CHAPTER SIX: TURNING EFFECTS OF FORCES – PHYSICS S1-S2 NOTES

6.1 TURNING EFFECTS OF FORCES

The turning effects of forces are seen in the following actions:

• Opening or closing a door, window, lid of a container e.g. a Jeri can, turning on a tap etc.
• Doing or undoing a nut using a spanner.
• Two kids playing on a see-saw.
• etc.

(a) Factors determining the effect of a turning force

There are two factors which determine the effect of a turning force. They include:

1. The magnitude of the force.
2. The perpendicular distance from the line of action of the force to the pivot.

The combined effect of the above factors is called moment.

(b) Moment

Moment of a force about a point is the product of the force and the perpendicular distance from the line of action of the force to the point.

(c) The S.I unit of moment

S.I unit of moment = Nm (Newton metre)

Consider the diagram in figure 6.1 below in which a uniform metre rule is balanced on a knife edge at its centre.

The moment about the pivot due to forces, F₁ and F₂, is got by using the formula:

Moment = Force × Perpendicular distance from the pivot to the line of action of the force

Moment due to F₁ = F₁ × y

= F₁y Nm (Clockwise moment)

Moment due to F₂ = F₂ × x

= F₂x Nm (Anti-Clockwise moment)

If the two moments are equal, i.e. F₁y = F₂x, the body is said to be in equilibrium according to the principle of moments.

6.11 The Principle of Moment

The Principle of Moment states that:

When a body is in equilibrium, the sum of clockwise moments about any point is equal to the sum of anti-clockwise moments about the same point.

(a) A Couple or Torque (Parallel forces)

A couple refers to two equal and opposite parallel forces whose lines of action do not meet.

A couple produces rotation and can only be stopped or balanced by an equal and opposite couple.

Where: r₁ = r₂ = radius of the circle, r₁ + r₂ = d = diameter of the circle

F₁ = F₂

Characteristics of a body under the action of a couple:

1. The resultant force on the body is zero.
2. The turning effects of the forces cause a rotational effect.
3. The forces act in opposite directions.

(b) Moment of a couple (Torque)

The moment of a couple is equal to the product of one force and the perpendicular distance between the two forces.

The moment due to the couple in figure 6.1 above is given by the formula:

Moment of a couple = One force × the perpendicular distance between the forces

= F₁ × d Nm

Or Moment of a couple = One force × the perpendicular distance between the forces

= F₂ × d Nm

Or Moment of a couple = (F₁r₁ + F₂r₂) Nm

6.12 Parallel Forces in Equilibrium

(a) Conditions for a body in equilibrium

The following are the conditions for a body to be in equilibrium when under the action of a number of parallel forces.

1. The sum of forces acting in one direction is equal to the sum of forces acting in the opposite direction. I.e. the net resultant force on the body is zero.
2. When a body is in equilibrium, the sum of clockwise moments about any point is equal to the sum of anti-clockwise moments about the same point.

Consider the diagrams in figures 6.3 and 6.4 showing a uniform body in equilibrium under the action of parallel forces below for conditions (i) and (ii) above.

In the diagram, three forces (F₃, F₄ and F₅) act downward and two forces (F₁ and F₂) act upward and since the body is in equilibrium, then

Using condition (i) above we have:

=

• (F₁ + F₂) = (F₃ + F₄ + F₅)

=

F₃y + F₄d = F₁s + F₂x

Using condition (ii) above we have:

R = F₁ + F₂ + F₃ + F₄

(b) Resultant Moment

When dealing with problems involving a number of moments acting on a body which is NOT in equilibrium, the following steps are taken:

1. Draw the sketch diagram indicating all the forces and their respective distances from the pivot (fulcrum).
2. Give a positive sign to anticlockwise moments and a negative sign to clockwise moment(s).
3. Add the moments algebraically. The numerical value of the answer gives the magnitude of the resultant moment and the arithmetic sign of the answer gives the direction of the resultant moment.

Example 1

Two forces 20 N and 50 N act on a body as shown in figure 6.5. If the 20 N is 0.1 m from the pivot and has clockwise effect while 50 N is 0.2 m from the pivot and has anticlockwise effect.

(a) Sketch the diagram showing all the forces and their respective distances from the pivot.

(b) Calculate the resultant moment on the body.

Solution: (a)

(b) Resultant moment = – (20 × 0.1) + (50 × 0.2)

= -2.0 + 10

= +8 Nm

Hence the resultant moment is 8 Nm acting in anti-clockwise direction.

2. Four forces, 10 N, 20 N, 40 N, act downward and 30 N acts upward on a body which is 40 cm long. The 10 N is hung at 0 cm mark, the 30 N acts at 10 cm mark, 20 N acts at 20 cm mark and 40 N acts at 40 cm mark. If the knife edge is placed at 30 cm mark, calculate the resultant moment on the body.

The sketch of the diagram showing the four forces acting on a body.

Calculations

Resultant moment = +10 × + 20 × – 30 × – 40 ×

= + 10 × + 20 × – 30 × – 40 ×

= + + +

= +3 + 2 – 6 – 4

= +5 – 10

= -5 Nm

Hence the resultant moment is 5 Nm acting in a clockwise direction.

6.2 Centre of Gravity

Every body may be regarded as being made up of a very large number of very tiny and equal particles (according to Dalton Atomic theory). Each of these particles is pulled towards the centre of the earth as shown in figure 6.6 below.

Thus, the earth’s pull on the body consists of very large number of equal parallel forces. The resultant of the forces is equal to the total force of gravity on the body and it acts through a point G called the centre of gravity.

Definition: The centre of gravity of a body is defined as the point of application of the resultant force due to the earth’s attraction on it.

Resultant force is a single force which has the effect of two or more forces acting on a body.

6.21 Methods to locate the Centre of gravity of a body

The method chosen to determine the centre of gravity of a body depends on the following factors:

• The nature and
• The shape of the body.

The shape may either be regular or irregular.

(a) A regular body

The centre of gravity of a regular body is found by using:

• Balancing method
• Intersection of diagonal method

(i) Balancing method

The centre of gravity of a long uniform object such as a metre rule may be determined by balancing method.

In this method, the metre rule may be balanced on a knife edge or hung from a loop of thread and then adjusted until it balances horizontally.

The point at the knife edge is the centre of gravity.

Note: A uniform metre rule or a uniform body is one in which the particles are uniformly distributed.

The centre of gravity of a uniform body is always at its centre.

(ii) Intersection of diagonals method

This method applies mostly to two-dimensional figures e.g. a rectangular lamina or cardboard.

Figure 6.7

The point of intersection G is the centre of gravity.

Irregular Body

The centre of gravity of an irregular body is determined by using the plumb line method.

Experiment 6.1 To determine the centre of gravity of irregular lamina

Apparatus

Retort stand/Clamp, a pin fixed to a rubber band, plumb, thin string, paper punch machine

Procedure

• Make three holes at well spaced intervals around the edge of the lamina and label them A, B and C.
• Tie one end of the string to the plumb and tie the other end to form a loop so that it can easily be suspended from the support pin.
• Using hole A, suspend the lamina and the plumb from the support pin and wait until the two come to rest.
• Mark two crosses with a thin pencil point on the plumb line, one near hole A and the other below it.
• Remove the plumb and the lamina.
• Join the two crosses on a straight line.
• Repeat procedures (c) to (f) for the remaining holes B and C.

Observation

All the three straight lines intersect at a common point G as shown in figure 6.8 below.

Figure 6.8

Conclusion: Since the centre of gravity lies on the straight lines drawn on the lamina, it must be situated at the point of intersection G.

Note:

(i) In the laboratory, the plumb can be replaced by a pendulum bob.

(ii) Not always that the centre of gravity of a body lies within the material; it may also be at a point in the air nearby.

The best examples are: tripod and a laboratory stool.

6.23 Applications of the Principle of Moments

The principle of moment is applied in the following:

1. The determination of mass or weight of a uniform metre rule.
2. The determination of mass or weight of an object.

Determination of the mass of a uniform metre rule using a standard (known) mass

• Suspend a uniform metre rule using a loop of thread from a support.
• Adjust the metre rule until it balances horizontally.
• Read and record the value at the pivot, say 50 cm.
• Suspend a standard mass, say 100 g, by means of a thread at 0 cm mark.
• Readjust the position of the metre rule until it balances horizontally again.
• Read and record the new value at the point of pivot, say 30 cm.
• Measure and record the respective perpendicular distances from 100 g mass and M from the pivot.

Consider the diagram in figure 6.9 below:

Applying the principle of moments:

Clockwise moment = Anti-Clockwise moment

M × (50 – 30) cm = 100 g × (30 – 0) cm

M × 20 cm = 100 g × 30 cm

=

M = 150 g

(b) Determination of the mass or weight of an object using a uniform metre rule and a standard mass or weight

Procedure

• Suspend a uniform metre rule using a loop of thread from a support.
• Adjust the metre rule until it balances horizontally.
• Read and record the cm mark at the pivot.
• By means of a thread, suspend a standard mass, say 200 g, and the object whose mass is to be determined on either side of the pivot.
• Readjust their distances from the pivot until the metre rule once more balances horizontally.
• Measure and record their respective perpendicular distances from the pivot.

Consider the diagram in figure 6.10 below:

Applying the principle of moments:

Clockwise moment = Anti-Clockwise moment

P × (90 – 50) cm = 200 g × (50 – 20) cm

P × 40 cm = 200 g × 30 cm

=

P = 150 g

NB: To get the weight (W) of the object, we use the formula:

W = mg

Where: g = gravity and

m = mass of the object in kg.

To get the weight, we use a standard weight directly instead of a standard mass.

Worked Examples

Note: When solving problems involving the law of moments, identify the force(s) which has/have clockwise moment(s) and anti-clockwise moment(s) so that you substitute them correctly. Experience has shown that many students tend to forget this and put clockwise moments under anti-clockwise moments and vice versa.

1. A uniform metre rule balances horizontally on a knife edge placed at 60 cm when 2 N weight is suspended at one end.
   1. With the help of a diagram, show at which end of the metre rule the 2 N weight must be suspended.
   2. Calculate:
      1. the weight,
      2. the mass of the metre rule,
      3. the reaction, R, at the knife edge. (Take g = 10 m/s²)

Solution: (a)

Since the metre rule is uniform, its weight W acts through centre, i.e. at 50 cm mark, then the 2 N weight must be suspended on the other side of the pivot so as to balance it.

(b) (i) Applying the principle of moments:

Clockwise moment = Anti-Clockwise moment

2 N × (100 – 60) cm = W × (60 – 50) cm

2 N × 40 cm = W × 10 cm

=

W = 8 N

(ii) W = 8 N, g = 10 m/s², m = ?

Using W = mg

8 = m × 10

m =

m = 0.8 kg

(iii) Applying the condition for a body in equilibrium:

(Upward force) = (Sum of downward forces)

R = W + 2 N

= 8 N + 2 N

R = 10 N

2. A uniform half-metre rule is freely pivoted at the 20 cm mark and it balances horizontally when a weight of 10 N is suspended at the 4 cm mark.
   1. Draw a sketch diagram showing all the forces acting on the metre rule.
   2. Calculate:
      1. the weight of the metre rule.
      2. the reaction at the knife edge.

Solution

(a) A sketch diagram showing the forces acting on a half-metre rule:

(b) (i) Applying the principle of moments:

Clockwise moment = Anti-Clockwise moment

W × (25 – 20) cm = 10 × (20 – 4) cm

W × 5 cm = 10 × 16 cm

=

= N

W = 32 N

(ii) Applying (Upward force) = (Sum of downward forces)

we have: R = W + 10 N

= 32 N + 10 N

R = 42 N

6.3 Stability and Equilibrium

(a) Stability

Stability, in physics and engineering, refers to the property of a body that causes it to return to its original position or motion as a result of the action of the restoring forces, or torques, once the body has been disturbed from a condition of equilibrium or steady motion.

In simple terms, we can define stability as the difficulty in causing a body to fall or topple over.

(b) Equilibrium

Equilibrium refers to the state of a body where the net force acting on the body is zero.

In the case of a stationary body, the large-scale property of the position of the body will remain unchanged over time.

6.31 Types of Equilibrium

Mechanical equilibrium can be of three kinds:

• Stable Equilibrium
• Unstable Equilibrium
• Neutral Equilibrium

Stable Equilibrium

Stable equilibrium is a state of a body in which on a slight displacement it returns to its original position.

Characteristics of a body in stable equilibrium

A body in stable equilibrium is characterized by or is associated with the following:

(i) The centre of gravity of the body is at the lowest position.

(ii) The base area is large.

Thus, on a slight displacement of a body in stable equilibrium, the centre of gravity is raised and the body returns to its original position after the displacement.

Consider a glass block resting on a table as shown in figure 6.11 below.

Explanation

The block of glass exerts force, W, (its weight) on the table and the table exerts an upward force called reaction force, R, equal to the weight of the block. Since the weight and the reaction are equal and opposite forces, the resultant force on the block is zero, so the block is stable (i.e. does not move).

When the block is tilted slightly, the two forces act as shown in figure 6.11 (b). The reaction, R, acts at the point of contact while the weight, W, still acts through G, the centre of gravity, but outside the point of contact. The result is a couple, two equal and opposite parallel forces, which tend to rotate the block in anti-clockwise direction and therefore, the block falls back to its original position.

Unstable Equilibrium

Unstable equilibrium refers to a state of a body in which on a slight displacement it does not return to its original position.

Characteristics of a body in unstable equilibrium

A body in an unstable equilibrium is associated with the following:

(i) The centre of gravity of the body is at the highest position.

(ii) The base area is small.

Thus, on a slight displacement of a body in unstable equilibrium, the centre of gravity lowers and the body does not return to its original position after the displacement; it topples over.

Explanation

In figure 6.12 (a), the block is in an unstable equilibrium. When it is tilted slightly, the two forces act as shown in figure 6.12 (b). The reaction, R, acts at the point of contact while the weight, W, still acts through G, the centre of gravity, but outside the point of contact. The result is a couple, two equal and opposite parallel forces which tend to rotate the block in clockwise direction and does not return to its original position, hence topples over.

Neutral Equilibrium

This is a state of a body in which on slight displacement its centre of gravity is neither raised nor lowered.

Characteristics of a body in neutral equilibrium

A body in a neutral equilibrium is associated with the following:

(i) The centre of gravity of the body is always at the same height and directly above the point of contact.

(ii) The area in contact is very small.

A slight displacement does not alter the position of the centre of gravity, thus the body is always at rest whichever position it is placed.

Example of a body in neutral equilibrium is any perfect sphere.

NB: Using the characteristics of a body in stable equilibrium, a body can be designed such that it is in stable equilibrium.

6.32 Making a body Stable

In engineering, the stability of a structure is increased by:

1. Making its base wide.
2. Keeping its centre of gravity as low as possible by putting more weight in the lower part than in the upper part.

Notable examples are seen in the following:

• Racing cars – which have both a low centre of gravity and a wide wheel base.
• Modern Isuzu/Scania coach buses – They have fairly wide wheel base and low centre of gravity which is achieved by packing load in boots.

Note: The old system of loading buses on the racks of buses raises the centre of gravity of the bus thus making it unstable on the road.

Self-Check 6.0

1. Figure 6.1 shows a uniform beam in equilibrium when a force R acts on it at one end. Find the weight W of the beam.

A. B. C. D.

2. The diagram in figure 6.2 shows a uniform half-meter rule suspended at point C. The mass of the rule is

A. 0.020 kg

B. 0.025 kg

C. 0.100 kg

D. 0.125 kg

3. A uniform wooden beam of weight W is pivoted at a distance 1/5 of its length from the end A and kept in equilibrium by applying a force of 30 N as shown in figure 6.3. The force exerted by the pivot on the beam is

A. 50 B. 40 C. 30 D. 20

4. A uniform rod 100 cm long pivoted at the 90 cm mark, balances horizontally when a mass of 200 g is suspended at the 100 cm mark as shown in the figure 6.3.

The mass of the rod is

A. 40 g B. 50 g C. 400 g D. 800 g

5. Which one of the following statements is true about two equal forces acting on a bar of length l, shown in the figure.

(i) The resultant force on the bar is zero.

(ii) The forces cause a rotational effect.

(iii) The forces act in opposite directions.

(iv) The forces produce different turning effects.

A. (i) only B. (i) and (ii) only C. (i), (ii) and (iii) only D. All

6. The figure below shows a uniform metre rule of mass 0.1 kg pivoted at the 80 cm mark. It balances horizontally when a mass P is hanging at the 95 cm mark. Find P.

A. 0.08 kg B. 0.2 kg C. 0.4 kg D. 1 kg

A uniform metre-rule is pivoted at its centre shown in the figure. If the rule is in equilibrium, find value of F.

A. 4 N B. 33.3 N C. 50 N D. 100 N

The shaft in an engine is subjected to two parallel but opposite forces of 500 N each as shown in the figure. The rotation is best stopped by applying

A. Two forces of 500 N acting at right angles to each other

B. A single force of 1000 N.

C. Two parallel but opposite forces of 500 N

D. A single force of 250 N.

9. The above figure shows a crank of a bicycle pedal. The force a cyclist exerts on the pedal varies from a minimum to maximum. When does the cyclist exert maximum turning effort?

A. crank makes 90⁰ with the foot push

B. crank makes 0⁰ with the foot push

C. cyclist is climbing a hill

D. cyclist is turning a corner

10. Find the weight, w, of a uniform metre rule if a force of 60 N at one end balances it as shown figure.

A. 24 N B. 40 N C. 90 N D. 100 N

SECTIONS B

11. (a) (i) Define moment of a force and state its SI unit.

(ii) State the principle of moments.

(iii) State the conditions for a body to be in equilibrium.

(b) A uniform meter ruler is pivoted at the 40 cm mark as shown in fig. 6.11. The meter ruler is in equilibrium under its weight W and a 20 N force acting at the 10 cm mark. Calculate:

(i) The weight W of the metre rule.

(ii) The reaction at the knife edge.

12. (a) What is meant by centre of gravity?

(b) (i) Describe an experiment to determine the centre of gravity of an irregular lamina.

(ii) Describe how you would measure the mass of a metre rule using a known mass and a knife-edge only.

(c) If the metre is in equilibrium when weights of 10 N, 2 N and 5 N are attached to it as shown in figure 6.12.

Calculate the: (i) Tension in the string. (ii) Normal reaction, R, at the wedge.

14. (a) State (i) the types of equilibrium.

(ii) the characteristics associated with the types of equilibrium you have stated in 5 (a) above.

• Explain giving reasons whether it is advisable to load a bus on the rack.

Describe how you would design a given structure to have a high stability.

(c) A uniform beam of weight 2.5 N is pivoted at its mid-point P, as shown in figure 6.14.

The beam remains in equilibrium when force R and S act on it. If R is 5 N, find the:

1. Value of S.

(ii) Reaction at the pivot