Question

A triangular set square of angle ${30}^{\circ },{60}^{\circ }$ and ${90}^{\circ }$ and of negligible mass is suspended freely from the right-angle corner and weights are hung at the two corners. If the hypotenuse of the set square adjust horizontally, then the ratio of the weights $\frac{W_1}{W_2}$ is $1:n$. The value of n is (integer only)

Answer 1

The hypotenuse will be oriented horizontally only when torque on the hypotenuse becomes zero.

Using, principle of moments,

Clockwise torque about point $\text{D}$ = Anticlockwise torque about point $\text{D}$

$W_2\left(BD\right)=W_1\left(AD\right)$

$\Rightarrow \frac{W_1}{W_2}=\frac{BD}{AD}.......\left(1\right)$


Now, From the diagram,

Let $\text{AB}=l$

$\text{AC}=l\cos {30}^{\circ };\text{BC}=l\cos {60}^{\circ }$

$\Rightarrow \text{AD}=\text{AC}\cos {30}^{\circ }=l{\cos }^2{30}^{\circ }$

$\Rightarrow \text{BD}=\text{BC}\cos {30}^{\circ }=l{\cos }^2{60}^{\circ }$

Putting this values in (1) we get,

$\therefore \frac{W_1}{W_2}=\frac{BD}{AD}=\frac{{\cos }^2{60}^{\circ }}{{\cos }^2{30}^{\circ }}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

Hence, $n=3$

Why this question ? This question involves idea of rotational equilibrium, when object is in rotational equilibrium, net clockwise torque and net anticlockwise torque about axis of rotation will be equal.