A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-hand end of the bar. What distance x from the left-hand end of the bar should an object with mass 0.100 kg be suspended so the bar is balanced in a horizontal position

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nuuk nuuk · Answer 1 · 2019-11-05T10:02:37+01:00

Answer:0.5 m

Explanation:

Given

length of Metal Bar L=5 m

mass of bar [tex]m=0.3 kg[/tex]

mass of another object [tex]m_2=0.1 kg[/tex]

Let this mass be hanged at a distance of y from Pivot so it cancel out the torque of weight of bar

Torque of weight of bar[tex]=W\times 0.5=0.3\times 10\times 0.5[/tex]

torque by mass[tex]=mg\times y=0.1\times 10\times y[/tex]

Equating both torques

[tex]0.1\times 10\times y=0.3\times 10\times 0.5[/tex]

y=1.5 from Pivot point so it is 0.5 m form left end