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(A) Let the linear density of wires $AB$ and $AC$ be $\lambda$. Then the linear density of wire $BC$ is $2\lambda$.Mass of $AB = 5\lambda$,Mass of $AC

Vedclass · Accepted Answer

$\frac{34}{11} \,cm$

Vedclass · Answer

(A) Let the linear density of wires $AB$ and $AC$ be $\lambda$. Then the linear density of wire $BC$ is $2\lambda$.Mass of $AB = 5\lambda$,Mass of $AC = 5\lambda$,Mass of $BC = 6 	imes 2\lambda = 12\lambda$.The height of the triangle $ABC$ from $BC$ to $A$ is $h = \sqrt{5^2 - 3^2} = 4\,cm$.Let $BC$ lie on the $x$-axis with $B$ at $(0,0)$ and $C$ at $(6,0)$. Then $A$ is at $(3,4)$.The centre of mass of $AB$ is at $(1.5, 2)$,$AC$ is at $(4.5, 2)$,and $BC$ is at $(3, 0)$.The $y$-coordinate of the centre of mass is $Y_{cm} = \frac{m_{AB}y_{AB} + m_{AC}y_{AC} + m_{BC}y_{BC}}{m_{AB} + m_{AC} + m_{BC}} = \frac{5\lambda(2) + 5\lambda(2) + 12\lambda(0)}{5\lambda + 5\lambda + 12\lambda} = \frac{20\lambda}{22\lambda} = \frac{10}{11}\,cm$.This is the distance from the base $BC$. The distance from $A$ is $h - Y_{cm} = 4 - \frac{10}{11} = \frac{44-10}{11} = \frac{34}{11}\,cm$.

$A$ thin uniform wire is bent to form the two equal sides $AB$ and $AC$ of triangle $ABC$,where $AB = AC = 5\,cm$. The third side $BC$,of length $6\,cm$,is made from a uniform wire of twice the linear density of the first. The distance of the centre of mass from $A$ is

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