Say you have two bodies connected by a "weightless" rope, one hanging from a ledge on a pulley (classic pulley problem).

If the body that is on the surface and being pulled by the object hanging freely has no friction with the ground then obviously we could just as well discard that body from our calculations, couldn't we?

I have homework and such a problem where a box is being dragged by another box hanging from a ledge on a pulley, except the box being dragged on a frictionless surface which doesn't make any sense to me.

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closed as off-topic by Brandon Enright, Abhimanyu Pallavi Sudhir, Waffle's Crazy Peanut, Dan, Qmechanic Feb 3 '14 at 0:51

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  • 1
    $\begingroup$ A picture would be really useful here but if the surface is slanted, then, even though there is no friction, gravity will still pull on both sides of the pulley. $\endgroup$ – John Habert Feb 1 '14 at 15:52
  • $\begingroup$ Picture is added. It's not slanted. $\endgroup$ – Paze Feb 1 '14 at 15:55
  • $\begingroup$ Assuming that isn't an artifact of how you made the picture, shading under a surface usually indicates friction is not $0$. $\endgroup$ – John Habert Feb 1 '14 at 15:57
  • $\begingroup$ It literally says in the problem: "The wagon is dragged with no significant friction" $\endgroup$ – Paze Feb 1 '14 at 15:58
  • $\begingroup$ @Paze One way to think about it: The wagon on the flat surface has momentum (even though it doesn't have friction). You need to include it in the FBD. $\endgroup$ – apnorton Feb 1 '14 at 16:15

You can't discard that body from your calculations, even in a frictionless case. To convince yourself of that you can just think of the fact that the box would accelerate without the need of any force, which is obviously absurd.

You are not considering the rope's tension, or you're not considering the fact that it's influenced both by the first and the second body.

You have to write down Newton's second law separately fro the two bodies keeping Tension into account.

$mg-T=ma$ for the "falling body"

$m_2a=T$ for the box.

You keep talking of friction but friction isn't a "requisite" for the box to slow down the falling object. In this case the falling object is slowed because it's pulling the box which has an INERTIA (a mass), and this means we have to exert a force to move it.

In this system you have the two object are accelerating at the same rate, right(because of the inextendibility of the rope)? So there MUST be a force responsible for the acceleration of the box and it is the tension of the rope. The force that keeps in tension the rope is given by the falling object wich "spends" a part of his acceleration to move the box. Try to understand the couple of equation I gave you and you will understand my argument.


How can you discard the mass even though the surface is frictionless. Lets say the freely hanging body has mass $m_1$ and the other one $m_2$ and let the wedge make angle $\alpha$ with the horizontal. Draw the FBD to get $m_1g - T = m_1a$ and $T - m_2g \sin \alpha = m_2a$ where $T$ is the rope tension. Eliminate $T$ to get the acceleration $a$ of the system.

  • $\begingroup$ I have considered a general wedge. The method remains the same. $\endgroup$ – Sandeep Thilakan Feb 1 '14 at 15:58
  • $\begingroup$ Can you find the tension of the rope given that m=1.5 kg? $\endgroup$ – Paze Feb 1 '14 at 15:59
  • $\begingroup$ The figure that you have drawn is the simplest possible case. Its equivalent to $\alpha = 0$ in my solution. So $a = \frac{m_1}{m_1 + m_2} g$ i.e. $a = 3/13 g$. $T = m_1(g - a) = 15/13 g.$ $\endgroup$ – Sandeep Thilakan Feb 1 '14 at 16:06
  • $\begingroup$ I realize that, but the tension of the rope is just the mg force of the lowered object.. I am then asked to find the mass of "m" if the acceleration of the system is 1,75m/s^2 and that's where a problem arises. They will have the same mass...They will always have the same values because of no friction. That's what I mean by 'discarding' the other object. $\endgroup$ – Paze Feb 1 '14 at 16:07
  • $\begingroup$ $m=1$ kg will produce a much smaller acceleration than $m=100$ kg. $\endgroup$ – John Habert Feb 1 '14 at 16:09

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