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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?

Respuesta :

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

[tex]80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)[/tex]

[tex]80=5(50-x)[/tex]

[tex]80=250-5x[/tex]

[tex]5x=170[/tex]

[tex]x=\frac{170}{5}[/tex]

[tex]x=34 cm[/tex]

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