Statics, crankshaft-like device. Need help.

[Femme_physics]

Homework Statement

http://img841.imageshack.us/img841/4725/device01.jpg

(I hope I translated it correctly, this has some technical terms...)

This device is used as a key for installing a disk on a stational shaft, O. The device is equipped with pins A and B, that goes into the holes of the disk. Screwing the disk is does by turning the handle BC clockwise, through the force F that's 90 degrees to the handle.
Fixing the disk to the shaft requires a rotating moment M, around the center of the disk O that equals 150 [N x m].
A) Calculate the force F you must apply at point C to get the required moment.
B) Calculate the forces acting on pins A and B and the force acting at pin D. The measurements are given at mm.

Homework Equations

Statics.
Answers are
F = 450 [N]
Fd = 1670.73 [N]
Fb = 1431.26 [N]

The Attempt at a Solution

Attached are my 2 attempts in solving this question.

 

[I like Serena]

I would suggest you start with a drawing with all the external forces and moments which would keep the system in static equilibrium.
This would consist of a moment of 150 Nm at O and a force at O.
And perhaps you could calculate the "arm" of force F relative to O?

 

[method_man]

I think you made a mistake by determening an angle.

 

[Femme_physics]

I would suggest you start with a drawing with all the external forces and moments which would keep the system in static equilibrium.

Attached, but I find myself with too many unknowns!

This would consist of a moment of 150 Nm at O and a force at O.

Force at O? Other than a moment? Mmm, my intuition told me that the design assures that the rotation happens without resistance, and since friction isn't mentioned...I figure there's only a force resisting F, but perhaps I'm wrong.

And perhaps you could calculate the "arm" of force F relative to O?

Too many unknowns. They could've at least given me F!

I think you made a mistake by determening an angle.

But if I didn't, I wouldn't know the distance from F to O.

 

[Femme_physics]

Aherm, forgot the attachment...

 

[I like Serena]

Attached, but I find myself with too many unknowns!

Well, the forces at A and B are internal forces, so those would be 4 unknowns that you do not have in this first equation.

Force at O? Other than a moment? Mmm, my intuition told me that the design assures that the rotation happens without resistance, and since friction isn't mentioned...I figure there's only a force resisting F, but perhaps I'm wrong.

Consider a doorknob.
If you push the doorknob down at the tip, you're effectively pushing the entire doorknob (and door) down. The door has to resist, otherwise the doorknob would fall on the ground. And because the door resists, the doorknob starts turning.
If we only look at the doorknob itself, the resisting force must be at the axis of rotation, and be equal and opposite to the force you are exerting at the tip of the doorknob.

Too many unknowns. They could've at least given me F!

Well, they want you to calculate F.
It wouldn't be right if they'd give you F then.

But if I didn't, I wouldn't know the distance from F to O.

In your sketch the distance from F to O is not quite right yet.
You included a factor of 0.35 for this distance, but this distance is slightly longer than the real distance. Can you work those angles?

Furthermore in your moment summation, you included A and B, but those do not play a part yet, being internal forces.
And you forgot to add the moment at O, which is 150 Nm, which must be added.
Note that the force at O is eliminated in the moment sum, because its distance to O is zero.

[EDIT]Btw, there must be some type of friction or resistance, because the problem specifies that you need a moment of 150 Nm to turn the "lock". This means you need to overcome some internal friction or resistance equivalent to this moment.
Note that in a doorknob, there is spring, that you turn against, which will make the doorknob come back to its original position.[/EDIT]

 

[Femme_physics]

*smacks forehead*

Yes, I realize my silly error with respect from the distance of F to O. I actually got it right if you look at one of my first attempts, but I included the force reaction at the pins. Let me rework that, I think I got it figured out.

[EDIT]Btw, there must be some type of friction or resistance, because the problem specifies that you need a moment of 150 Nm to turn the "lock". This means you need to overcome some internal friction or resistance equivalent to this moment.
Note that in a doorknob, there is spring, that you turn against, which will make the doorknob come back to its original position.[/EDIT]

Strange, they didn't say anything about friction...I'll make another crack at the problem with the newfound knowledge and be back with the results :) thanks.

 

[I like Serena]

Yes, I realize my silly error with respect from the distance of F to O. I actually got it right if you look at one of my first attempts, but I included the force reaction at the pins. Let me rework that, I think I got it figured out.

Strange, they didn't say anything about friction...I'll make another crack at the problem with the newfound knowledge and be back with the results :) thanks.

Note that in your first attempt you did calculate with the angles.
But you only took Fy in the moment sum, and not Fx.
(And in the sum of vertical forces, you left By out.)

I'd suggest you try to calculate the distance from F to O perpendicular to F.

 

[Femme_physics]

I'm 3.2 [N] shy of the answer. Is it rounding error? Maybe from the author's part, I paid a lot of attention for the significant digits in this case.

 

[I like Serena]

I'm 3.2 [N] shy of the answer. Is it rounding error? Maybe from the author's part, I paid a lot of attention for the significant digits in this case.

The vertical distance of Fx to O is not 50 mm, it is a little more...

[EDIT]I can already tell you that the proper answers are slightly different from the solution manual, but the deviations are less than 1.[/EDIT]

 

[Femme_physics]

Oh, I went with the assumption that C and A are on the same line, but that seems frivilous of me now. Looks like some trigonometry in play here. Let me see if I can work this out... thanks

 

[Femme_physics]

I got F :) It actually came out exactly 450 [N]. It was just painful trigonometry that they make you jump through, can never have a lazy mind in mechanics! I'll see about A and B now, and will upload the solution tomorrow (scanner's hijacked) so it'll be here for the record. Thanks a bunch Serena!