Question

A uniform metre rule of mass 100gf is balanced on a fulcrum at mark 40cm by suspending an unknown mass m at the mark 20 cm.

(I) Find the value of m.

(II) To which side the rule will tilt if the mass m is moved to the 10 cm mark?

(III) What is the resultant moment now?

(IIII) How can it be balanced by another mass of 50 g?

Answer 1

Mass of the ruler = 100g = 0.1 kg

Length of the ruler l = 1m

Moment of the unknown mass of the fulcrum is equal to mg(0.2)

Moment of the mass of the ruler of the fulcrum is equal to (0.1)g(0.1)

Because the system is in equilibrium so mg(0.2) = (0.1)g(0.1)

=> m = 0.05kg = 50g



The 50g mass is moved to mark 10cm

The moment of the fulcrum because of the above stated mass is (0.05)(g)(0.3)
= 0.147 Nm

The movement of the ruler about the fulcrum is (0.1)g(0.1) = 0.098 Nm

The ruler will tilt towards the mass placed at 10cm mark

To make the system equal let another mass 50g or 0.05kg be placed
at xm from the fulcrum


Moment of the 50g mass at 10 cm mark is equal to moment of the ruler about the fulcrum + moment of the new 50g mass about the fulcrum

=>0.147 = 0.098 + (0.05)(g)(x)

=> 0.049 = 0.49x

=> x = 0.1m = 10cm from the fulcrum

Thus, the new 50g is placed at 50cm mark of the ruler

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