# Homework Help: Statics question

1. Jun 9, 2006

### Rumpelstiltzkin

Here's a statics problem that I'm unsure about. The system is in static equilibrium, and I'm supposed to find the reaction forces.

{convention: right and up are taken as positive for x and y respectively, and counter-clockwise is positive for moment}

$$\sum F_{y} = 0$$
$$B_{y} = -1000N$$

$$\sum F_{x} = 0$$
$$A_{x} + B_{x} = 0$$

Taking the sum of moment at the centre of the circular thingy
$$\sum M_{z} = 0$$
$$0 = -0.05(A_{x}) + 0.05(B_{y}) - 0.25(1000)$$

$$A_{x} = -6000 N$$
$$B_{x} = 6000 N$$

I solved this problem assuming $$A_{y} = 0$$, by inspecting the diagram. Is that alright? In cases where it's not as obvious, how do I determine which reactions are not supposed to be considered? I think I must be neglecting something small.

Last edited: Jun 9, 2006
2. Jun 9, 2006

### dav2008

You can't just randomly consider a reaction force to be 0. I'm not sure how you inspected the diagram to come to that conclusion.

It comes down to you having 4 unknown reaction forces ($$A_x,A_y,B_x,B_y$$) and only 3 equations of equilibrium ($$\sum F_{y} = 0$$, $$\sum F_{x} = 0$$,$$\sum M = 0$$) Like you said, you have to find some other equation based on other information.

By looking closer at the diagram it appears as if point B is not physically attached to the circle. That would suggest that it is free to move vertically so $$B_y=0$$. It's hard to say from that diagram though.

Does anyone have a second opinion?

(By the way, welcome to PF and thanks for showing your work and making it very clear)

Edit: Also, when you summed the forces in the y direction I think you missed a negative sign.

Last edited: Jun 9, 2006
3. Jun 9, 2006

### Hootenanny

Staff Emeritus
I don't think you can assume that, infact I think the opposite, for the system to remain static Ay must be non zero. Perhaps I am wrong, could you post your thought processes.
I would agree with you there dav. We can say that By = 0 if we assume that the system is frictionless, which seems like a reasnable assumtion as we are not given any coefficents. I am also assuming ofcourse that the spanner is light.

So now you have three unknowns and three equations. Happy Days I also would like to comend you on your post, all the required information was provided and it is presented in a clear logical fashion, a pleasure to read.

Last edited: Jun 9, 2006
4. Jun 9, 2006

### Pyrrhus

Definetly agree with the replies, but i will also want to request the problem statement, maybe there's more info there?