1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why is there no work for someone walking on flat plane

  1. Feb 3, 2015 #1
    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.
     
  2. jcsd
  3. Feb 3, 2015 #2

    rock.freak667

    User Avatar
    Homework Helper

    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 ?
     
  4. Feb 3, 2015 #3
    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...
     
  5. Feb 3, 2015 #4

    Drakkith

    User Avatar

    Staff: Mentor

    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.
     
  6. Feb 3, 2015 #5
    There would be no horizontal force for a rolling boll on a flat surface ehh?
     
  7. Feb 3, 2015 #6
    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
     
  8. Feb 3, 2015 #7

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

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

    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.")
     
  9. Feb 3, 2015 #8

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

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

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

    [tex] \vec F = F_x \hat x + F_y \hat y + F_z \hat z [/tex]
    [tex] \vec d = d_x \hat x + d_y \hat y + d_z \hat z [/tex]

    And the vector dot product is

    [tex] W = \vec F \cdot \vec d = F_x d_x + F_y d_y + F_z d_z [/tex]

    And [itex] F_x d_x + F_y d_y + F_z d_z [/itex] is mathematically equivalent to [itex] |F||d| \cos \theta [/itex]. So that's another way to approach the problem.
     
  10. Feb 5, 2015 #9

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  11. Feb 7, 2015 #10
    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
     
  12. Feb 7, 2015 #11

    Drakkith

    User Avatar

    Staff: Mentor

    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.
     
  13. Feb 7, 2015 #12
    then describe how the net force is 0 in the horizontal direction please.
     
  14. Feb 7, 2015 #13

    Drakkith

    User Avatar

    Staff: Mentor

    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.
     
  15. Feb 7, 2015 #14
    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...
     
  16. Feb 7, 2015 #15

    Drakkith

    User Avatar

    Staff: Mentor

    What are you talking about? That was my reason.
     
  17. Feb 7, 2015 #16
    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.
     
    Last edited: Feb 7, 2015
  18. Feb 7, 2015 #17

    Drakkith

    User Avatar

    Staff: Mentor

    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.
     
  19. Feb 7, 2015 #18
    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
     
  20. Feb 8, 2015 #19

    Drakkith

    User Avatar

    Staff: Mentor

    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.
     
  21. Feb 8, 2015 #20

    Mark44

    Staff: Mentor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why is there no work for someone walking on flat plane
  1. Why is energy flat? (Replies: 36)

Loading...