Question

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

Answer 1

Answer:0.5 m

Step-by-step explanation:

Given

length of Metal Bar L=5 m

mass of bar
`m=0.3 kg`

mass of another object
`m_2=0.1 kg`

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
`=W* 0.5=0.3* 10* 0.5`

torque by mass
`=mg* y=0.1* 10* y`

Equating both torques

`0.1* 10* y=0.3* 10* 0.5`

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