# Simple statics

1. May 26, 2006

### TSN79

I'm trying to solve the problems on this page:

http://physics.uwstout.edu/StatStr/statics/StatII/statp21f.htm

The second last one is a bit hard for me, but I think I've got it. Just need to get my approach confirmed as correct. I first found the external support reactions, that went fine.

Then, to determine the force in CD, I first drew an FBD (free body diagram) of member ABC placing two uknown forces at point B (Bx and By) and the one given force at C (should I have placed two unknown forces here also, Cx and Cy ?). Solving this I found the values for Bx and By.

I now drew a new FBD for member BD, placing two new unknown forces (Dx and Dy) in point D along with the given force in the middle, the known force E, and Bx and By (which I found earlier) in point B. Solving for Dx I got the correct answer of 9000 lb, but is my approach valid, or was this just coincidence?

Any help will be appreciated...

Last edited by a moderator: Apr 22, 2017
2. May 26, 2006

### Pyrrhus

Which problem is this? the third?

3. May 27, 2006

### TSN79

No, fourth.

4. May 27, 2006

### Pyrrhus

Yes, member ABC ha Cx and Cy on its FBD.

So you took member BD, and put the forces at B (Bx and By), the forces at D (Dx and Dy, 4000 and the E force)??, plus the 5000 force in the middle??

5. May 27, 2006

### haynewp

You're doing too much work. You can solve for CD by just balancing the free body of member ABC now that you have the reactions at A. CD is a 2 force force member so it only has axial force (in its X direction), but you still have to apply the 10,000lb load to joint C in the Y direction when balancing the free body of ABC. Correctly sum moments on the free body of ABC and the force in CD can be found.

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