Question

Two $L$ shaped wires are kept over each other as shown. The wire on the left is fixed and the wire on the right is movable on the left wire without any friction. The whole system in a horizontal plane. Now we make a soap film of surface tension $0.1\text{N/m}$ in between the wires such that it covers the common quadrilateral area. If the right wire has a mass of $250\text{gm}$, what is its initial acceleration (in ${\text{m/s}}^2$ ) on being released?

Answer 1

Let $l_1,l_2$ be lengths of limbs of movable L shaped wire and $S$ be surface tension.
As soap film is introduced in between L shaped wires,
Force on limb of length $l_1$ of movable L shaped wire is $2S{l}_1$
Force on limb of length $l_2$ of movable L shaped wire is $2S{l}_2$
As limbs are perpendicular to each other, net force is
$F_{net}=\sqrt{(2S{l}_1{)}^2+(2S{l}_2{)}^2}=ma$

Given that $l_1=1.5\text{m}{l}_2=2\text{m}S=0.1\text{N/m}m=0.25\text{kg}$

$a=\frac{2S\sqrt{l_1^2+l_2^2}}{m}=\frac{2\times 0.1\times \sqrt{{1.5}^2+2^2}}{0.25}$
$a=2{\text{m/s}}^2$