A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 10 cm by suspending an
A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 10 cm 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 mark 10cm?
(iii) what is the resultant moment now?
(iv) How can it be balanced by another mass of 50g?
1 Answers
(i) From the principle of moments,
Clockwise moment = Anticlockwise moment
100g × (50 − 40) cm = m × (40 − 20) cm
100g × 10 cm = m × 20 cm = m = 50 g
(ii) The rule will tilt on the side of mass m (anticlockwise), if the mass m is moved to the mark 10cm.
(iii) Anticlockwise moment if mass m is moved to the mark 10 cm = 50g × (40−10)cm = 50 × 30 = 1500 g cm
Clockwise moment = 100g × (50 − 40) cm = 1000g cm
Resultant moment = 1500g cm − 1000g cm = 500g cm (anticlockwise)
(iv) From the principle of moments,
Clockwise moment = Anticlockwise moment
To balance it, 50g weight should be kept on right hand side so as to produce a clockwise moment. Let its distance from fulcrum be d cm. Then,
100g × (50 − 40) cm + 50g × d = 50g × (40 − 10)cm
1000g cm + 50g × d = 1500 g cm
50 g × d = 500g cm
So, d =10 cm
By suspending the mass 50g at the mark 50 cm, it can be balanced.