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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?
[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.