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tnich
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Right. Now what is the mass of the part of the board hanging over?Richie Smash said:F=m*g
Right. Now what is the mass of the part of the board hanging over?Richie Smash said:F=m*g
Say the mass of the board is m. How much of that mass is hanging over the edge?Richie Smash said:The mass is... m?
What is the fraction of the board hanging over?Richie Smash said:m/10 Kg
No, that is still not right.Richie Smash said:Might this be on the right track?
Let's try it this way. What is the mass per unit length of the board?Richie Smash said:Sorry that I can't figure this out,
I know the fraction of the board in length hanging over is x/l and I suppose for the mass it would be x/m?
OK. If x meters of board are hanging over, how many kilograms would that be?Richie Smash said:That would be 5kg per metre
No. x is a measure of length in meters. You have the mass per length of the board as 5kg/m, so how would you find the mass of x meters of board?Richie Smash said:x Kg
Correct.Richie Smash said:Hello, I have figured out that it would be 5x Kg.
You seem to be just guessing. Try figuring it out.Richie Smash said:Would the average position be 5/x?
I hope I am not judged for having so many posts in one question, this concept is foreign to me.
Very good! It earned you a like. From old days you may remember the see-saw that is horizontal if both loads are the same and at the same distance. No toppling if moments clockwise and anticlockwise cancel. And the support is in the center (something else you can discover from the books experiment: the main contact force in the extreme situation is at the edge of the supporting book). You could draw a free body diagram of the forces that act for the top book and check force and moment balances.Richie Smash said:I did it with two books just now and it seems to be about half the length of book with length of about 25cm is the point of max equilibrium.
That cannot be right. (5x*x)/5x simplifies to just x. That would put the center of gravity of the right half of the board exactly at the tip of the board.Richie Smash said:cg = ((m1 * x1) + (m2 * x2))/ (m1 + m2).
I found this on a website, but I was only asked to find the COG for the right side, so perhaps I would exclude m2 and x2
and write that the average position of the COG on the right side is (5x *x)/5x
Yes!Richie Smash said:Oh I didnt know that!
Then it would simply be x/2 yes?
The full equation for what? The torque due to gravity acting on the right half of the top board?Richie Smash said:Ah hold on mr J briggs, I'm on to something, the centre of gravity is a the point where the total weight is acting upon it, we know the mass.. we know the length..
The full equation is x *5x *x/2?
Yes.Richie Smash said:Ah yes I see you're right, I understand now finally... wow..
So would you confirm 5x2/2 is onto something?
Yes! Now you have the moment acting on the right side of the board, according to the diagram in terms of the length x of that side.Richie Smash said:Ah yes I see you're right, I understand now finally... wow..
So would you confirm 5x2/2 is onto something?
Oops, sorry, missed the last two posts. Yes, you have the right expression for the moment on the other side.tnich said:Yes! Now you have the moment acting on the right side of the board, according to the diagram in terms of the length x of that side.
What is the length of the other side?
Good. That looks entirely correct.Richie Smash said:the mass of the left half would be 5(l-x) and the postion of the center of gravity would be (l-x)/2
So the Moment would be (5l2-10lx+5x2)/2
Yes.Richie Smash said:Yes
(5x2-10lx+5l2)/2=5x2/2
And i know l =2
So x ultimately is 1?