Question

physics:- forces

A uniform metre rule of mass 100g is balanced on a fulcrum at 40 cm mark by suspending

an unknown mass m at 20 cm mark. Find the value of m. To which side the rule will tilt if

the mass is shifted to 10 cm mark. How can it be balanced by another mass of 50g ?



if u know then only answer this sum and pls show all the steps ​

Answer 1

Given :-

• A uniform metre rule of mass 100g
• balanced on a fulcrum at 40 cm
• unknown mass m at 20 cm mark
• the mass is shifted to 10 cm mark.
• another mass of 50g

To Find :-

• principle of moments,
• Find the value of m.
• To which side the rule .
• How can it be balanced

(i) From the principle of moments,

Clockwise moment = Anticlockwise moment

• 100g × 10 cm = m × 20 cm = m = 50 g
• 100g × (50 − 40) cm = m × (40 − 20) cm

(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.

Additional Information :-

• Clockwise moment = m2× d2

• Anti-clockwise moment = m1× d1