Question

The diagram below shows a metal rod of length 120 cm and weight 420 gf
balanced on a pivot. How much minimum weight needs to be suspended on it if
it has to be balanced as its centre?

Answer 1

Answer:

minimum weight needs to be suspended on it is 70 gf

Explanation:

As we know that the rod is initially balanced at the position of x = 50 cm mark

So here center of mass of the rod must exist

Now we wish to balance the rod at its mid point

So the torque of its weight must be counter balanced by the external weight suspended on other side

So here we will have

$\tau_{rod} = \tau_{weight}$

so we have

$(420 g)(60 - 50) = m(120 - 60)$

so we have

$4200 = 60 m$

$m = 70 gf$

#Learn

Topic : Torque Balance

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