Question

A merter stick pivoted at its center has a 150gm mass suspended at its 20cm mark. a) Where should an 100gm mass be placed to produce equilibrium? b) What mass placed at the 90cm mark is needed to produce equilibrium?

Answer 1

Answer:1) 100 gm mass should be placed at 95 cm mark.

2) Mass of 112.5 gm should be placed at 90 cm mark.

Explanation:

For equilibrium of the meter stick the sum of the moment's generated by the masses should be equal and opposite

Answer to part b)

Since a meter stick is 100 cm long and it is pivoted at it's center i.e at 50 cm

Thus

1) Moment generated by 100 gm mass about center = $M_{1}=m_{1}g\times r_{1}\\\\M_{1}=0.15\times 9.81\times 0.3=0.44145Nm$

Let a mass 'm' be placed at 90 cm mark thus moment it generates equals

$M_{2}=m\times 9.81\times 0.4=3.924m$

Equating both the moments we get

$0.44145=3.924m\\\\\therefore m=\frac{0.44145}{3.924}\times 1000grams\\\\\therefore m= 112.5grams$

Answer to part a)

Let the 100 grams weight be placed at a distance 'x' right of center

Moment generated by 100 grams weight equals

$M_{1}=0.1\times 9.81\times x$

equating the moments of the forces we get

$0.1\times 9.81\times x=0.15\times 0.3\times 9.81$

$\therefore x=\frac{0.4415}{.981}=45centimeters$

thus the mass of 100 gm should be placed at 95 cm mark in the scale.