A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 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 10 cm ?
(iii) What is the resultant moment now ?
(iv) How can it be balanced by another mass of 50 g ?
A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 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 10 cm ?
(iii) What is the resultant moment now ?
(iv) How can it be balanced by another mass of 50 g ?
Mass of the ruler, M = 100 g = 0.1 kg
The center of gravity of the ruler is at 50 cm mark. Therefore, distance of the center of gravity of the ruler from Q = 10 cm = 0.10 m
Let ‘m’ be the mass of the block at 20 cm. Distance of block from the point of suspension = 20 m = 0.20 m
1. For the system to be at equilibrium, the moment about O due to the weight at B must cancel the moment about O due to the weight of the ruler at A (at the center of the ruler).
m × g × 0.2 = 0.1 × g × 0.1
⇒ m = 0.05 kg = 50 g
2. If the mass ‘m’ is shifted towards 10 cm mark, the moment due to the weight of ‘m’ will increase because the distance between the point of force and the axis will increase. So, the ruler will tilt downwards on the side of the mass ‘m’.
3. The net moment when the ‘m’ is brought to 10 cm mark can be found as:
New distance between ‘m’ and O is = 30 cm = 0.3 m
Moment about O due to ‘m’ = 0.05 × g × 0.3 = 0.015 × g
Moment about O due to the weight of the ruler = 0.1 × g × 0.1 = 0.01 × g
So, the net moment about O will be = 0.015 × g – 0.01 × g = 0.05 Nm (considering g = 10 ms-2)
4. Let another mass of 50 g = 0.05 kg is placed at a distance ‘X’ from O on the right side of the ruler.
Now, the net moment about O due to weights on the left side = 0.05 × g × 0.3 = 0.015 × g
The net moment about O due to weights on the right side = 0.1 × g × 0.1 + 0.05 × g × X
For equilibrium
0.015 × g = 0.1 × g × 0.1 + 0.05 × g × X
⇒ X = (0.015 – 0.01)/0.05 = 0.005/0.05 = 0.1 m = 10 cm
Thus, the new mass should be placed 10 cm to the right of O, i.e., at 50 cm mark.
Mass of the ruler, M = 100 g = 0.1 kg
The center of gravity of the ruler is at 50 cm mark. Therefore, distance of the center of gravity of the ruler from Q = 10 cm = 0.10 m
Let ‘m’ be the mass of the block at 20 cm. Distance of block from the point of suspension = 20 m = 0.20 m
1. For the system to be at equilibrium, the moment about O due to the weight at B must cancel the moment about O due to the weight of the ruler at A (at the center of the ruler).
m × g × 0.2 = 0.1 × g × 0.1
⇒ m = 0.05 kg = 50 g
2. If the mass ‘m’ is shifted towards 10 cm mark, the moment due to the weight of ‘m’ will increase because the distance between the point of force and the axis will increase. So, the ruler will tilt downwards on the side of the mass ‘m’.
3. The net moment when the ‘m’ is brought to 10 cm mark can be found as:
New distance between ‘m’ and O is = 30 cm = 0.3 m
Moment about O due to ‘m’ = 0.05 × g × 0.3 = 0.015 × g
Moment about O due to the weight of the ruler = 0.1 × g × 0.1 = 0.01 × g
So, the net moment about O will be = 0.015 × g – 0.01 × g = 0.05 Nm (considering g = 10 ms-2)
4. Let another mass of 50 g = 0.05 kg is placed at a distance ‘X’ from O on the right side of the ruler.
Now, the net moment about O due to weights on the left side = 0.05 × g × 0.3 = 0.015 × g
The net moment about O due to weights on the right side = 0.1 × g × 0.1 + 0.05 × g × X
For equilibrium
0.015 × g = 0.1 × g × 0.1 + 0.05 × g × X
⇒ X = (0.015 – 0.01)/0.05 = 0.005/0.05 = 0.1 m = 10 cm
Thus, the new mass should be placed 10 cm to the right of O, i.e., at 50 cm mark.
