Auniform 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

0.5 m

Explanation:

Given

length of Metal Bar L=5 m

mass of bar

mass of another object

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

torque by mass

Equating both torques

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

volts times amps 6 x 0.5 3w

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