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Work Done & Frictional Forces

  1. May 4, 2015 #1
    1. The problem statement, all variables and given/known data
    We did an experiment in class where a marble was released down a ramp and traveled a short distance before entering a hole inside an upside down cup, causing the cup to get displaced. Different masses were placed on top of the cup and the resulting displacements were measured. We have to discuss the physics behind the experiment. We graphed distance against mass on top of the cup.

    2. Relevant equations
    1. Friction force = u N
    2. Work Done = force x distance

    3. The attempt at a solution
    The marble has GPE equal to height of ramp x 9.81 x mass of marble. As it rolls down the ramp and strikes the back of the cup, this energy results in work being done on the cup which is equal to GPE minus any frictional losses on the way. The bigger the mass on the cup the bigger the static frictional force according to the above equation. I'm struggling to explain in rigid physics sense why the cup moves a smaller distance with more mass on it. The energy available to be transferred into work done is fixed (i.e. total GPE) so what determines how much is transferred into work done? Will the force in equation 2 be the same each time?
     
  2. jcsd
  3. May 4, 2015 #2
    Since greater mass is used every time, the normal force on cup i.e. N increases which in turn increases the frictional force. So since GPE is fixed, work done by friction is also fixed but it requires small d (distance) to dissipate the same amount of GPE.
     
  4. May 4, 2015 #3
    Thanks. How do you explain why the cup doesn't move when the mass is too big. What happens to GPE when the distance is zero? I would guess that all the GPE must just get dissipated as heat and sound. Also, is the following argument valid in this experiment: The force the marble exerts on the cup is fixed, therefore, according to F = ma the acceleration will be less as mass increases? I have my doubts as the F = ma refers to the net force (in this case along the horizontal plane) which is not fixed?
     
  5. May 4, 2015 #4
    Snapshotcup.jpg
    Does this accurately represent the situation, Jimmy?

    Yes, I think that is a valid statement. The normal force exerted on the cup by the marble should be approximately constant.
    But as Halo CX pointed out

    Increasing the mass should result in a smaller distance travelled by the cup/mass, which your experiment confirmed.

    Do you know what static friction is, and how it differs from kinetic friction?
     
  6. May 4, 2015 #5
    Thanks for the reply and your help. Yes that diagram is correct. Even though the normal force from the marble onto the cup is constant I thought F = ma only refers to the net force which wouldn't be constant because a bigger mass would mean that same force from the marble will cause the resultant force to be less? Out of interest, how could you measure the force the marble exerts on the cup?

    We have only dealt with static friction which matches the applied force until its limit is reached and this limit depends on a coefficient (u) and the normal reaction force.
     
  7. May 4, 2015 #6
    Yes, I believe that is correct too. Do you understand why? Draw a free body diagram and make a sum of forces equation for the horizontal. Examine the signs of the terms, and keep in mind that friction only ever opposes the direction of sliding. Which is also to say that if the friction force is greater than the force causing the sliding, there will not actually be any sliding.

    Exactly! And how does what you are varying in your experiment affect the normal force or the coefficients of friction?

    Additionally, what if the normal force exerted from the marble is less than the static frictional force opposing it?
     
    Last edited: May 4, 2015
  8. May 4, 2015 #7
    Thanks again. So, does F = ma therefore not apply if I understand you correctly. The force from the marble is fixed, however the net force (which is used in F = ma) is not fixed because if you double m for example then 'a' won't go down by a factor of 2 because Fnet is not constant - it will be smaller as the opposing frictional force against that same force from the marble will result in a small Fnet. Is that right?

    If you plot a graph of distance moved by the cup against the varied mass 'm' then the graph is non-linear - is there an equation that shows why this is the case?

    To answer your question, if the normal force exerted by the marble is less than the static frictional force opposing it then it will not move, hence no work is done on the cup. Will all the GPE from the marble just therefore go into heat and sound?
     
  9. May 4, 2015 #8
    No, F = ma most certainly applies. However, it is actually much more beneficial to consider the situation from a linear momentum standpoint. Have you learned about momentum/impulse yet? I should also clarify that Fn in the force equation is the instantaneous force exerted by the marble on the cup. The marble actually exerts a force over the course of the cup/mass and marble collision, which takes place over a chunk of time Δt.

    We do not know the details of the collision of the marble and the cup/mass, so there is a gray area here. However, there is an equation that explains why the distance traveled by the cup/mass as a function of the varied mass is non-linear regardless of the elasticity of the collision. It requires you to analyze the situation from a linear momentum standpoint.

    The first step is to consider the forces acting on the cup/mass after the collision. Can you tell me which forces act on the cup, and whether or not the acceleration of the cup would be constant?

    Correct!

    A certain amount of the gravitational potential energy that the marble has at the top of the ramp would become rotational kinetic energy, which would not affect the cup/mass collision so much. Really, the most meaningful quantities to consider here would be the translational kinetic energy of the marble and its linear momentum.

    Additionally, what if the marble bounces back from the cup? That certainly isn't sound or heat.
     
    Last edited: May 5, 2015
  10. May 5, 2015 #9
    Sorry for the wall of text.

    This is real physics you are doing: interpreting and explaining experimental results. I find this to be the most difficult thing, at least for myself.
     
  11. May 5, 2015 #10

    haruspex

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    There is another loss. The marble will have rotational KE. Where did that come from and where does it go?
    Not really. The proportion of energy transferred in a collision depends on the mass ratio.
    Another complication: If the cup+mass is light compared to the marble, the marble will keep going forward, but more slowly than the cup at first. Friction will decelerate the cup much more than the marble, so the marble may well catch up again and give it another nudge.
    You do have to be very careful discussing force in a collision. In many collisions, you have large forces acting for short times. The graph of force against time is often unknowable (without details of the materials concerned). You can only safely work with momentum and, if you know it's elastic, energy.
    Yes, if the static frictional force can be great enough, it may be that the cup does not move at all, but I suggest it would have to be very large. More likely, it does move very slightly, and you did not detect it.
     
  12. May 5, 2015 #11
    Do you think it would be unreasonable to approximate the graph of force against time for each of the heavier mass trials as being the same? Only the heavier mass trials, because, as you described
     
  13. May 5, 2015 #12

    haruspex

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    I don't think there is any need to get involved with what the graph of force against time looks like. Just work with momentum and energy.
    The exception is if you have to explain a case where it really does not move at all. Conservation of momentum then implies the collision has a duration. E.g. a certain amount of elasticity in the wall of the cup spreads the momentum over a short time, keeping the peak force below that available from static friction. Not sure what that means for the graph of mass against distance moved in the vicinity of this case... some sort of smooth transition I would guess.
    I suggest that the more interesting analysis is around the question of whether the marble rebounds or keeps going forwards. That depends on the two masses in a very simple way. If we ignore rolling losses, if the marble keeps going forwards it will strike the cup again, and again.... so that in effect the two coalesce. This will produce a distinct bend in the graph.
     
  14. May 6, 2015 #13
    Wait now I'm confused. So are you saying that the work done by friction is not constant? I have had a good read of all the threads and I will write what I think and could someone please kindly confirm if this is correct.

    The marble has GPE equal to its mass x height of release x 9.81. This is converted into KE as it rolls down the slope. It strikes the back of the cup and exerts force which can be measured as the rate of change of momentum of the cup at that particular instant. The work done by friction is not always the same because it depends on the amount of energy transferred from the marble to the cup which changes as the mass of the cup increases. The NET force exerted on the cup is the rate of change of its momentum it experiences at one instant minus the opposing frictional force. This instantaneous net force causes the cup to accelerate until dynamic friction causes it to come to a rest. This instantaneous NET force changes because both the instantaneous force (determined by the rate of change of momentum) can vary and so can the frictional force (as it is proportional to the normal force). Could someone kindly write a few key points that I can use in my experiment write up. We were asked to discuss work done and frictional forces. I'm happy with frictional forces increasing as more mass is added but I'm now quite confused on how to go about discussing the work done correctly.
     
  15. May 6, 2015 #14

    haruspex

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    Yes, but it will become important to distinguish rotational and linear KE.
    It is not fruitful to discuss force during impact, unless you are going to get into details of the process of impact, such as flexing of the wall of the cup. I don't think that is appropriate to the task at hand. For the impact itself, just discuss momentum and energy.
    Yes.
    Are you thinking of static friction here? Static friction never does work since, by definition, it never acts over a distance.

    Think of it in these stages:
    1. the impact, which will almost always generate enough peak force to overcome static friction; if so, since the static friction does no work, just treat it as any old collision and ignore friction at this stage. Analyse it using momentum conservation. Again, since the impact is taken as happening over negligible time, friction does not affect momentum in this stage.
    Remember, for the purpose of momentum here, it's only the linear speed of the marble that counts.
    If static friction is enough to match the peak force, it will be like the marble bouncing off a wall. But if you treat it as static friction being overcome then the kinetic friction bringing the cup to a halt very quickly, I don't think it will make much difference to the result.

    2. Post impact, unless the impact was completely inelastic (it never is), the cup will be moving faster than the marble. Kinetic friction now does work to bring the cup to rest, but....

    3. After the impact, the marble may be continuing on, at a slower speed, or may have rebounded. Which of these happens will depend on the relative masses. Either way, its rotation will no longer match its linear motion, i.e. it will be sliding, not rolling. This will accelerate it forwards. If it rebounded, that forward acceleration may or may not be enough to restore it to a forward velocity. If it manages to continue on, either by being massive enough not to have rebounded or by having enough forward spin after rebounding to go forwards again, it will almost surely catch up to the cup and knock it on. In principle, this could happen repeatedly, but it would very soon become indistinguishable from marble and cup moving forward together. Thus, the frictional force will do work not just to bring the cup to rest but the marble too.
     
  16. May 8, 2015 #15
    Thanks for all of that info - it has really helped. I have been thinking about this hard over the past few days and done lots of background reading. Could you tell me why it is not good to think about the force the marble exerts on the cup? If the marble is rolling down the same ramp and covering the same distance will it not always exert the same force on the cup? Could you therefore not make the following argument:

    The force the marble exert on the cup is the same. If the cup has more mass on it then the time of the collision will be reduced (I think) which means the change in momentum/Impulse (force x time) will be less. How do you actually explain why the cup without mass on it goes further? I know when there is more mass, kinetic friction does more work because the normal force is bigger, is that correct? Or does the normal force only relate to static friction. Is there an equation that can relate the energy gained by the cup and why the frictional forces bring it to a halt quicker?
     
  17. May 8, 2015 #16
    Looking at the collision from a forces perspective is really only useful for explaining the special case in which the cup/mass does not move at all.

    These assumptions are not particularly useful for your purposes.

    The amount of work friction will have to do to stop the cup/mass will depend upon how much kinetic energy is transferred to the cup from the marble. How much kinetic energy is transferred from the cup to the marble depends upon their mass ratio, which Haruspex pointed out. Since you are changing this ratio by varying one of the masses as part of your experiment, the work done by friction will not necessarily be constant.

    Do you mean "is there an equation that can express the energy gained by the cup and explain why the frictional forces bring the cup/mass to rest over a shorter distance as the mass is increased?"
    If that is what you mean, then yes, there are certainly equation(s) to explain these things. Work with conservation of linear momentum and conservation of energy. Don't forget to take into account that a portion of the marble's initial GPE will become rotational kinetic energy.
     
  18. May 8, 2015 #17

    haruspex

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    Force is something that applies at some instant of time. The affect of the force depends how it varies over time, and what affect the force has over some period may affect the magnitude of the force over the next. This can get very complicated, and really is not important at the macroscopic level. The bottom line is that the law of action and reaction, combined with F=ma, leads to the conservation of momentum. You can apply that without having to worry about how long the collision lasted or how the force varied during it.
     
  19. May 8, 2015 #18
    Thanks guys. Would it be correct to say that the bigger the mass of the cup the less kinetic energy transferred and also the higher the kinetic friction (due to N increasing). Both these things contribute to the smaller distance traveled by the more massive cup? Please could you give me and example in the context of this experiment where you can apply action, reaction, combined with F = ma to show conservation of momentum as you say? Say if the kinetic energy of the marble was 12J just before impact, what equations would show how much is transferred to a particular mass of the cup and how does momentum link into it? I've had a look to try and find equations relating momentum and energy -would you use E = p^2/2m?
     
  20. May 8, 2015 #19

    haruspex

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    This is something you should determine for yourself using the conservation equations. Model it as a simple collision of a moving mass against an initially stationary one. Conservation of momentum gives you one equation. For the other, you can use the fact that the KE will not increase, or you can use Newton's Experimental Law.
     
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