Force and Laws of Motion

In the previous chapter, we learned how to describe motion using position, velocity, and acceleration. Now, we'll explore what causes motion. Why do things speed up, slow down, or change direction? What is the nature of this cause?

For centuries, understanding motion was a puzzle. It seemed like objects naturally come to rest. A ball rolling on the ground eventually stops, suggesting that rest is its "natural state." Galileo Galilei and Isaac Newton changed this understanding with a new approach.

Example

Fig. 8.1: Pushing, pulling, or hitting objects change their state of motion.

In everyday life, we know it takes effort to start something moving or to stop it. We push, hit, or pull. This idea of a push or pull is the basis of force. What exactly is a force? We can't see, taste, or feel it directly, but we can see its effects. Pushing, hitting, and pulling are all ways to apply a force and cause motion.

A force can change the magnitude of velocity of an object (making it move faster or slower) or to change its direction of motion. A force can also change the shape and size of objects.

Example

Fig. 8.2: (a) A spring expands on application of force; (b) A spherical rubber ball becomes oblong as we apply force on it.

8.1 Balanced and Unbalanced Forces

Example

Fig. 8.3 shows a wooden block on a horizontal table. Two strings X and Y are tied to the two opposite faces of the block as shown. If we apply a force by pulling the string X , the block begins to move to the right. Similarly, if we pull the string Y , the block moves to the left. But, if the block is pulled from both the sides with equal forces, the block will not move.

When equal forces are applied from opposite directions, they are called balanced forces. Balanced forces do not change the state of rest or motion of an object.

Now, imagine applying unequal forces to the block from opposite sides. The block will move in the direction of the larger force. These are unbalanced forces. An unbalanced force acting on an object brings it into motion.

Example

Fig. 8.4: What happens when some children try to push a box on a rough floor? If they push the box with a small force, the box does not move because of friction acting in a direction opposite to the push [Fig. 8.4(a)]. This friction force arises between two surfaces in contact; in this case, between the bottom of the box and floor's rough surface. It balances the pushing force and therefore the box does not move. In Fig. 8.4(b), the children push the box harder but the box still does not move. This is because the friction force still balances the pushing force. If the children push the box harder still, the pushing force becomes bigger than the friction force [Fig. 8.4(c)]. There is an unbalanced force. So the box starts moving.

Friction is a force that opposes motion between two surfaces in contact. When the pushing force is small, friction balances it, and the box doesn't move. Only when the pushing force overcomes friction does the box start to move.

When we stop pedaling a bicycle, it slows down because of friction. To keep it moving, we need to keep pedaling to overcome friction.

It might seem like objects need a continuous force to keep moving. However, an object moves with a uniform velocity when the forces acting on it are balanced, and there is no net external force. An unbalanced force is needed to change its speed or direction. The change continues as long as the unbalanced force is applied. If the unbalanced force is removed, the object continues moving at the velocity it had already acquired.

8.2 First Law of Motion

Galileo observed objects moving on inclined planes. He realized that objects move at a constant speed when no force acts on them.

Example

Fig. 8.5: (a) the downward motion; (b) the upward motion of a marble on an inclined plane; and (c) on a double inclined plane.

When a marble rolls down an inclined plane, its velocity increases. This is because of gravity, an unbalanced force. When the marble rolls up, its velocity decreases.

Imagine a marble on a frictionless plane inclined on both sides. If released from the left, it rolls down and up the other side to the same height. If the angle of the right side is decreased, the marble travels further to reach the same height. If the right side is horizontal, the marble would continue forever, trying to reach the original height. In this case, the unbalanced forces are zero.

This suggests that an unbalanced force is required to change the motion, but no force is needed to sustain uniform motion. In reality, friction makes it difficult to achieve zero unbalanced force.

Newton studied Galileo's ideas and presented three fundamental laws of motion. The first law of motion states:

An object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force.

In other words, objects resist changes in their motion. This tendency is called inertia. The first law of motion is also known as the law of inertia.

Experiences in a car illustrate this law. We stay at rest until the car starts moving. When the brakes are applied, we tend to continue moving forward due to inertia. Safety belts apply a force to slow our forward motion.

Similarly, when a bus starts suddenly, we fall backward. Our feet move with the bus, but the rest of our body resists the motion due to inertia.

When a car turns sharply, we are thrown to the side. We tend to continue in a straight line, and the car's force changes our direction, making us slip to the side.

The following activities illustrate inertia:

Activity 8.1

Aim

To demonstrate inertia using carom coins.

Materials Required

Carom coins
Striker
Table

Procedure

Make a pile of carom coins on a table.
Use another coin or the striker to hit the bottom coin horizontally.

Observation

If the hit is strong enough, the bottom coin moves out quickly, and the other coins fall vertically onto the table.

Conclusion

The inertia of the coins makes them resist the horizontal force, causing them to fall straight down when the bottom coin is removed.

Example

Fig. 8.6: Only the carom coin at the bottom of a pile is removed when a fast moving carom coin (or striker) hits it.

Activity 8.2

Aim

To demonstrate inertia using a coin, card and tumbler.

Materials Required

Five-rupee coin
Stiff card
Empty glass tumbler
Table

Procedure

Set the coin on the card covering the tumbler.
Flick the card sharply with a finger.

Observation

If done fast, the card shoots away, and the coin falls into the glass.

Conclusion

The coin's inertia resists the horizontal motion, causing it to fall into the tumbler when the card is removed.

Example

Fig. 8.7: When the card is flicked with the finger the coin placed over it falls in the tumbler.

Activity 8.3

Aim

To demonstrate inertia using a water filled tumbler on a tray.

Materials Required

Water-filled tumbler
Tray

Procedure

Place a water-filled tumbler on a tray.
Hold the tray and turn around as fast as you can.

Observation

The water spills.

Conclusion

The water spills because it resists the change in motion due to its inertia.

Note

A groove in a saucer prevents a tea cup from toppling over during sudden jerks.

8.3 Inertia and Mass

The activities above show that objects resist changes in their motion. This resistance is inertia. Do all bodies have the same inertia?

It's easier to push an empty box than a full one. A kicked football flies away, but a stone of the same size barely moves, and might even injure your foot! A smaller force is needed to perform Activity 8.2 with a one-rupee coin than with a five-rupee coin. A small cart can easily pick up a large velocity, but the same force will produce a negligible change in the motion of a train.

The train has more inertia than the cart. Heavier, more massive objects have larger inertia. Mass measures an object's inertia.

Inertia is the natural tendency of an object to resist a change in its state of motion or of rest. The mass of an object is a measure of its inertia.

8.4 Second Law of Motion

The first law tells us that an unbalanced force changes an object's velocity, causing acceleration. Now, let's see how acceleration depends on force and how we measure force.

A table tennis ball hitting a player doesn't hurt, but a fast cricket ball can. A parked truck is harmless, but a moving truck, even at low speeds, can be deadly. A bullet fired from a gun can kill.

These observations suggest that the impact depends on mass and velocity. To accelerate an object, a greater force is needed for a greater velocity. There's a quantity that combines mass and velocity, called momentum, introduced by Newton.

The momentum $p$ of an object is defined as the product of its mass $m$ and velocity $v$ . That is,

$p = m v$

Momentum has both direction and magnitude. Its direction is the same as the velocity $v$ . The SI unit of momentum is kilogram-metre per second ( $\text{kg m s}^{-1}$ ). Since an unbalanced force changes velocity, it also changes momentum.

Imagine pushing a car with a dead battery to get it started. A sudden push might not work, but a continuous push over time gradually accelerates the car. The change in momentum depends on both the force and the time it's applied.

The force needed to change an object's momentum depends on the rate at which the momentum is changed.

The second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.

8.4.1 Mathematical formulation of SECOND LAW OF MOTION

Consider an object of mass $m$ moving along a straight line with initial velocity $u$ . A constant force $F$ is applied, uniformly accelerating it to velocity $v$ in time $t$ . The initial and final momentum are $p_1 = m u$ and $p_2 = m v$ , respectively.

The change in momentum is:

$\propto p_2 - p_1$

$\propto m v - m u$

$\propto m \times (v - u)$

The rate of change of momentum is:

$\propto \frac{m \times (v - u)}{t}$

Or, the applied force,

$F \propto \frac{m \times (v - u)}{t}$

$F = \frac{k m \times (v - u)}{t}$

$= k m a$

Here $a = [(v - u) / t]$ is the acceleration, the rate of change of velocity. The quantity $k$ is a constant of proportionality. The SI units of mass and acceleration are $\text{kg}$ and $\text{m s}^{-2}$ , respectively. The unit of force is chosen so that $k$ becomes one. One unit of force produces an acceleration of $1 \text{ m s}^{-2}$ in an object of $1 \text{ kg}$ mass. That is,

$1 \text{ unit of force} = k \times (1 \text{ kg}) \times (1 \text{ m s}^{-2})$

Thus, $k = 1$ . From the equation above:

$F = m a$

The unit of force is $\text{kg m s}^{-2}$ or newton, symbol N. The second law gives us a way to measure force as the product of mass and acceleration.

The second law is seen in everyday life. When catching a cricket ball, a fielder pulls their hands back. This increases the time during which the ball's velocity decreases to zero, reducing the acceleration and the impact. Stopping the ball suddenly would result in a large force and could hurt the fielder's palm. High jumpers land on cushioned beds to increase the time of their fall, decreasing the force.

Example

Fig. 8.8: A fielder pulls his hands gradually with the moving ball while holding a catch.

The first law can be derived from the second law:

$F = m a$

$F = \frac{m(v - u)}{t}$

$F t = m v - m u$

When $F = 0$ , $v = u$ for any time $t$ . The object continues moving with uniform velocity $u$ . If $u$ is zero, $v$ is also zero, and the object remains at rest.

Example

Example 8.1 A constant force acts on an object of mass 5 kg for a duration of 2 s . It increases the object's velocity from

3 \text{ m s}^{-1}

7 \text{ m s}^{-1}

. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s , what would be the final velocity of the object?

Given

Mass, $m = 5 \text{ kg}$
Initial velocity, $u = 3 \text{ m s}^{-1}$
Final velocity, $v = 7 \text{ m s}^{-1}$
Time, $t = 2 \text{ s}$

To Find

(i) Magnitude of the applied force, $F$ (ii) Final velocity if the force is applied for 5 s, $v$

Formula

$F = \frac{m(v - u)}{t}$

$v = u + \frac{F t}{m}$

Solution

(i) Calculate the magnitude of the force

Substitute the given values into the formula

$F = \frac{5 \text{ kg} \times (7 \text{ m s}^{-1} - 3 \text{ m s}^{-1})}{2 \text{ s}}$

$F = \frac{5 \text{ kg} \times 4 \text{ m s}^{-1}}{2 \text{ s}} = 10 \text{ N}$

Answer for part (i) = $10 \text{ N}$

(ii) Calculate the final velocity after 5 seconds

Substitute the given values into the formula

$v = 3 \text{ m s}^{-1} + \frac{10 \text{ N} \times 5 \text{ s}}{5 \text{ kg}}$

$v = 3 \text{ m s}^{-1} + \frac{50 \text{ kg m s}^{-1}}{5 \text{ kg}} = 3 \text{ m s}^{-1} + 10 \text{ m s}^{-1} = 13 \text{ m s}^{-1}$

Answer for part (ii) = $13 \text{ m s}^{-1}$

Example

Example 8.2 Which would require a greater force - accelerating a 2 kg mass at

5 \text{ m s}^{-2}

or a 4 kg mass at

2 \text{ m s}^{-2}

Given

Mass 1, $m_1 = 2 \text{ kg}$
Acceleration 1, $a_1 = 5 \text{ m s}^{-2}$
Mass 2, $m_2 = 4 \text{ kg}$
Acceleration 2, $a_2 = 2 \text{ m s}^{-2}$

To Find

Which force is greater, $F_1$ or $F_2$

Formula

$F = m a$

Solution

Calculate the force for the first mass

$F_1 = m_1 a_1 = 2 \text{ kg} \times 5 \text{ m s}^{-2} = 10 \text{ N}$

Calculate the force for the second mass

$F_2 = m_2 a_2 = 4 \text{ kg} \times 2 \text{ m s}^{-2} = 8 \text{ N}$

Since $F_1 > F_2$ , accelerating a 2 kg mass at $5 \text{ m s}^{-2}$ would require a greater force.

Final Answer Accelerating a 2 kg mass at $5 \text{ m s}^{-2}$ would require a greater force.

Example

Example 8.3 A motorcar is moving with a velocity of

108 \text{ km/h}

and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg .

Given

Initial velocity, $u = 108 \text{ km/h} = 30 \text{ m s}^{-1}$
Final velocity, $v = 0 \text{ m s}^{-1}$
Time, $t = 4 \text{ s}$
Mass, $m = 1000 \text{ kg}$

To Find

Force exerted by the brakes, $F$

Formula

$F = \frac{m(v - u)}{t}$

Solution

Substitute the given values into the formula

$F = \frac{1000 \text{ kg} \times (0 \text{ m s}^{-1} - 30 \text{ m s}^{-1})}{4 \text{ s}}$

$F = \frac{1000 \text{ kg} \times (-30 \text{ m s}^{-1})}{4 \text{ s}} = -7500 \text{ N}$

The negative sign indicates that the force is opposite to the direction of motion.

Final Answer The force exerted by the brakes is $-7500 \text{ N}$ .

Example

Example 8.4 A force of 5 N gives a mass

m_1

, an acceleration of

10 \text{ m s}^{-2}

and a mass

m_2

, an acceleration of

20 \text{ m s}^{-2}

. What acceleration would it give if both the masses were tied together?

Given

Force, $F = 5 \text{ N}$
Acceleration 1, $a_1 = 10 \text{ m s}^{-2}$
Acceleration 2, $a_2 = 20 \text{ m s}^{-2}$

To Find

Acceleration, $a$ , if both masses are tied together.

Formula

$m = \frac{F}{a}$

$F = ma$

Solution

First, find the individual masses

$m_1 = \frac{F}{a_1} = \frac{5 \text{ N}}{10 \text{ m s}^{-2}} = 0.50 \text{ kg}$

$m_2 = \frac{F}{a_2} = \frac{5 \text{ N}}{20 \text{ m s}^{-2}} = 0.25 \text{ kg}$

Calculate the total mass

$m = m_1 + m_2 = 0.50 \text{ kg} + 0.25 \text{ kg} = 0.75 \text{ kg}$

Calculate the acceleration of the combined mass

$a = \frac{F}{m} = \frac{5 \text{ N}}{0.75 \text{ kg}} = 6.67 \text{ m s}^{-2}$

Final Answer The acceleration of the combined mass is $6.67 \text{ m s}^{-2}$ .

Example

Example 8.5 The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in Fig. 8.9. How much force does the table exert on the ball to bring it to rest?

Example

Fig. 8.9

Given

Mass, $m = 20 \text{ g} = 0.02 \text{ kg}$
Initial velocity, $u = 20 \text{ cm s}^{-1} = 0.2 \text{ m s}^{-1}$
Final velocity, $v = 0 \text{ cm s}^{-1} = 0 \text{ m s}^{-1}$
Time, $t = 10 \text{ s}$

To Find

Force exerted by the table, $F$

Formula

$a = \frac{v - u}{t}$

$F = m a$

Solution

First, calculate the acceleration

$a = \frac{0 \text{ m s}^{-1} - 0.2 \text{ m s}^{-1}}{10 \text{ s}} = -0.02 \text{ m s}^{-2}$

Now, calculate the force

$F = (0.02 \text{ kg}) \times (-0.02 \text{ m s}^{-2}) = -0.0004 \text{ N}$

The negative sign indicates that the force is opposite to the direction of motion.

Final Answer The force exerted by the table is $-0.0004 \text{ N}$ .

8.5 Third Law of Motion

The first two laws tell us how force changes motion and how to measure force. The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object.

In football, when players collide while kicking the ball, both feel hurt because each applies a force to the other. There is a pair of forces, not just one. The two opposing forces are known as action and reaction forces.

Example

Fig. 8.10: Action and reaction forces are equal and opposite.

Consider two spring balances connected together. The fixed end of balance B is attached to a wall. When a force is applied to the free end of spring balance A, both balances show the same readings. The force exerted by spring balance A on balance B is equal but opposite to the force exerted by balance B on balance A. Either force can be called the action, and the other the reaction.

The third law can be stated as: to every action there is an equal and opposite reaction. Remember that action and reaction always act on two different objects simultaneously.

When you walk, you push the road backward. The road exerts an equal and opposite force on your feet, making you move forward.

Action and reaction forces are always equal in magnitude, but they may not produce equal accelerations because each force acts on a different object with a different mass.

When a gun is fired, it exerts a forward force on the bullet. The bullet exerts an equal and opposite force on the gun, causing it to recoil. The gun's mass is much greater than the bullet's, so the gun's acceleration is much less.

Example

Fig. 8.11: A forward force on the bullet and recoil of the gun.

When a sailor jumps out of a boat, the force on the boat moves it backward.

Example

Fig. 8.12: As the sailor jumps in forward direction, the boat moves backwards.

Activity 8.4

Aim

To demonstrate the third law of motion.

Materials Required

Two carts
Sandbag or other heavy object

Procedure

Have two children stand on separate carts.
Give them a sandbag and ask them to play catch.
Observe the motion of the carts.

Observation

Each child experiences an instantaneous force when throwing the sandbag.

Conclusion

This demonstrates the action-reaction forces. When one child throws the bag, they exert a force on it (action), and the bag exerts an equal and opposite force on them (reaction), causing their cart to move.

Example

Fig. 8.13

Note

Placing two children on one cart and one on another cart demonstrates the second law of motion, as the same force will produce different accelerations due to the different masses.

Note

The carts can be constructed from plywood with ball-bearing wheels.

What you have learnt

First law of motion An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.
The natural tendency of objects to resist a change in their state of rest or of uniform motion is called inertia.
The mass of an object is a measure of its inertia. Its SI unit is kilogram (kg).
Force of friction always opposes motion of objects.
Second law of motion The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
The SI unit of force is $\text{kg m s}^{-2}$ . This is also known as newton and represented by the symbol N. A force of one newton produces an acceleration of $1 \text{ m s}^{-2}$ on an object of mass 1 kg .
The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. Its SI unit is $\text{kg m s}^{-1}$ .
Third law of motion To every action, there is an equal and opposite reaction and they act on two different bodies.

Chapter Notes

Force and Laws of Motion

8.1 Balanced and Unbalanced Forces

8.2 First Law of Motion

Activity 8.1

Aim

Materials Required

Procedure

Observation

Conclusion

Activity 8.2

Aim

Materials Required

Procedure

Observation

Conclusion

Activity 8.3

Aim

Materials Required

Procedure

Observation

Conclusion

8.3 Inertia and Mass

8.4 Second Law of Motion

8.4.1 Mathematical formulation of SECOND LAW OF MOTION

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

8.5 Third Law of Motion

Activity 8.4

Aim

Materials Required

Procedure

Observation

Conclusion

What you have learnt

Congratulations! You've completed this chapter