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(clamp is perpendicular to the wall in the top right pulley system ) I have seen some diagrams similar to these . What are the forces on the massless and frictionless pulley in these situations ? As far as I know the clamp should exert a force outward or inward .But then how is the net force on the massless pulley 0. In these figures force by clamp cannot balance the force by rope on pulley . Am I missing something or these diagrams are inaccurate ?

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  • $\begingroup$ Take one body at a time and forget all others. In the dual pulley configuration, let's focus on only the left pulley. The things that are physically interacting with this pulley are 1) the cord, which is trying to pull it down and to the right, and 2) the clamp. The pulley is however stationary. So the only way this can happen is if both the vertical & horizontal forces cancel. Which would mean... ? $\endgroup$
    – jayann
    Commented Jun 16, 2015 at 3:08
  • $\begingroup$ @jayann the pulley doesn't necessarily need to be stationery since it is massless.But the point i want to make is that there is no force to balance the horizontal force on the pulley .So the only way it's possible is that tension must be 0 .But again that's not the case. $\endgroup$
    – Robin Hood
    Commented Jun 16, 2015 at 5:33

2 Answers 2

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You should imagine that the pulley is fixed in place by a bracket that can transmit any necessary force to the connected object. Any force from the rope on the pulley (in any direction) is countered by the bracket.

Do not think of it like a rope that only pulls in one direction.

The net force on the upper pulley is zero and that means that the bracket is supplying a force with both horizontal and vertical components.


The simpler assumption is that the bracket is rigidly attached and can transfer any force from the axis to the holding object. You can see that this bracket will hold the axis in one position regardless of the angle that the rope makes.

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But if it's not rigidly attached, it will simply pivot so that the force vector necessary is along the attachment point.

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These pulleys are not rigidly attached, but when the forces are static, it doesn't matter because they move to a position that opposes all forces from the rope.

Both assumptions yield the same solution.

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  • $\begingroup$ Yes i am thinking of the bracket(which i am referring to as a clamp) like a rod (not rope) , that can provide force only inward or outward.Can you give me a little bit more practical intuition as to why it can provide all necessary force or this is just assumed for simplicity from the point of the person who made the question ? $\endgroup$
    – Robin Hood
    Commented Jun 16, 2015 at 11:31
  • $\begingroup$ Why do you think a bracket can only provide force inward or outward? If you push on it left or right, it will resist such motion, and therefore provide a force that is neither inward nor outward. In general if you clamp something to a heavy object, it will provide a resistive force against any applied force. $\endgroup$
    – BowlOfRed
    Commented Jul 20, 2015 at 5:17
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There is just a slight mistake in your concept @Robin Hood. The clamp is not limited to apply force only inward or outward. It can apply force in any possible direction. Just imagine clamp as a very small rod. If it can apply force only inward or outward,why would we have to apply force at all to bend it in a U shape. Rest of your concept is correct bro.

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  • $\begingroup$ i am not able to get "why would we have to apply force at all to bend it in a U shape" plz elaborate. $\endgroup$
    – Robin Hood
    Commented Jun 16, 2015 at 16:56
  • $\begingroup$ When a force is applied on a rod it tries to apply a counter force to prevent change in its shape and dimension by virtue of the property of elasticity.Supposing a rod can apply force only inward and outward.So if you apply a force on it in a direction horizontal to its axis, it will not be able to counter it.So it would be very easy to bend it in a L shape(pardon the u shape, but before making U u would surely have to make a L) $\endgroup$
    – user247855
    Commented Jun 16, 2015 at 17:12
  • $\begingroup$ @Robin Hood.I hope that I presented in a more better way this time. $\endgroup$
    – user247855
    Commented Jun 16, 2015 at 17:21

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