# Centre of mass question

a triangle ABC with AB=BC=2a and AC=2a rt(2) is drawn on a uniform lamina. A semi-circle is drawn on BC as diameter,on the opposite side of BC from A and the area enclosed by the triangle and the semi-circle is cut out.

The resulting lamina is suspended freely from B,show that AB makes an angle X,st

tan(X)=(8+3∏)/4

This part i got. i found,using B as (0,0), the centre of mass to be

(-4a/(3∏+12) , (8+3∏)a/(3∏i+12)

A point P is taken on AB and the triangle APC is cut off. if the remaining lamina hangs with BC vertical show the distance BP is a√(2)

for this i have said if BC hangs down,the com of the shape must have x-cordinate 0 so the com lie on BC.

now let BP=P

centre of mass of the triangular bit APC left can be found

using com of triangle APB and ABC.

i find x-cord of com of triangle is

-2a/3[2a^2-P]/area of triangle

this leads to x-cord for new shape:

-4a/3(∏+4){area of original shape) - {-2a/3[2a^2-P]/area of triangle}{area of trianlgle}

which can clearly be seen to be wrong as i do not have a p^2 term and i want p=a√2

where am i going wrong? am im correct to say x-cord=0?

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I did not look at the first part as you say that you got that.

For the second part I can't follow what you are trying to show, if you add in the intermediate steps it should help me.

Yes the center-of-mass for the new shape must lie along the line BC in the second part as we are told that the line BC is vertical and we already know that BC passes through the point of suspension for the new shape. I'm not sure where you are going with the x-cord = 0 statement as it is always the case that in equilibrium in a uniform gravitational field that the center-of-mass of an object lies vertically below the point of suspension.

I did get the answer BP = a√(2) quite quickly so here's a question for you: what is the angle ABC?

I did not look at the first part as you say that you got that.

For the second part I can't follow what you are trying to show, if you add in the intermediate steps it should help me.

Yes the center-of-mass for the new shape must lie along the line BC in the second part as we are told that the line BC is vertical and we already know that BC passes through the point of suspension for the new shape. I'm not sure where you are going with the x-cord = 0 statement as it is always the case that in equilibrium in a uniform gravitational field that the center-of-mass of an object lies vertically below the point of suspension.

I did get the answer BP = a√(2) quite quickly so here's a question for you: what is the angle ABC?
ok,ill try and outline my idea a bit better/.
consider the shape with APC cut out.
i know BG (G=com) is vertical if shape is suspended at B.
if BC is vertical then G lies on BC
then the shape has com (0,y bar) if i measure from B

so i need to find the com off the new shape and set the x-cordinate = to 0

the above is what my thought process was to try and involve the length BP in an equation.

to find the com of the shape i found the com of the triangle APC and i know com of the original shape from part 1 of the question.

then
(x-cordinate of the com of new shape)(mass of new shape)=

(x-cord of com of original shape)(mass of original)-(x-cord of com of APC)(mass of APC)

which =0 by above .

this is what i tried to do. if you have the answer easily,can you explain your way of getting it? and if possible, why my method does not work?

anyone else want to join in and lend a hand? i seem to have lost nopoke

my own algebra has let me down :(

back to the question I asked you: what is the angle ABC? The answer makes dealing with the triangle PBC much easier.

As you know that you want the com to be along the line BC then moments about that line are easily equated as the shapes are now a triangle on one side and a semicircle on the other. You don't care where the com is along BC just that it is on BC.

my own algebra has let me down :(

back to the question I asked you: what is the angle ABC? The answer makes dealing with the triangle PBC much easier.

As you know that you want the com to be along the line BC then moments about that line are easily equated as the shapes are now a triangle on one side and a semicircle on the other. You don't care where the com is along BC just that it is on BC.
i know ABC is 90. the triangle left is not a nice one which is why i am saying com of APC can be worked using com of APC and ABP which as they are right angled i know com is 1/3 of the distance of base and height from right angle.

then as i dont care where on BC just that it is on BC i let x bar= 0. this does not get me a quadratic in P though...

PBC is a right triangle too, with BC vertical and PB horizontal. Can you add a sketch of what you think this new shape looks like?

I ended up with a linear equation when solving for the position of P.

PBC is a right triangle too, with BC vertical and PB horizontal. Can you add a sketch of what you think this new shape looks like?

I ended up with a linear equation when solving for the position of P.
i am such an idiot!! I put the point on BC and not AP!! your post made me scratch my head when you mention a right angle triangle! this led me to discover i had the wrong point!! DOH

i now have answer as i get the quadratric i was after :)

thanks