How do I calculate the Tension in a rope going over a smooth pulley?

[carlcla]

Homework Statement

Calculate the tension in the rope AD over the smooth pulley B

Relevant Equations

Moment at E = 0

I really cant wrap my head around how I can find the tension i in the rope going from A to D via the smooth pulley B.

P=4.91N
Distances are in mm

I assume I cant use the triangle EBD as it can be dynamic without the triangle ABC changing. I know I can find the y component of the tension in point A by doing moment at C=0. But I dont get how i can convert that y component into annything usefull without know any of the angles of the rope at A.

 

[BvU]

Hello and ,

I think you are correct in assuming you don't have enough information. Did you overlook something in the instructions?

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[berkeman]

Are the joints at E and C hinged, or rigid?

 

[carlcla]

Are the joints at E and C hinged, or rigid?

E is fixed to the ground with bolts and C is pinned. So C need to have moment = 0

 

[berkeman]

E is fixed to the ground with bolts and C is pinned. So C need to have moment = 0

Sorry, I'm not understanding your reply. By "hinged" I mean free to rotate. If E and C are both not free to rotate, the rope is not needed.

 

[carlcla]

Sorry, I'm not understanding your reply. By "hinged" I mean free to rotate. If E and C are both not free to rotate, the rope is not needed.

Im sorry for the confusion. E is fixed and C is free to rotate.

 

[Lnewqban]

Im sorry for the confusion. E is fixed and C is free to rotate.

It seems to me that it is the other way around for both, the photo and the diagram.

Note guides reinforced with gussets in junction C, as well as small gap between horizontal and vertical links, that prevents rotation.

Also, the problem wording specifies "Moment at E = 0", which it would only be possible if junction E is a hinge.

If the above is correct, the solution is possible.

 

[carlcla]

Alright lets assume you are correct, that c is rigid and e is a hinge. I still cant see how to solve for the tension. I still end up with too many unkown variables. How would you proceed?

 

[berkeman]

How would you proceed?

One way to do it is to measure your drawing to get more numbers so you can calculate the rope angle at the left. It's not ideal, but if that's all you've got, then at least you can get an answer to within a few percent.

 

[jbriggs444]

Alright lets assume you are correct, that c is rigid and e is a hinge. I still cant see how to solve for the tension. I still end up with too many unkown variables. How would you proceed?

If C is rigid then the rope on the left side is irrelevant. The leftward-extending arm AC is already rigidly attached. The rope there is not supporting anything. All you have is a needlessly complicated way of tieing off the rope.

Also, if this interpretation (with the hinge at E) is correct then measuring the angle of AB will tell you nothing important.

If the interpretation with the hinge at C is correct then measuring the angle will confirm what you can see by eye: the thing is unstable and no possible tension can hold the assembly in equilibrium.

There is a possible third interpretation. If a hinge at C leaves rod AC free (within the small limits imposed by the guides within which it loosely lies) and if rods BC and CE are rigidly joined as a single anchored vertical post then the rope on the left is relevant and supports rod AC. A measurement can be used to determine the angle of the left hand rope segment AB and thereby infer the rope tension. I do not believe that this is the intended interpretation.

 

[Lnewqban]

Alright lets assume you are correct, that c is rigid and e is a hinge. I still cant see how to solve for the tension.

As moment at E = 0, and the armature remains in equilibrium, the counterclockwise rotation of the BE link, that is induced by the three masses (times their horizontal distances to E), must be exactly compensated by the tension in the rope BD.

 

[carlcla]

As moment at E = 0, and the armature remains in equilibrium, the counterclockwise rotation of the BE link, that is induced by the three masses (times their horizontal distances to E), must be exactly compensated by the tension in the rope BD.

Yes let me see if I understand you correctly. The the loads times thier horizontal distance will be equal to the x component of the rope BD correct? As the y component goes through point E and excerts no moment. What i still dont understand is how to inerprete a smooth pulley wheel such as B. When a rope goes over a smooth pulley that is pointed verticaly such as in this problem. Can the rope excert also an x component on that pulley or only y component (Normal to the pulley)

 

[Steve4Physics]

I'll jump in if I may (working on the assumpion that the only pivot is at E)...

Yes let me see if I understand you correctly. The the loads times thier horizontal distance will be equal to the x component of the rope BD correct?

The 'x component of the rope' doesn't mean anything. I think you mean 'the moment about E produced by the x-component of the tension in BD'. Then you would be correct.

Note that the tension in AB is irrelevant (as already explained by @jbriggs444). The tension in AB acts as an internal force so it doesn't affect the (external) balancing moments. The pulley could be removed and the rope (BD) simply tied-on at B; the tension in BD would be the same as when using the pulley!

As the y component goes through point E and excerts no moment. What i still dont understand is how to inerprete a smooth pulley wheel such as B. When a rope goes over a smooth pulley that is pointed verticaly such as in this problem. Can the rope excert also an x component on that pulley or only y component (Normal to the pulley)

For a rope going over a smooth pulley, the direction of the force on the pulley is radially inwards - from the centre of the contact-arc to the centre of the pulley. The force can have both x and y components. For example, if a rope’s direction changes from horizontal to vertical when passing over the pulley, the direction of the force on the pulley is at 45$^o$ to the vertical. But this is not a useful in this question.

 

[carlcla]

I'll jump in if I may (working on the assumpion that the only pivot is at E)...


The 'x component of the rope' doesn't mean anything. I think you mean 'the moment about E produced by the x-component of the tension in BD'. Then you would be correct.

Note that the tension in AB is irrelevant (as already explained by @jbriggs444). The tension in AB acts as an internal force so it doesn't affect the (external) balancing moments. The pulley could be removed and the rope (BD) simply tied-on at B; the tension in BD would be the same as when using the pulley!


For a rope going over a smooth pulley, the direction of the force on the pulley is radially inwards - from the centre of the contact-arc to the centre of the pulley. The force can have both x and y components. For example, if a rope’s direction changes from horizontal to vertical when passing over the pulley, the direction of the force on the pulley is at 45$^o$ to the vertical. But this is not a useful in this question.

Alright thanks, so to see if I understand you. If I were to draw a free body diagram of the structure, can i ignore the fact that there is a pulley in b and Just draw rope BD as an external force pulling at the angle calculated by the distances given on the right side?

And then calculate the moment around E by the loads P and the x component of the force i drew representing the rope?

 

[Lnewqban]

Yes let me see if I understand you correctly. The the loads times thier horizontal distance will be equal to the x component of the rope BD correct?

You are incorrectly comparing one moment to a force.
Did you mean equal to the x component of the rope BD times its vertical distance to pivot E?

What i still dont understand is how to inerprete a smooth pulley wheel such as B. When a rope goes over a smooth pulley that is pointed verticaly such as in this problem. Can the rope excert also an x component on that pulley or only y component (Normal to the pulley)

Yes, it can.
If the tensions on both sides of the pulley that is pointed vertically are not symmetrical regarding angles.

 

[carlcla]

You are incorrectly comparing one moment to a force.
Did you mean equal to the x component of the rope BD times its vertical distance to pivot E?


Yes, it can.
If the tensions on both sides of the pulley that is pointed vertically are not symmetrical regarding angles.

Yes I meant BD times the distance from E.

If I do the loads times thier horizontal distance from E; must that be equal to the x component of the tension BD times the distance EB?

On annother note: If i were to draw the free body diagram of the structure. Can I write the tension BD as an external force pulling at the angle calculated by the distances given. Can I ignore the fact that it is a pulley at B and act as if the rope is just tied in B on the free body diagram?

 

[jbriggs444]

If I do the loads times thier horizontal distance from E; must that be equal to the x component of the tension BD times the distance EB?

Yes.

This is most easily seen as the result of a torque balance about the pivot at point E, looking at all the external forces and their associated torques. The total torque must be zero. The only non-zero individual torques are from the x component of rope tension and from the loads.

 

[Lnewqban]

Yes I meant BD times the distance from E.

Excellent!
Yes and yes.

 

[Steve4Physics]

Alright thanks, so to see if I understand you. If I were to draw a free body diagram of the structure, can i ignore the fact that there is a pulley in b and Just draw rope BD as an external force pulling at the angle calculated by the distances given on the right side?

And then calculate the moment around E by the loads P and the x component of the force i drew representing the rope?

Yes. And to add to what @jbriggs444 said...

The system behaves just the same as if you had a single rigid object (blue object in diagram below) and wanted to find the tension in BD.

The internal structure (of the blue object) doesn't affect the behaviour of the object when external forces and torques act.

Be careful when using this idea though! It wouldn't work if there were a pivot at C because the object wouldn't be rigid. You would need to consider the requirement to have zero net torque about C.

 

[carlcla]

Yes. And to add to what @jbriggs444 said...

The system behaves just the same as if you had a single rigid object (blue object in diagram below) and wanted to find the tension in BD.
View attachment 350920
The internal structure (of the blue object) doesn't affect the behaviour of the object when external forces and torques act.

Be careful when using this idea though! It wouldn't work if there were a pivot at C because the object wouldn't be rigid. You would need to consider the requirement to have zero net torque about C.

Excellent!
Yes and yes.

Alright! Thanks alot guys, get it now. I appreciate everything so much!