- #1
jiboom
- 91
- 0
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 I am correct to say x-cord=0?
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 I am correct to say x-cord=0?