Question

A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick should a 5.00-g tack be placed so that the stick will balance at the 50.0 cm mark?

Answer 1

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

$80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)$

$80=5(50-x)$

$80=250-5x$

$5x=170$

$x=\frac{170}{5}$

$x=34 cm$