CBSE Physics Experiment To Measure the Mass of a Body on a Metre Scale Using the Principle of Moments

State the Principle of Moments - The algebraic sum of moments in the clockwise direction is equal to the algebraic sum of moments in the clockwise direction when several forces are acting on a rigid body, and the body is in equilibrium. In other words, the total algebraic sum of all the forces' moments is zero. 

Some daily life examples of the principle of moments are opening up a door, transporting materials in a wheelbarrow, using a screwdriver to remove a screw, using a knife to cut open a tin, a ride on a seesaw.

Table of content

• Aim

• Theory

• Observation

• Result

Aim

To measure the mass of a body on a metre scale using the principle of moments.

Apparatus Required

• A load of unknown mass

• A wooden or metal wedge with a sharp edge

• A weight box

• A thread that is nearly 30 cm long

• A spirit level

• A raised platform that is about 20 cm high

• A wooden metre scale with uniform thickness can also be used

Theory

The sum of the clockwise moments is equal to the sum of the anticlockwise moments for a body free to spin about a fixed axis in equilibrium. \[\] 

\[{M_2}{L_2} = {M_1}{L_1}\] 

if M1 is the known mass suspended at a distance L1 on one side of a beam's centre of gravity and M2 is the unknown mass suspended at a distance L2 on the other side, and the beam is in equilibrium.

Procedure

1. On a table, build a raised platform. To execute this, one can utilize a metal or wooden block. However, the platform needs to be strong such that you can lay a wedge on it with a sharp edge. 

2. Alternatively, the wedge can be fixed to a lab stand at a height of roughly 20 cm above the table top. Set the level of the wedge horizontally using a spirit level.

3. Make two thread loops that will be used to suspend the weights from the metre scale and the unknown mass (beam). Place the loops 10 cm in from each side of the metre scale's edge.

4. Place the metre scale on the wedge using the thread loops, then tweak it until it is balanced. In the area above the wedge where the scale is balanced, mark two points on the scale. The equilibrium position can be identified by connecting these two points with a straight line, even if the scale tips over from the wedge for whatever reason. This line runs across the scale's centre of gravity.

5. Pick the unknown weight in one hand and with the other hand pick another weight from the weight box, which appears equal to the unknown weight while holding in hand.

6. Suspend the unidentified mass from one of the two thread loops connected to the metre scale. Using the other loop, suspend the known weight (Image A).

7. Move the loop to change the known weight's position until the metre scale is balanced on the acute wedge. Make sure the line drawn in Step 3 is exactly above the wedge in the balanced position and that the thread of the two loops crossing the scale is parallel to this line.

8. The distance between the loops' current locations and the line drawn in Step 3 should be measured. Keep a record of your findings.

9. Repeat the activity at least twice using a slightly lighter and heavier weight. Take note of each case's unknown mass and weight offsets from the step 3 line.

Experimental Diagram

Observations

Position of the centre of gravity=________cm.

Calculations

The moment of the force on one side of the wedge will be equal to the moment of the force on the other side in the balanced position of the meter scale.

Moment of the force due to known weight = \[\begin{array}{l}({M_1}.{L_1})g\\\end{array}\] 

Moment of the force due to unknown weight =\[{M_2}.{L_2}(g)\] 

In balanced position,

\[\begin{array}{l}{M_1}.{L_1} = {M_2}.{L_2}\\{M_2} = \dfrac{{{M_1}.{L_1}}}{{{L_2}}}\end{array}\] 

Average mass of unknown load = ______g

Result

Mass of the given body = _____g

Precautions

1. The wedge must be sharp and always perpendicular to the scale's length.

2. The length of the scale should be perpendicular to the thread loops.

3. Thin, light, and strong thread should be used for the loops.

4. Reduce air currents as much as possible.

Lab Manual Questions

1. What is the term for the point on the scale when it is horizontally balanced on the wedge?

Ans: Pivot

2. How is the accuracy of mass measurement limited by the metre scale's least count?

Ans: Therefore, 0.1 cm or 1 mm is the least count on the metre scale.

3. Explain how the principle of moments are used in physical balances.

Ans: According to the Principle of Moments, the total clockwise and anticlockwise moments about a given point are equal when a body is balanced. That is why the principle of moments is used in physical balances.

4. What issues might air currents bring about with this activity?

Ans: Air currents will make an issue while balancing the meter scale on the wedge.

Practical Based Questions

1. How does the moment principle relate to the beam's balance?

Ans: If all of the moments acting on an item during static equilibrium are equal to zero, the object will not rotate. The sum of the clockwise moments should equal the sum of the anticlockwise moments if the net sum of the moments of the forces is zero.

2. Describe the moment of force.

Ans: The product of the force's magnitude and the line of action's perpendicular distance produces the moment of force. Therefore, force x perpendicular distance of force from point O equals moment of force about a given axis passing through point O.

3. How do we write a dimensional formula for the moment of force?

Ans: The moment of force is the tendency to cause the rotation of an object about its axis.

Mathematically, the definition of a moment of force is 

\[force \times dis\tan ce\] 

• The Force’s dimensional formula is \[{M^1}{L^1}{T^{ - 2}}\] .

• The distance’s dimensional equation is \[{M^0}{L^1}{T^0}\] 

4. What is the distance between the point of application of the force and the point if the moment of force of 20 N around a point is specified as 4 Nm?

Ans: We know that,

\[\begin{array}{l}{\rm{moment of force = force}} \times {\rm{distance}}\\{\rm{4 Nm = 20 N}} \times {\rm{r}}\\{\rm{r = 0}}{\rm{.2 m}}\end{array}\] 

5. What does a force moment look like in everyday life?

Ans: The best illustration of a moment of force in our daily life is a see-saw that we swing on in a park. The swing remains in a balanced position when weight is applied to both sides of a see-saw to balance the moment, but the swing does not remain in a balanced position if greater weight is applied to one side. This is referred to as an imbalanced moment, as a result. The idea of the moment of force is evident in many more instances throughout our daily lives.

6. Calculate the force's moment if a 30 N force acts at a 0.20 m perpendicular distance from the turning point.

Ans: we know that,

\[\begin{array}{l}Moment = f \times d\\moment = 30N \times 0.20m\\moment = 6Nm\end{array}\] 

7. List out 4 applications of the principle of moments.

Ans: Four applications of principle of moments are:

• Opening up a door.

• Transporting materials in a wheelbarrow.

• Using screwdriver to remove a screw.

• A ride on a seesaw.

8. From the diagram below, determine the resultant moment:

Ans:

\[\begin{array}{l}{\rm{Perpendicular distance = 1}}{\rm{.25cos(25) = 1}}{\rm{.13m}}\\{\rm{Moment = 625}} \times {\rm{1}}{\rm{.13 = 708 Nm}}\end{array}\] 

9. Explain why it would be preferable for a forklift truck to transport its cargo near to the ground rather than higher up in the air using your understanding of the centre of mass.

Ans: The centre of mass of the load and the vehicle move closer to the earth as the mass of the load is lowered toward the ground. This increases the load's and forklift truck's stability.

Viva Questions

1. What exactly is a moment?

1. A force's turning effect.

2. A point's mass of an object.

3. The distance between a mass and a pivot.

4. When a system is balanced.

Ans: A) A force’s turning effect

2. What is the formula for size of the moment?

1. \[M = f \times d\] 

2. \[F = M \times d\] 

3. \[M = \dfrac{F}{d}\] 

4. \[d = M \times f\] 

Ans: A) \[M = f \times d\]

3. What does the symbol d in the previous equation mean?

1. The parallel distance between the action line and the pivot.

2. The perpendicular distance between the pivot and the action line.

3. Distance between the two planes.

4. Distance of the plane.

Ans: B) The perpendicular distance between the pivot and the action line.

4. What must the total clockwise moment be about any pivot if the item is not rotating, as opposed to the total anti-clockwise moment?

1. The magnitude of the clockwise moment is double than that of the anticlockwise moment.

2. The amount of the clockwise moment is three times that of the anti-clockwise moment.

3. Anti-clockwise moment has a half-size compared to clockwise moment.

4. The anti-clockwise moment and the clockwise moment have exactly the same magnitude.

Ans: D) The anti-clockwise moment and the clockwise moment have exactly the same magnitude.

5. What would happen to the moment if the point of force's application was shifted further from the pivot?

1. Due to the greater distance, the moment would be bigger.

2. The force hasn't altered, thus the moment would stay the same.

3. Since it is further away, the moment would be less and have less of an impact on the pivot.

4. In order to adequately respond to this question, further details are required.

Ans: A) Due to the greater distance, the moment would be bigger.

6. Which of the following illustrates the use of the moments principle?

1. A rock falling

2. Lifting up a book

3. A USB stick 

4. A crowbar being used to lift a drain cover

Ans: D) A crowbar being used to lift a drain cover

7. Which option is correct for state of equilibrium in principle of moment 

1. Resultant moment should be zero

2. The tension must be equal on both sides

3. Both objects' weights should balance out on the beam.

4. The two objects should be the same distance apart in the beam balance.

Ans: A) Resultant moment should be zero

8. According to the principle of moments, the motion of the centre of mass depends upon

1. Total internal forces

2. Total external forces 

3. Both A and B

4. None of the above

Ans: B) Total external forces

9. On a line, where does the centre of mass of a two particle lies,

1. At any angle on this line

2. Joining the particles

3. Perpendicular to the line joining the particles

4. None of the above

Ans: B) Joining the particles

10. Where does the centre of mass of a system of particles does not depend upon

1. Mass of the particle

2. Symmetry of the body

3. Position of the particle

4. None of the above

Ans: D) none of the above

Conclusion

From this experiment, we can see that when two pans are filled with weights that are slightly different from one another, a beam balance will come to rest at an angle to the horizontal. The beam is supported by a pivot at a single fixed point. The net torque produced by the two weights around this single fixed point is not zero at the equilibrium position. The system stops rotating around the fixed point because the moment is balanced.