Centre of mass
A uniform triangular lamina XYZ has XY=XZ and the perp. distance of X from YZ is h. Prove, through integration, that the centre of mass of the lamina is a distance 2h/3 from X.
I've done this.
A uniform triangular lamina ABC has AB=AC=5a, BC=8a and D is the centre of mass of the lamina. (Think a triangle with another triangle cut out with the same base leaving an arrow shape and the inner point of the smaller triangle being the centre of mass (see http://upload.wikimedia.org/wikipedi...w_Down.svg.png))
Worked out the centre of mass from A as 5a/3.
The plate is of mass M, has a particle of mass M attached at B. The loaded plate is suspendend from C and hangs in equilibrium.
b) Prove that in this position CB makes an angle of arctan(1/9) with the vertical.
How do I do this please?
what is wrong with ?
And therefore finding the centre of mass of the new shape?
But from that step, how do I find the angle?
[Question 30 M3, Review Exercise 2] found it! that makes it easier.
So i am guessing that you got part a)
for part b) I would use B as my "origin" you have two masses, one at B (0,0) of mass M and the other of mass M at position ( 4 a ,4/3 a) then just use the standard method to find the new potion of the centre of mass.
Sorry for not giving the page reference, I'm using the old M2 text book when there were only 2 modules.
So if I draw up the table and work out the x and y bar (2a and 2a/3 I think?) what do I do to work out the angle?
have you ever done this before?
Originally Posted by Sarah.
Draw a line from the center of mass to C and you need to find the acute angle this line makes with the line BC. You can easily work out the tangent of this angle using basic trig.
If you really stuck I'll draw a diagram.
No, I'm only first year.
Thanks for the offer but I've got it.
Thanks lots for all your help. x