Statics - When 2 unknowns and 2 equations cancel each other out

[Femme_physics]

I hate it when that happens, but whenever I make 2 equations and 2 unknowns, setting the sum of all torques at the known force (rather than the support) and do sum of all forces equation (either of x or y) I get complete canceling of terms and 0 = 0. What does it mean?

[PLAIN]http://img638.imageshack.us/img638/9334/spring62009.jpg

EXAMPLE:

Homework Statement

A and B are immobile support. W = 400 [N]

The Attempt at a Solution

I first found out that Ax = 920 and Bx = 920. That's true according the book. Attached here only are my equations which led to my confusion with respect to my terms cancelling out. (I know that I can isolate the beams but I want to see why a perfectly legimate method doesn't seem to work)

 

[Studiot]

It's a pity you've made a new thread, my head is getting dizzy swapping back and fore.
No wonder I'm confused.
Perhaps some kind Mod will combine the threads.

I'm confused because

I can't find X in the diagram.
You say you have calculated moments at B, but you also say that rotation is allowed at B ie B is pinned?
Also I am not clear about the connection at C.

 

[SteamKing]

Equations not attached. Unable to examine your work.

 

[Femme_physics]

Sorry - now they're attached.

It's a pity you've made a new thread, my head is getting dizzy swapping back and fore.

Sorry, I felt it would help for the sake of order and neatness.

I'm confused because

I can't find X in the diagram.
You say you have calculated moments at B, but you also say that rotation is allowed at B ie B is pinned?
Also I am not clear about the connection at C.

It's all pins, A B and C. Movement on the X and Y axises is not allowed, rotation is.

What do you mean you can't find X? It's actually a 2D problem if youve notice. X is the horizontal axis.

 

[I like Serena]

I don't quite follow the steps in your calculation.
But I can say that you can't just take the moment relative to W.

To calculate with moments, the bodies involved need to be "rigid".
The cable is not rigid.

The "trick" would be to first eliminate the cable and the wheel as well.
Then the problem reduces to a rigid-body problem.

Can you say what the forces will be at E and at D due to the cable with weight W attached?

[edit]Picture attached[/edit]

 

[Femme_physics]

But I can say that you can't just take the moment relative to W.

To calculate with moments, the bodies involved need to be "rigid".
The cable is not rigid.

The "trick" would be to first eliminate the cable and the wheel as well.
Then the problem reduces to a rigid-body problem.

Ah, I see. But if it were a force apply on a rigid surface, can I calculate the sum of all moments to that?

Can you say what the forces will be at E and at D due to the cable with weight W attached?

Ex = Dx
Dy = W

 

[I like Serena]

Ah, I see. But if it were a force apply on a rigid surface, can I calculate the sum of all moments to that?

Yes.

Ex = Dx
Dy = W

There's an additional rule about cables.
In a cable the size of the force (called "tensile force") is the same at any point in the cable.
More specifically, Ex=W.

[edit]So can you set up the sum-of-moments in the rigid system now?[/edit]

 

[Femme_physics]

There's an additional rule about cables.
In a cable the size of the force (called "tensile force") is the same at any point in the cable.
More specifically, Ex=W.

Ah. I see, now I understand why in the solution paper they sometimes use that neat little fact instead of using the beam isolation method.
Thanks a bunch. Always very helpful Serena. :)

 

[Studiot]

Grrr just lost a long post here, hit the submit button and it vanished.

OK so there is more to be learned from this question, based on the fact that if a body is in equilibrium under the action of three coplanar forces, the forces are either parallel or concurrent (meet at a point somewhere).

This frame is under the action of three external forces, the weight, W, and the reactions at A & B ( not the components the reactions themselves).

To use this theorem I have introduced two auxiliary points F and G.
G is the point of concurrency, not necessarily in the frame.
F is the point at the end of the pulley where the cable turns vertical.

Draw a vertical line through F. Since this is the line of action of the vertical load, G must be somewhere on this line.

Prop AC has no other reactions than the pin reactions at A & C so the reaction at A, R_A must pass through A and C.
So draw AC produced to meet the vertical line through F at G. This fixes point G.

Member BCF has non nodal loading at F so is also subject to bending so the reaction at B will not be axial.
However by the concurrency theorem the reaction at B must pass through B and G
So draw BG as the line of action of R_B

Introducing angles alpha and beta it is a matter of applying some simple trigonometry and the conditions of vertical and horizontal equilibrium at G to solve the frame.

You can check my sums if you wish, my arithmetic is atrocious.

Two further issues arise here.

Firstly
ILS commented that the tension in the cable is constant throughout.
This is only true if there is no friction in the pulley and no power is transmitted by the cable.
Knowing when this condition applies is particularly important in machinery where power is transmitted by belt drives and it is the difference in tension which allows the power transmission.

Secondly
You mentioned torque.
The physics/engineering world is remarkably sloppy about terminology here.
Although all turning effects are ultimately the same, there are three distinct and significantly different associated circumstances.
This is again particulary important in machinery where torque is used to represent the result of torsion and measured in rpm or angle or shaft output.

You may wish to review post#16 of this thread

https://www.physicsforums.com/showthread.php?t=406626&highlight=torque

go well

 

[Femme_physics]

Studiot, thank you! This is incredible insight that was missing from my know-how in the first semester, it'll definitely come in handy :D I'll review the post and try to learn more about concurrent forces. They seem to be an ignored issue, but they're factually important, I think!

 

[Studiot]

Concurrent forces appear pretty often in machinery theory. And you can often reduce them to 3 forces.