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.(adsbygoogle = window.adsbygoogle || []).push({});

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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# Centre of mass question

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