Understanding Tension on a Pulley

  • #1

Homework Statement


Problem image: https://prnt.sc/gvhjso
In this case I have to find the reaction forces at the point E (the Fx, Fy, and the Moment at point E) by using the given data. The 20kN forces at the load AC are concentrated, and are 1.8m far from each other. The tension of the cable is 150kN

Homework Equations


Fx = Fcos( θ w/ respect to x-axis) or;
= Fsin( θ w/ respect to y-axis)
Fy = Fsin( θ w/ respect to x-axis) or;
= Fcos( θ w/ respect to y-axis)
Moment = Fd (where d is the perpendicular distance between the point of moment and the force)

The Attempt at a Solution


This problem has the solution already given, which is this:
https://prnt.sc/gv93sy

Now my problem is I do not understand why the solution did not include the force components at the BD part of the rope, only factoring the force components of the DF part of the rope/string. Why is it not included in solving the missing reaction forces at point E?
 
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Answers and Replies

  • #2
kuruman
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According to the FBD in the solution, the system consists of the vertical and horizontal booms plus the rope. The BD part of the rope is part of this system and the system cannot exert a force on itself. Only the parts of the world that are external to the system can exert a force or torque on it. These are Earth (gravity acts "at a distance") and the ground at points E and F.
 
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  • #3
According to the FBD in the solution, the system consists of the vertical and horizontal booms plus the rope. The BD part of the rope is part of this system and the system cannot exert a force on itself. Only the parts of the world that are external to the system can exert a force or torque on it. These are Earth (gravity acts "at a distance") and the ground at points E and F.

So you mean to say that the BD part of the rope still exerts a force on its own but that force is not relevant to the whole system? I find it hard to accept because should the force exerted by the BD part of the rope become too low, won't the boom AC fall down, assuming its weight is too heavy for the support at point C to handle alone?
 
  • #4
haruspex
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So you mean to say that the BD part of the rope still exerts a force on its own but that force is not relevant to the whole system? I find it hard to accept because should the force exerted by the BD part of the rope become too low, won't the boom AC fall down, assuming its weight is too heavy for the support at point C to handle alone?
As kuruman wrote, the system being considered is the frame plus the cable. Internal forces are irrelevant because they always consist of equal and opposite pairs acting on that system.
E.g. we could choose to consider the cable BD as a separate system. The tension in BD results in a force on the rest of the frame+cable system at B and an equal and opposite force on that subsystem at D, and these are in the same line, so no net force and no net moment.
 
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  • #5
So it's like this: There is a tension from point B and there is also a tension from point D and they both cancel each other out? So this subsystem is also considered an equilibrium system on its own? Is this assumption right?
 
  • #6
haruspex
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There is a tension from point B and there is also a tension from point D and they both cancel each other out?
Yes.
So this subsystem is also considered an equilibrium system on its own?
It can be. You are free to subdivide or combine into whatever systems you wish. If all is static then each must be in equilibrium in its own right.
 
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  • #7
Alright, thanks for answering, kuruman and haruspex :D
 

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