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A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the stick and a 320.0 gram object is hung at the 75.0 cm mark. Drawing is approximate. The meter stick is unbalanced. Determine the cm-mark on the meterstick that a 400 gram object needs to be hung to achieve equilibrium. A) 10.0 B) 40.0 C) 60.0 D) 90.0 E) none of the above is within 10% of my answer

Respuesta :

Answer:C

Explanation:

Given

mass [tex]m_1=400\ gm[/tex] is at [tex]x=20\ cm[/tex] mark

mass [tex]m_2=320\ gm[/tex] is at [tex]x=75\ cm[/tex] mark

Scale is Pivoted at [tex]x=50\ cm mark[/tex]

For scale to be in equilibrium net torque must be equal to zero

Taking ACW as positive thus

[tex]T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)[/tex]

[tex]T_{net}=0.12g-0.08g=0.04g[/tex]

Therefore a net torque of 0.04 g is required in CW sense which a mass [tex]400\ gm[/tex] can provide at a distance of [tex]x_o[/tex] from pivot

[tex]0.04g=0.4\times g\times x_o [/tex]

[tex]x_o=0.1\ m[/tex]

therefore in meter stick it is at a distance of [tex]x=60\ cm[/tex]

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