Weightless beam AO with vertical force and wire tensions

[Femme_physics]

Homework Statement

http://img84.imageshack.us/img84/1715/given2.jpg

A weightless beam AO is held in a horizontal position (at the xy plane) as described. At its tip O the ball is attached to a ball joint, and at points C and A its tied via wires. horizontal wire CD and slanted wire AB, whose in the plane parallel to xz. At point E acts on the beam vertical force Q. The shape OKAD is a rectangle (look at the upper view in the drawing).

Given:


http://img35.imageshack.us/img35/9106/given1.jpg

The Attempt at a Solution

http://img215.imageshack.us/img215/2376/per1v.jpg

http://img845.imageshack.us/img845/9779/per2.jpg

They give out Oz as 0.86666 [kN]. I doubt this could just be a rounding error?

 

[I like Serena]

Hey, I recognize this problem!

They give out Oz as 0.86666 [kN]. I doubt this could just be a rounding error?

I agree, but I get 0.9 kN as well...

 

[Femme_physics]

Hmm...my Tcd is wrong too according to the 1.732 gotten by the manual. Can we say we found another manual mistake or am I pulling the wrong wires? (excuse the pun )http://img861.imageshack.us/img861/9152/isitisit.jpg

http://img827.imageshack.us/img827/4494/xyxy2.jpg

 

[vela]

It looks like you got the components of T_CD switched in your moment equation.

Hint: You can calculate the moment about O due to T_CD more simply since you know the angle between the lever arm and the force.

 

[Femme_physics]

It looks like you got the components of TCD switched in your moment equation.

Did I really? What is the angle from Tcd to the Y axis? It's 60! I reckon I got it right.

Look, I made it more clearly on powerpoint so you'll see exactly what I'm seeing

http://img11.imageshack.us/img11/7708/metersandeverything.jpg

According to the axes, I used the right distances and angles

Hint: You can calculate the moment about O due to TCD more simply since you know the angle between the lever arm and the force.

but I don't have to use the diagonal distance.

 

[vela]

Did I really? What is the angle from Tcd to the Y axis? It's 60! I reckon I got it right.

According to the axes, I used the right distances and angles

Yeah, I think you are right. It's just a calculation error then. I haven't been able to reproduce your answer from your equation. I get $T_{CD}=\sqrt{3}~\mathrm{kN}$.

but I don't have to use the diagonal distance.

I have no idea what you mean. The torque τ is just τ = rF sin θ = (0.6 m)T_CD sin 60.

 

[Femme_physics]

Thanks I "think" I solved it. I'll post my full solution this evening scanned. Only Oz is different. I noticed the solution manual makes it out to be 1.5 kN, whereas it really should be 0.9 kN