Question

A uniform meter rule placed on a fulcrum at its mid point O having a weight 40 gf at the 10 cm mark and a weight of 20 gf at the 90 cm mark. 1-Is the meter rule in equilibrium? If not, how will the rule turn 2.)How can the rule be brought in equilibrium by using an additional weight of 40 gf

Answer 1

Anticlockwise moment = 40 × (50 – 10)

= 1600 dyne cm

Clockwise moment = 20 × (90 – 50)

= 800 dyne cm.

Since anticlockwise moment is greater than the clockwise moment. So, for equilibrium suspend the 40gf away from 50 cm mark.

Then, 1600 = 800 + 40x

⇒ 40x = 800



The 40 g has to be suspended at (50 + 20) = 70 cm.

Answer 2

Answer:

it will placed pn 70 cm marks.

add the right side 50+20.