TURNING EFFECT OF FORCES, CENTRE OF GRAVITY AND STABILITY

Keywords

• centre of gravity
• centre of mass equilibrium
• force
• pivot
• stable, unstable and neutral equilibrium.
• turning effect and torque

By the end of this chapter, you will be able to:

(a) Understand the turning effect of forces and its applications.

(b) Understand and apply the concept of centre of gravity.

INTRODUCTION

What happens when children are playing using a seesaw? Why is it easier to unscrew a nut using a longer spanner than a shorter one? Why is walking more stable?

 To understand these, you need to understand the concept of turning effects or moments of forces. In this chapter, you will be able to investigate the relation between the turning effect of forces and the stability of bodies.

TURNING EFFECT OF A FORCE

 The turning effect of a force is called the moment of a force or torque (ꚍ). It is defined as the product of the force and the perpendicular distance of the line of action of the force from the turning point (pivot). From this definition, we can deduce the formula below; r – force x perpendicular distance. That is, ꚍ = FXd

From the formula, we derive the SI units for the moment of a force to be Nm. Pivot Turning effect produced by force

 Figure 2.1: A spanner Force here

The turning effects can be demonstrated in many ways, and one of these is seen in Figure 2.1 as a screw driver is being turned. A force is applied at one point of the screw driver at a certain distance from the screw. This turns the screw in a clockwise direction.

Example 2.1

Richie applies a force of 12.0N to open their gate at home for his mother to drive out. The gate handle is at 3.5m from the hinges. Calculate the torque created by his force.

 Data:

Force is 12.0N and distance is 3.5m

But ꚍ = F.d

Therefore, ꚍ =- 12.0 x 3.5=42.0Nm

Example 2.2

While opening a screw, the torque used is 150.9Nm when the force used is 38.2N. What is the perpendicular distance of the force from the turning point?

 Since the question requires us to determine the distance, we shall use

PRINCIPLE OF MOMENTS

 The principle of moments states that when an object is in equilibrium, the sum of anticlockwise moments about any point equals the sum of clockwise moments about the same point. This means that net torque on a system in equilibrium at any point is zero.

Activity 2.1 verifying the principle of moments using a suspended metre rule and attached weights

What you need

• A uniform metre rule, retort stand, boss and clamp, two 100g mass hangers and 12, 100g slotted masses, a G – clamp, a string (or thread) Support

Figure 2.2: Verifying principle of moments using a metre rule

What to do

1. Suspend the metre rule at the 50cm mark so that it is balanced horizontally. The ruler is said to be in equilibrium.
2. Suspend a mass, m₁, from one side of the ruler at distance, d₁, from the pivot. Read the distance d₁, in cm, from m₁, to the pivot. Record in a suitable table. Record the value of mass m₁, in kg in the table, too.
3. Suspend a second mass, m₂, from the other side of the pivot. Carefully move this mass backwards and forwards until the ruler is once more balanced horizontally. Read the distance d₂, in cm from the mass m₂, to the pivot. Record d₂ , in cm, in the table, along with the mass m₂ , in kg.
4. Repeat several times using different masses and distances.
5. Calculate the turning forces, F₁ and F₂, using W = mg.
6. Calculate the clockwise and anticlockwise moments.
7. Tabulate your results in the table of the format indicated below:

Safety

• Place an obstacle, such as a stool, to keep one’s feet from beneath the metre rule, to make sure the mass hangers do not fall on someone’s foot.
• Safety glasses should be worn in case the metre rule swings and hits someone in the eye.

From the above activity performed in groups, state the principle of moments.

 Assessment 2.1

 In groups, using materials in the laboratory carry out research on how you can use the principle of moments and equilibrium to determine an unknown mass using a metre rule of known mass.

Example 2.3

 The diagram shown below is in equilibrium. Calculate;

1. the value of force (F)
2. the normal reaction (R) at the pivot

Figure 2.3 Equilibrium position (1) (ii)

Solution

At equilibrium, the principle of moments applies, because there is no rotation. So, the sum of clockwise moments about the pivot is equal to the anticlockwise moments about the same pivot.

1. F₁xd₁₌-F₂xd₂, So, 20 x 12 -=F x 8a and this gives F = 30N
2. Since the system is now moving in any way in the vertical plane , that means the summation of all forces going vertically upwards ( normal reaction R ) is equal to the summation of all forces in the direction downwards ( 20 + F )

 R = 20 + F = 20 + 30 = 50N

Example 2.4

Figure 2.4: Normal reactions at supports A and B

 Figure 2.4 shows two cars of masses 4,500kg and 1,100kg resting on a uniform simple beam of mass 20,000kg at distances 4.0m and 6.0m from the opposite ends, respectively, Calculate the normal reactions at the supports A and B.

 Solution:

Let us remember that since the beam is uniform, the centre of gravity is in the middle (8m from either end) and that is where the weight concentrates as 200,000N.

 Also, note that while calculating moments, we must convert the masses of each into weight by multiplying with the value of acceleration due to gravity.

 Let us use symbols and to represent the reactions at the ends A and B respectively , and taking moments about A , the forces which pass through this point like , have no turning effect about that point , therefore , using the principle of moments , we obtain the equation :

Hint

Weight = Mass x Acceleration due to gravity

W = M x 10ms^–¹

 R_Bx 16=45,000 (4) +200,000 (8) +11,000 (10)

So, R_B = 118,125N

To calculate the normal reaction at A, we can take moment about point B, and for this case, the force has no moment about point B, therefore

  R_A (16) -11,000 (6) +200,000 (8) +45,000 (12)

 So, R_A = 137,875N

The alternative way to calculate the second normal reaction is to use the other condition of equilibrium which gives the equation:

 R_A + R_B= 45,000 + 11,000 + 200,000

 R_A + 118,125 = 256,000

R_A = 137,875N

Exercise 2.1

1. A girl weighing 200N sits on a seesaw and is balanced by a boy weighing 300N. Where should the boy sit to make the seesaw balance?
2. . Which of these sportsmen most needs to have a low centre of gravity? A high jumper, a snooker player, a weightlifter and a hurdler. Explain your answer
3.  When forcemeter A was used on its own on the spanner to loosen a nut, it read 200N. Forcemeter B was then used on its own to loosen the nut and it read 300N

What was the distance X in the diagram?

4. Kawa is an engineer who wants to use a ladder to climb the wall to replace a faulty electric lamp. As he places the ladder on the vertical wall and against a horizontal floor, his friend Moana advises him to make sure both the points of contact of the ladder with the wall and the floor are very rough. Why do you think Moana makes such a statement? How do you think such advice applies in moments and equilibrium?

2.1 Centre of Gravity

Why is it that some objects are stable while others are unstable and can fall off easily?

 Bodies are stable and balance off at a particular point called centre of gravity.

The centre of gravity is a point from which the total weight of a body or system may be considered to act.

Activity 2.2 determining the centre of gravity of a metre rule

 Key Question: How can you determine the centre of gravity of a metre rule?

What you need

• A metre rule and a knife edge.

What to do

 (a) Working in groups and using a metre rule and knife edge (pivot), balance the metre rule horizontally on the knife edge as shown below:

Figure 2.5: Balancing metre rule

(b) Read and record the position on the metre rule where it balances and name it O.

1. Discuss in groups and come up with a name you can give position O.
2. Share with other groups to see the values obtained as O, for the other metre rule.
3. Have you noticed any difference in the values? If yes, discuss with the other groups why you think it is so. (iv)
4. What do you think are the best names that describe a metre rule whose position is at 50.0cm mark and the one whose position O is not at 50.0cm mark?

Exercise 2.2

In pairs, investigate and make a report on how the centre of gravity has implications for the design of objects or structures. Present your findings to your Physics teacher and to the whole class.

2.2 Centre of Gravity of Regular Shapes

 It is the average position of all the parts of the system where the mass of the body balances.

For simple rigid objects with uniform density, the centre of gravity is located at the centroid. For example, the centre of gravity of a uniform disc shape would be at its centre. Sometimes the centre of mass does not fall anywhere on the object. The centre of gravity of a ring, for example, is located at its centre, where there is no material, as illustrated below.

Figure 2.6: Centre of gravity of different shapes

Activity 2.3 locating the centre of gravity of regular object like a piece of paper

What you need

• A piece of paper (A4), a ruler, a pen or a pencil.

What to do

(a) Place your paper on the table.

(b) Draw diagonals from one point of the paper to another and also from one end to another.

 (c) Where do those lines meet?

(d) Basing on your observations, where do you think is the centre of gravity of your paper?

(e) (Note down your observations.

(f) Share your observations with the rest of the members of your class

From the activity, you will realize that the lines will meet at the centre of the paper, hence being the centre of gravity of this paper.

Example 2.3

 Determine the centre of gravity of a uniform solid of dimensions 10cm by 6cm by 12cm.

Since the solid is uniform, the centre of gravity is exactly in the middle at a point which is half each dimension.

Experiment: Determining centre of gravity of an irregular shape

What you need

• cardboard , pairs of scissors , a pin or a 20cm piece of thread , a pen , a small weight to hang from the thread ( eg . a washer , a key or a small stone ) .

 What to do

1. Attach a small fixed weight to the end of your thread. This is your plumb line it follows the direction of the gravitational pull
2. Either using chalk or a pen, cover the thread with chalk or trace the line using a pen.
3. Hang the plumb line on the pin; Pierce the pin anywhere along the edge of the shape so that your shape is free to swing.
4. Hold the pin and wait until the shape and the plumb line have settled. Mark where the thread crosses the shape by tracing the thread’s path using a pen.
5. Remove the pin and plumb line from the shape. Pierce them through another point along the shape’s edge. It should not be too close to the last hole you made.
6. Repeat steps (a) to (e) three more times.
7. Mark the spot where the lines intersect. Make a general comment about this point.
8. Discuss / present your findings to the rest of the class.

Figure 2.7: An irregular lamina

Types of equilibrium

When a body or system is in equilibrium, it may be in either stable unstable or neutral equilibrium.

Activity 2.4 Demonstrating types of equilibrium

What you need

• Two cones (you can use funnels from the school laboratory) and a ball.

What to do

1. Explain how the object talked about in each case behaves when displaced.
2. Basing on your explanation in a) above, what type of equilibrium is shown by each body?
3. Name any other two examples of each of the above demonstrated states of equilibrium as observed in your day – to – day life.

Assessment 2.2

 As stated, stable, unstable and neutral equilibrium are the three types of equilibrium.

(a) In groups, with relevant diagrams, using the internet and other textbooks, carry out research, write some notes and explain each of the three types of equilibrium.

(b) In each case, suggest any two day – to – day examples where it is practically experienced.

c) Present the rest of the findings to the class.

2.3 Equilibrium of a Mechanical System

Figure 2.8: Tall buildings

 The diagram above shows illustrations of some of the tallest structures in the world that are supported by metals and reinforced concrete to keep them in total equilibrium even if there is an earthquake as earthquakes are more severe in Asia .

 Did you know that equilibrium is the state of a body at rest or in uniform motion when the resultant of all forces acting on the body is zero?

 CONDITIONS FOR A MECHANICAL SYSTEM TO BE IN EQUILIBRIUM

• The sum of all external forces acting on the body is zero.
• The sum of all external torques from external forces is zero.

Note

These two conditions must be simultaneously satisfied in equilibrium. If one of them is not satisfied, the body is not in equilibrium.

 Explanation

The condition “net force is zero” must be true for both static equilibrium where the object’s velocity is zero, and for dynamic equilibrium, where the object is moving at a constant velocity.

 Figure 2.9 shows a motionless person in static equilibrium. The forces acting on him add up to zero. Both forces are vertical in this case. The normal reaction acting on the person is denoted by the symbol N and his weight is denoted by W.

 In the diagram below, the normal reaction is denoted by the symbol N and weight by W.

 Figure 2.9: Free body

In figure 1.10, the car is in dynamic equilibrium because it is moving at constant velocity. There are horizontal and vertical forces, but the net external force in any direction is zero. The applied force between the tyres and the road is balanced by air friction, and the weight of the car is supported by the normal forces, here shown to be equal for all four tyres.

 The second condition necessary to achieve equilibrium involves avoiding accelerated rotation (maintaining constant angular velocity). A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it

Figure 2.10: Moving at constant velocity

Chapter Summary

• In this chapter, you have learnt that:
• The centre of gravity is a point from which the weight of a body or system may be considered to act. Uniform gravity is the same as the centre of mass
• The moment of a force is also called torque and is calculated as the product of the force and the perpendicular distance of the force from the turning point.
• For a mechanical system to be in total equilibrium the net torque must be equal to zero and the net force must be equal to zero.
• Stability is determined greatly by the position of the centre of gravity and the size of the base. In other words , structures with a wide base may be much stable than the ones with a narrow base , and in order to make a structure more stable , we must make its base heavier , thereby lowering the centre of gravity