Question

A $198\,cm$ tall girl lies on a light board which is supported by two scales one under the top of her head and one beneath the bottom of her feet. The two scales read respectively $36$ and $30\,kg$. What distance is the centre of gravity of this girl from the bottom of her feet?

A) $99\,cm$

B) $90\,cm$

C) $108\,cm$

D) $82\,cm$

Answer 1

Here we have to use the relation of balancing torque to get the answer. We have to balance the torques applied on both the weights.

Complete step by step solution:

Length of the girl $= 198\,cm$

Let O be the centre of gravity.
Let the distance between the feet and the centre of gravity be $x$

Therefore, the distance between the centre of gravity and head is $198 - x$ .
Now we have to balance the torques to get the answer.
Let the $36\,kg$ be in the clockwise direction and $30\,kg$ be in the anti-clockwise direction, then-
$36g\left( {198 - x} \right) = 30gx \\\ 1188 - 6x = 5x \\\ x = 108\,cm \\\$

Hence, option C is correct.
Thus, the distance from the bottom of the feet to the centre of gravity is $108\,cm$ .

Additional information:
-Torque is a force measure that can cause an object to rotate around an axis. In linear kinematics, just as force is what causes an object to accelerate, torque is what causes an object to achieve angular acceleration. The torque vector orientation is dependent on the direction of the force on the axis.
-In angular motion, torque is also known as the parallel of force. The tuning effect is thus algebraically strictly proportional to both the force magnitude and the perpendicular distance between the force vector and the hinge commonly referred to as the fulcrum in which angular motion is to take place.
-For an unconfined object, or for a confined object, the product of the force and the perpendicular distance to the centre of gravity is referred to as the torque or moment.

Note:
The centre of gravity lies in the middle of the body and the torque acts in the clockwise or anti-clockwise direction. So, the body is unbalanced. To balance the body we need to equate the torque on both sides.