Torque Problem

[triden]

I have never really ever understtod the idea of torque correctly. I just cannot grab the concept. In this question the board is not moving, so the torque CCW must = torque CW...but how? as you can see I tried it, I only got 2/6 for marks on this question. What are the steps that I should take?

http://upload.cybermart.ca/2004/torquerpboem.gif

[Pyrrhus]

This is obviously an equilibrium problem of rigid bodies.

You need to apply this two conditions

\sum_{i=1}^{n} \vec{F}_{i} = 0

\sum_{i=1}^{n} \vec{\tau}_{i} = 0

This is a simple 2 condition solution, if you pick as your origin the point where R1 is you can find R2, and then you could use as your origin R2, so you can find R1.
Remember the weight of the board is at its center of gravity.

You can pick any point as your origin for the 2nd condition and the torque sum will always be 0, there should be a proof of this in your text book, if you still want to know why, i can get into detail.

[triden]

This is obviously an equilibrium problem of rigid bodies.

You need to apply this two conditions

\sum_{i=1}^{n} \vec{F}_{i} = 0

\sum_{i=1}^{n} \vec{\tau}_{i} = 0

This is a simple 2 condition solution, if you pick as your origin the point where R1 is you can find R2, and then you could use as your origin R2, so you can find R1.
Remember the weight of the board is at its center of gravity.

You can pick any point as your origin for the 2nd condition and the torque sum will always be 0, there should be a proof of this in your text book, if you still want to know why, i can get into detail.


Ok got that. Now waht Force would I use for R1 and R2. Would I add the person and the weight of the board together? Like say to get R1, Torque at R1 = F * LengthPerpendicular, so would I use 760N (the board and the person)?

[Pyrrhus]

No, you don't add the weight of the person and the board together...

You find each of the torque's for all the forces at the R1 force position as your origin, give each of the torques its sign by using the right hand rule, and add them algebraicly.

Remember

\vec{\tau} = \vec{r} \times \vec{F}

[seiferseph]

this appears to be a simple question, you can take torques at R1, and use the condition that Tcw = Tccw to solve. you know Fbeam + Fperson + R2 = R1, so after you get R1 or R2 you can easily solve for the other