A triangular set square of angle 30∘,60∘ and 90∘ 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 W1W2 is 1:n. The value of n is (integer only)
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Solution
The hypotenuse will be oriented horizontally only when torque on the hypotenuse becomes zero.
Using, principle of moments,
Clockwise torque about point D = Anticlockwise torque about point D
W2(BD)=W1(AD)
⇒W1W2=BDAD.......(1)
Now, From the diagram,
Let AB=l
AC=lcos30∘;BC=lcos60∘
⇒AD=ACcos30∘=lcos230∘
⇒BD=BCcos30∘=lcos260∘
Putting this values in (1) we get,
∴W1W2=BDAD=cos260∘cos230∘=1434=13
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.