Rigid bodies question help?

Tips to help you. (1) This is a moments question. (2) Make a scale copy of the situation to help you visualise. (3) Draw and label a diagram with the letters and where it is sensible to take moments. (4) If you are still stuck, report (1), (2) & (3) back here for further advice.

http://m.imgur.com/enP8nTO
Im not really sure where to take moments from. Is it even possible to take moments as i dont know where the normal force or frictional force are acting in terms of measurements.

If the lamina is going to topple over, which point is it going to rotate about? And that's the place to take your moments from.

PS: Good diagram.

Apparently my diagram is wrong, the frictional force is supposed to be acting in the other direction, why si that the case? i figured if it toppled over to the right the frictional force would have to act towards the left. OH WAIT IS IT BECAUSE THE BOTTOM WILL ROTATE IN CLOCKWISE DIRECTION SO IT WILL ACTUALLY PIVOT TO THE LEFT SO THE FRICTION FORCE WILL ACT ON THE RIGHT OHHHHHHHHHHHHHHHHHHHHHHHHHH AAAAAGGGGHHHHHHHHHHHHHHHHHHHHHHH

The centre of gravity of the lamina is to the right of the base (AE), so the mass will topple to the right, pivoting about the point E.

Friction is irrelevant (but as it topples to the right, the base will tend to move to the left, so friction acts to the right to stop that). Whichever direction friction acts in, it's line of action will go through the pivot E. So it's moment about E will be zero.

would that be the case of the normal force as well? yes it would.

But i am confused on how to find the minimum force required. taking moments the perpendicular distance to e from the line of action would have to be at max, and apparently that is when it is perpendicular to ec. Why is that the case? Couldn't we make the angle even smaller for even greater distance?

Yes. When it's on the point of toppling, all the reaction will be through E, and hence no moment about E.



Making the angle smaller would make the distance smaller. The maximum distance for the line of action through C, is when that line of action is perpendicular to the line connecting C and E.

Try drawing a few cases, or do it algebraically if you wish expressing the perpendicular distance in terms of CE and the angle

You are a legend

and i will remember you on my deathbed

I would hope you had more meaningful things to remember by then, but thanks for the thought.

sorry to bother but got into bit of a pickle. This is the question

"
•Whatis the moment of inertia about the centre of mass of a rod of length 1.5 m,with bodies of mass 2 kg and 3 kg respectively, on either end of the rod(ignore the mass of the rod itself)? What is the kinetic energy of thisassembly if it rotates about its centre of mass at 30 revolutions per minute(RPM)?"

http://imgur.com/rz0y4af

Well this was my attempt and apparently i drew drew it wrong. C of G is apparently closer to 2kg and i got Xg and 1.5-Xg the wrong way around. I am not sure why though, shouldn't C of G be closer to 3kg as it's the heavier weight?

You have the centre of mass correct. Why do you think it is wrong?

Because this is the correct solution

So I'm just wondering why is the C of G closer to 2kg instead of 3, and how am i supposed to know which distance is Xg in a question like this.