Question

The diagram below shows a uniform bar supported at the middle point O. A weight of 40gf is placed at a distance 40cm to the left of the point O. How can you balance the bar with a weight of 80gf?

Answer 1

Hint: At first we need to find the direction of moment and also the magnitude of the moment due to 40gf weight this will be equation 1. After that we need to find the equation for the moment that is just opposite to the moment caused by the 40gf weight and consider the point at which the 80gf bar is present as x. Then equate both the equations to get the final result.

Formula used: $\tau =rF\sin \theta $
anticlockwise moment = clockwise moment.

Complete step by step answer:
The question says that there is a uniform bar which is supported in the middle.
There is a weight of 40gf at a distance 40cm to the left of the point o.
Now, how can we balance the bar with the weight of 80 gf.
We know that torque is, $\tau =rF\sin \theta $.
So, we can understand that a anticlockwise moment is already working on the bar due to the weight of 40gf, so the magnitude of the moment will be,
$\tau =40\times 40N$(Due to angle 90, sin90 will be 1)
Or,
$\tau =1600N$.
Now, we are asked to place the 80gf bar in such a way that the bar remain balanced, so we have to find the point where the force will be acting,
Let us consider the point where the 80gf bar is placed as x.
Now, as the bar should be balanced so, anticlockwise moment = clockwise moment.
That is,
$1600=80\times x$,
Which gives us the value as,
$x=20cm$.
So the weight of 80gf must be placed at 20cm from the point o on the right side of the point to balance the bar.

Note: Students must check that all the values that are given are in SI or not. After that they must know how to calculate the moment due to a weight. They must analyze which weight will cause which moment in which direction.