier in the thread. Re-read them and if you have questions on what's been said, ask. Don't just throw out an explanation and then demand that we explain why it's wrong. That's a bad way to learn.

 

[Jewish_Vulcan]

Because of the reasons stated earlier in the thread. Re-read them and if you have questions on what's been said, ask. Don't just throw out an explanation and then demand that we explain why it's wrong. That's a bad way to learn.

I did ask if that explination was correct on a question about work on an inclined plane, chestermiller and brainpushups replied to my conclusion and said it was correct, that is why I am agrueing with you. I have not even graduated general physics yet so maybe you should talk to them about why you think they are wrong
here is the thread about the different but similar subject: https://www.physicsforums.com/threa...s-for-work-problem.794786/page-2#post-5000178

 

[Drakkith]

I did ask if that explination was correct on a question about work on an inclined plane, chestermiller and brainpushups replied to my conclusion and said it was correct, that is why I am agrueing with you. I have not even graduated general physics yet so maybe you should talk to them about why you think they are wrong
here is the thread about the different but similar subject: https://www.physicsforums.com/threa...s-for-work-problem.794786/page-2#post-5000178

You've misunderstood. In the example in the other thread you've specifically stated that the person was already at rest and incorporated that into the example. But that needn't be the case if we just want to analyze the work done only when the person is climbing the inclined plane. What happens before or after has no relevance to what happens while on the plane.

 

[Mark44]

I did not see a scientific reason that stated why there is no work in the horizontal direction for say a person walking forward on a flat surface, please restate your reason of why there is no force in the horizontal direction. So far the only valuable information I have obtained from you is "What happened to the person before or after encountering the inclined plane doesn't matter when it comes to understanding the work done.", You could be correct but I would need other members to agree with you to convince me that you are correct. Furthermore that qoute does not explain why or how there is a net force of 0 in the horizontal direction.

You have asked your question and it has been answered. As has already been explained, a person walking on a flat surface does have to apply a force to keep moving, and as a result, the person does perform work to keep moving. A simpler example that might be easier to understand involves a smooth ball rolling on a flat, frictionless surface. Since the ball doesn't have to overcome any forces (including friction and gravity), no work is performed.

 

[A.T.]

I know knew that all along but it seems like there are still forces acting in the horizontal direction like the legs being pushed forward maybe...

Yes they are horizontal forces, but neglecting air resistance they average to zero over time, if initial and final speeds are equal.

 

[zanick]

I think what the OP is looking for, is how there can be work, if there is motion in the horizontal plane with no gravity or drag forces due to air.
I think a simple answer is that there is a force to get a ball, or person moving, regardless of the friction. the force applied over the distance (or time where both together would be = velocity) is the and the rate of doing this work, is the power. the time the of the power used, is the energy. (or unit measure of work = joules = power=seconds)
So, force is applied, the ball or person accelerates to some speed... force is removed and that object continues to travel at that terminal speed until its acted upon by another force, right? it has the energy stored in itself, in the form of kinetic energy. work was performed, energy was absorbed by the body and it follows the same rules as someone lifting a bowling ball on earth up a distance over some period of time.
This is my non -math interpretation of an answer to the question.. How far off am I?

as a side note.... as far as the "act of walking" horizontally, vs vertically, its a bit more complex and that's due to the human body's inefficiency of walking. if we had roller blades for feet, the concept is a little easier to understand, as we would generally only be fighting rolling and air resistance for horizontal movement. But still, force over any distance is work. by walking, which is much more inefficient, starting and stopping the mass of our legs Is the main purpose of the energy required and is responsible for the work done.

 

[haruspex]

the answer is: the force in the horizontal direction has a net of 0, when they say for a certain distance like 10m they make you "assume" that the person came to a stop after the 10 m, for him to come to a stop an equal but opposite force must act against the motion. because w=fd, w=0(10)=0,

I'm not at all clear what you are saying here, and maybe others have misinterpreted it.
First, there's no quoted context, so I don't know whether this is the inclined plane model or the horizontal plane model.
Re " when they say for a certain distance like 10m ", it would help if you were to give a complete example of such problem.
I'll suppose you mean something like "A man weighing 700N walks on a horizontal path for 10m in 15 seconds. How much work does the man do?"
I would first criticise the problem statement for not specifying whether the man was already walking before the 10m started, nor whether, as you say, he came to a stop at the end of the 10m.
Suppose it's rest to rest. As mentioned by others, even human gait (quite efficient really) is not 100% efficient, so it takes work just to walk at a steady pace. Leaving that aside, he has to accelerate from rest (steadily for 5m, say), then decelerate to a halt. During the deceleration, work is done on the man, technically, but the man is not able to turn that back into useful energy. It all ends as heat, which the man needs to lose. So whether the net work is zero or positive depends whether you wish to include the heat output as negative work for the man. The word 'work' takes on different meanings in different technical contexts.

A simpler example that might be easier to understand involves a smooth ball rolling on a flat, frictionless surface

Umm... if it's rolling, why do you care about friction?

 

[Mark44]

A simpler example that might be easier to understand involves a smooth ball rolling on a flat, frictionless surface.

Umm... if it's rolling, why do you care about friction?

I guess if the surface is frictionless, the ball would just slide. The frictionless restriction was to eliminate the consideration of any horizontal forces impeding the motion of the ball.

 

[Jewish_Vulcan]

Thanks to the many productive replies, I understand why there is no work in the horizontal direction. Maybe this thread should be closed now.

i'll suppose you mean something like "A man weighing 700N walks on a horizontal path for 10m in 15 seconds. How much work does the man do?"
I would first criticise the problem statement for not specifying whether the man was already walking before the 10m started, nor whether, as you say, he came to a stop at the end of the 10m.

that is why I posted this thread, in the general physics homework problems they did not specify the before conditions and after conditions so I did not see why the net force in the horizontal direction was 0. your information was helpful and verified my understanding.

 

[Jewish_Vulcan]

Thank you for all the educated responses, I think I got my question answered long ago by chestermiller and brainpushups but most of these recent posts verified my understanding. Maybe this thread should be closed now.

 

[georgir]

you've got all the math in the world by other answers, but i think they all missed to clarify something - perhaps they all consider it too obvious.
or it's already said but i didn't spot it above, i only skimmed.

anyway, i think it all comes down to which force's work are you actually looking at.
there is of course positive work done by whatever is pulling the object, as well as negative work done by the friction forces slowing it back down.
there is no work done by the force of gravity.

 

[Jewish_Vulcan]

you've got all the math in the world by other answers, but i think they all missed to clarify something - perhaps they all consider it too obvious.
or it's already said but i didn't spot it above, i only skimmed.

anyway, i think it all comes down to which force's work are you actually looking at.
there is of course positive work done by whatever is pulling the object, as well as negative work done by the friction forces slowing it back down.
there is no work done by the force of gravity.

yes that agrees with my conclusion, thank you for the reply.

 

[zanick]

yes that agrees with my conclusion, thank you for the reply.

and a force is a force... whether it is used to lift a body, accelerate a body or move the body against friction, its all a force and , times a distance , you get "work".
You cant move something without a force , even if there is no friction or if it isn't being lifted against gravity. Just as if something was lifted and put back down... It's the same as if something was accelerated and decelerated and it came back to rest on the same horizontal plane as far as the "work" done. power is the rate of doing that work.

 

[zanick]

Umm... if it's rolling, why do you care about friction?

rolling can still have friction.. "rolling without friction" allows you to calculate the work used strictly to move the object from one point to another.

 

[haruspex]

rolling can still have friction.. "rolling without friction" allows you to calculate the work used strictly to move the object from one point to another.

If it's rolling, the presence or absence or degree of friction with the surface makes no difference. You may be thinking of rolling resistance.

 

[zanick]

If it's rolling, the presence or absence or degree of friction with the surface makes no difference. You may be thinking of rolling resistance.

Yes, im speaking of rolling resistance..... is there any other kind that wouldn't make a difference? :)
also, I guess you could consider the extra rotational inertia of the rotating elelments as well ( WR^2)

 

[haruspex]

Yes, im speaking of rolling resistance..... is there any other kind that wouldn't make a difference? :)

If there's rolling resistance, that will apply equally with or without friction. The difference is that with friction both rotational and translational speeds will fall (in step); without friction only the rotational speed is affected - but as soon as that happens it is no longer considered rolling.

 

[Jewish_Vulcan]

If there's rolling resistance, that will apply equally with or without friction. The difference is that with friction both rotational and translational speeds will fall (in step); without friction only the rotational speed is affected - but as soon as that happens it is no longer considered rolling.

never hear about this type of stuff, would it only be affected without friction if the ball was not a perfect sphear? if not how would it lose any rotational speed?

 

[haruspex]

never hear about this type of stuff, would it only be affected without friction if the ball was not a perfect sphear? if not how would it lose any rotational speed?

It's not the shape, more the rigidity.
Real world objects are not completely rigid. A tyre deforms as it rotates, producing a flat area in contact with the road. This constant deformation takes work. But rolling resistance is not limited to that. It might be mostly that, but the term is a catch-all encompassing other losses as such axle friction.