# Problem with a "missing" force in x-direction...

1. Nov 23, 2015

### ozmac

Hey guys,

So it's been a while since I've done F.B.D's, and what seems like a simple problem is causing me grief.

I have this scenario below, where there are two rigid bodies, first one is ABC which pivots at fixed point B.
This is connected with a pin to CDE, where D permits horizontal motion only, and where E is a wall which has stopped all rotation.

The distances are only shown roughly to show you the logic i was using to try solve it.
First I looked at ABC, and dedcided Fc would need to balance the external force "T", so I got Fc=25N
Then I looked at CDE, and decided that Fe would need to balance Fc, and I got Fe=71.4N.

But I think i've missied something, as if I sum the forces on CDE, I only have the one force in the X direction, i.e. Fe =71.4N.
D cannot impose any force in the x direciton (ignoring friction), so this tells me I didn't calculate the forces at join 'C' properly.

Can someone tell me what I've done wrong here? I seriously was looking at this all day and was just scratching my head.

Thanks!

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2. Nov 23, 2015

### BvU

There is a horizontal component to Fc that seems to be missing in your treatment ?

3. Nov 23, 2015

### ozmac

Hmmm ok, so if instead of Fc, I should designate a Cx and a Cy? then solve the two equations simultaneously?
For arguments sake, lets say that B and C are not horizontal, i.e. C is a little below B, meaning you can't ignore Cx when calculating moment about B.

4. Nov 23, 2015

### ozmac

Please guys I'm really stuck here.
Am I suppose to work with a Cx and Cy instead?
I tried that, and added a few equations, for example solving the x forces and y forces to be zero on ABC, and then I took a moment about D considering both bodies, (so forces E, Bx, By, T acting on the relevant normal distance to the joing at D), and was able to simultaneously solve all equations (MESSY!) but it came up with a ridiculous answer. It seems I'm still missing something. anyone?

5. Nov 24, 2015

### BvU

Agree. The torque around B due to T is compensated by the torque due to the vertical component CDE exercises on ABC.
Don't follow: I can understand that the wall may be smooth so that CDE can only exercise a horizontal force on E. So that gives a torque of 1.4 m x FE. And the vertical component of the force by ABC on C gives 4 m x FC. But ABC also exercises a horizontal force on C that we haven't established yet. To compensate the only horizontal force in CDE: FE.

By the way, it is strange that the lever arm should be 4 m if the distance CD is only √10 m
I can't pin it down exactly, either. But my instinct tells me the angle between CE and the horizontal plays a role. And that angle isn't given.

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6. Nov 25, 2015

### Gobi

Yes, you should work with Cx and Cy. Start with FBD for each body separately. You will get Bx and By reactions in point B and Cx, Cy reactions in point C for left body. For right body, Cx and Cy should be of opposite directions than for the left one. D and E reactions are purely vertical and horizontal, respectively. Then write down two force and one torque equation for each body. For torque sumation, I would choose point B for left body and point C for right body. So, you have 6 equations and 6 unknown parameters (Bx, By, Cx, Cy, D and E). This should be solvable either by hand or with excel or something.

If I may ask, how many equations did you use and how did you know the answer is ridiculous ?