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Newtons 3rd law - how can this rope move?

  1. May 29, 2013 #1
    Hi I'm having trouble understanding this situation. Appologies if I posted in the wrong place, this is pretty simple but it's not really a homework question.

    A horse is attached to a large stone by means of a rope (modelled as light and inextensible). Let the mass of the horse be m and the mass of the boulder be m'. The horse applies a force (on itself) and the system begins to accelerate. As I see it, the forces involved are as follows (ignoring any resistive forces):
    The is a force of F on the system, causing it to accelerate according to F=(m+m')a.
    The forces on the horse are F and a tension from the rope, say Ft, so the resultant is F-Ft=ma.
    The force on the boulder must be from the tension in the rope, so Ft'=m'a.
    The rope acts on both the horse and the boulder so a reaction force from both of these acts on the rope, so there are two forces on the rope in opposite directions, and force on rope is equal to Ft-Ft'.

    It can be shown algebraiclally and is consistent with the model of tension in a rope that I am familiar with that Ft=Ft'. So my question is how can the rope accelerate if there is no resultant force acting on it?

    Thanks:)
     
  2. jcsd
  3. May 29, 2013 #2
    Because the rope has negligible mass "a rope (modelled as light and inextensible)", so that the mass times acceleration of the rope is virtually zero. So, if the mass of the rope is vanishingly small, even though it is accelerating, the forces on both ends are virtually equal. Looking at it another way, if they weren't equal, the acceleration of the rope would be infinite.
     
  4. May 30, 2013 #3

    SteamKing

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    Even if the rope has an appreciable mass, motion is possible.

    Suppose that the horse was pulling a heavy chain which was connected to the rock. Would the fact that the chain was heavier than the rope change the fact that motion is possible? Granted, the horse would be required to pull harder to set the rock in motion, but that is beside the point.

    The OP assumes that because the force at one end of the rope is equal and opposite to the force at the other end, the rope is incapable of movement. However, the rope in this case does not act in isolation: It is attached to a rock and a horse.

    If the rock were rigidly fixed to the ground, the horse would pull the rope taut and further movement would be impossible because the horse cannot exert enough force on the rock to overcome the reaction of the ground on the rock.

    If the rock is not rigidly fixed to the ground, but its motion is resisted only by friction, the the horse must exert enough pull to exceed the friction of the rock, so that F(horse) - F(friction) > 0. Because we have a horse, a rope, and a rock, we cannot simply say that because the net force on the rope is zero, motion of the rope is impossible. The entire system must be analyzed.
     
  5. May 30, 2013 #4

    rcgldr

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    Take the case of a non-massless rope. Instead of accelerating the rope, imagine that the rope is simply hanging by the top end of the rope. The tension at both ends of the rope is not the same. The tension at the top of the rope where it connects to whatever supports the rope equals the weight of the rope while the tension at the very bottom of the rope is zero.
     
  6. May 30, 2013 #5

    D H

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    Newton's second law doesn't have much to say about the rope in the case that the rope is massless. The net force must be zero lest the acceleration be infinite, and any acceleration satisfies F=ma when both F and m are zero. A massless rope is essentially a geometrical constraint. Since the net force is zero, the force at one end of the rope must be equal and opposite to the force at the other end and the tension in the rope must be uniform throughout the length of the rope.

    What if the rope isn't massless? Now the net force on the rope cannot be zero if the rope is accelerating. The force at one end is not equal but opposite to the force at the other end, and tension in the rope is not uniform.

    What if the mass of the rope is negligibly small compared to that of the horse or the boulder? Now the difference in the magnitude in the force at one end of the rope versus that at the other is also negligibly small, and it's safe to treat the rope as massless in this case.
     
  7. May 30, 2013 #6
    Thanks for explaining - that really helps :)

    I have another slightly similar situation that I can't get my head round:

    If an object A rests on a table B, the forces on A are its weight and a contact force from B, which is equal to its weight. By Newton 3, A must also exert a force with the same magnitude on B. What I can't make sense of is how A can exert a force on B if there is no resultant force on A?

    I know this is a bit silly but its really bugging me...

    Thanks!
     
  8. May 30, 2013 #7

    arildno

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    In cases where the rope experiences a non-zero frictional force, say by sliding across a pulley there will, however, be a difference in the magnitude of the tension between the two free ends.

    I agree, however, that the general constraint imposed upon the forces acting upon the rope is essentially analogous to the constraint placed on the normal force from a surface, in which the kinematic constraint that relative velocity should be zero "Mirrors" the dynamic constraint of the rope that the sum of forces acting upon it should be zero.
     
  9. May 30, 2013 #8

    D H

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    The upward normal force and downward gravitational force on the book are not third law pairs. Just because two forces happen to be equal but opposite does not mean they result from Newton's third law. For one thing, the two forces must be acting on two different bodies. That's not the case here as both forces are acting on the same body, the book. This book on a table the is not an example of Newton's third law.

    The third law interactions in this example are
    • The surface of the table exerts an upward normal force on the book, and therefore the book exerts an equal but opposite downward force on the surface of the table.
    • The Earth as a whole exerts a downward gravitational force on the book, and therefore the Earth as a whole is gravitationally attracted toward the book.

    Newton's second and third laws talk about very different things. The third law talks about individual forces between pairs of objects. Newton's second law talks about net forces on a single object. The superposition principle provides the connection between the individual forces acting on a single body and the net force acting on that body. Some introductory physics instructors go so far as to call the superposition principle "Newton's fourth law".

    Let's look at your book example in a bit more detail. The Earth is rotating, and thus the book is undergoing uniform circular motion about some remote axis (the Earth's rotation axis). The book is not at rest! It is accelerating. This means that the net force on the book is necessarily non-zero, which in turn means that the upward normal force exerted by the table on the book cannot be equal but opposite to the downward gravitational force exerted by the Earth as a whole on the book.

    Unless you have the book in a vacuum chamber, the book is surrounded by air that buoys the book upwards a tiny bit. This once again means that the upward normal force cannot be equal but opposite to the downward gravitational force.

    The upward normal force is an example of a constraint force. The direction and magnitude of this constraint force is dictated by the kinematics (F=ma) and the known non-constraint forces (gravitation and buoyance). The normal force is whatever is needed to satisfy both the superposition principle and the net force that is known to be exerted on the book via kinematics.
     
  10. May 30, 2013 #9

    SteamKing

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    You are confusing yourself.

    Ask yourself a question: What causes object A to have weight?
    Ans: The weight of A is the force of attraction due to gravity between the mass of A and the mass of the earth.

    If object A were held above the earth and released, A would fall toward the earth because of gravitational attraction. The force of attraction due to gravity between Object A and the earth would be F = mass of A * g, or the weight of A.

    If object A is placed on Table B, then no motion by A toward the earth is possible. How to reconcile this fact? There must be an equal and opposite force acting on the weight of A such that the net force acting on A is zero. If there is zero net force, there is zero motion. This equal and opposite force is called the normal force.
     
  11. May 30, 2013 #10
    Yeah I agree.

    I think what has confused me about Newton 3 is that when an object 'acts' on another, the magnitude of the force that it exerts is determined by a sort of mathematical necessity that occurs due the the boundries that Newtons laws have put in place. The intuative trail of thought would be, I think, that an object with a force on it exerts part of that force onto an adjacent object, which then pushes back on it. By Newtons model its sort of the other way round: that an object has a force on it, but when it is pushed up against another object the must be a on it force in the other direction not for an inuative reason, but simply to satisfy the mathematics of Newtons 2nd law. I am realising more and more how much of a genius Newton was to figure all this out!

    Does what I have said above make sense, and do people agree with it?

    Thanks :)
     
  12. May 31, 2013 #11
    So a constraint force is a force that is determined entirely by the necessity for a situation to satisfy F=ma, as opposed to a force due to gravity which is determined just by the masses of objects and their positions?

    Also, is there anything in beginning mechanics that tells us that tells us that an object should rest on the surface of another and not pass through it, or at this stage should I just take that for granted?

    Thank you so much, all the help is really appreciate
     
  13. May 31, 2013 #12

    SteamKing

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    Is there anything in your experience which tells you that things should normally pass through solid objects if they are at rest?
     
  14. May 31, 2013 #13
    Of course not, I just wondered if there was something in Newtons model that said that, or if it is so obvious that we can just take it for granted.
     
  15. May 31, 2013 #14

    WannabeNewton

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    If you constrain an object to move tangential to a certain prescribed path (e.g. along an inclined plane) then an associated constraint (normal) force will naturally come out of the equations of motion that remains orthogonal to the prescribed path throughout the *relevant* trajectory. Classical mechanics does not, however, fully explain the origin of the normal force if that is what you are asking.
     
  16. May 31, 2013 #15

    SteamKing

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    Even Newton had to pre-suppose certain things when he developed his natural philosophy. He took it for granted that his laws must conform to Nature as it was observed and experienced.
     
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