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How Forces Affect Motion Class 9 Science Notes

How Forces Affect Motion

In our daily lives, we see things move, stop, speed up, or change direction. This chapter explores the fundamental question: what causes these changes in motion? The answer lies in the concept of force. We will investigate what forces are and how they are governed by Newton's three laws of motion.

The Concept of Force

A force is a push or a pull on an object. Applying a force can have several effects:

It can make a stationary object start moving.
It can change the speed of a moving object.
It can change the direction of a moving object.
It can change the shape of an object (like squeezing a lemon).

Force is a physical quantity that requires both a magnitude (how strong the force is) and a direction. This is similar to other quantities you've studied, like velocity and acceleration.

The SI unit of force is the newton, and its symbol is N.

Note

The effect of a force changes if its magnitude, its direction, or both change.

Measuring the magnitude of a force

The magnitude of a force can be measured using a spring balance. You may have used one to measure the weight of an object, which is the gravitational force the Earth exerts on that object. A spring balance can measure any type of pulling force. When you pull on its hook, the scale indicates the magnitude of the force you are applying.

Balanced and Unbalanced Forces

Rarely does only a single force act on an object. Usually, multiple forces are at play. The overall effect on the object's motion depends on the combination of all forces acting on it.

Balanced Forces: When two forces are equal in magnitude but act in opposite directions on an object, they are called balanced forces. Balanced forces cancel each other out, and there is no change in the object's state of motion. The object will either remain at rest or continue to move at a constant velocity.
Unbalanced Forces: When the forces acting on an object are not equal and opposite, they are unbalanced. Unbalanced forces result in a net force, which is a non-zero total force that causes a change in the object's motion (i.e., it causes acceleration).

Calculating Net Force:

Forces in the same direction: The net force is the sum of the individual forces, acting in the same direction.
Forces in opposite directions: The net force is the difference between the magnitudes of the two forces, acting in the direction of the larger force.

Example

Example Two forces of 10 N and 6 N are acting on a block lying on the table as shown. What is the magnitude and the direction of the net force acting on the block in each case? (a) Both forces act towards the right. (b) The 10 N force acts right, and the 6 N force acts left. (c) The 10 N force acts left, and the 6 N force acts right.

Given

Force 1, $F_1 = 10 \text{ N}$
Force 2, $F_2 = 6 \text{ N}$

To Find

The magnitude and direction of the net force in three different cases.

Solution

(a) Both forces act towards the right

The forces are in the same direction, so we add their magnitudes. Net force $= 10 \text{ N} + 6 \text{ N} = 16 \text{ N}$

Answer for part (a) = $16 \text{ N}$ , acting towards the right side.

(b) 10 N force to the right, 6 N force to the left

The forces are in opposite directions, so we subtract the smaller magnitude from the larger one. The direction is that of the larger force. Net force $= 10 \text{ N} - 6 \text{ N} = 4 \text{ N}$

Answer for part (b) = $4 \text{ N}$ , acting towards the right side.

(c) 10 N force to the left, 6 N force to the right

The forces are in opposite directions. The larger force is to the left. Net force $= 10 \text{ N} - 6 \text{ N} = 4 \text{ N}$

Answer for part (c) = $4 \text{ N}$ , acting towards the left side.

The Force of Friction: Often Overlooked but Always Present

Have you ever tried to push a heavy box and found it didn't move until you pushed hard enough? This is because of the force of friction, which is a force that arises between surfaces in contact and acts in the direction opposite to the motion or attempted motion.

To start an object moving, the applied force must be greater than the force of friction.
Once an object is moving, if you stop applying a force, it is the force of friction that slows it down and eventually brings it to a stop.
To keep an object moving at a constant velocity, a continuous force must be applied to counteract the force of friction.

The force of friction depends on the nature of the surfaces in contact. Smoother surfaces generally have less friction than rougher surfaces.

Note

Multiple forces may act on an object, but its motion depends only on the net force.

Newton's First Law of Motion

For centuries, it was believed that a force was necessary to keep an object moving. Galileo Galilei, through thought experiments, argued that if all opposing forces like friction were removed, an object in motion would continue to move indefinitely. Isaac Newton built on this idea, which he called inertia—the tendency of an object to resist any change in its state of rest or of uniform motion.

This concept is formalized in Newton's first law of motion:

An object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless a net force acts upon the object.

In simpler terms, if the net force on an object is zero, its acceleration is zero. This means it won't start moving if it's at rest, and it won't change its speed or direction if it's already moving.

Note

Constant velocity means that both the speed and the direction of motion do not change. An object moving with constant velocity travels in a straight line at a constant speed.

Example

Example A person is exerting a force on a moving box in the forward direction which is equal to the force of friction acting between the bottom surface of the box and the floor. Will the box continue moving or will it come to rest after some time?

Solution

The force exerted by the person is in the forward direction. The force of friction acts in the backward direction. Since the two forces are equal in magnitude and opposite in direction, they are balanced.

Therefore, the net force acting on the box is zero. According to Newton's first law of motion, an object in motion will continue to move with a constant velocity if the net force is zero.

Final Answer The box will continue moving with a constant velocity.

Example

Example Draw (i) position-time, and (ii) velocity-time graphs for an object on which no net force is acting.

Solution

When no net force acts on an object, there are two possibilities: the object is at rest, or it is moving with a constant velocity.

Case 1: Object is at rest (i) Position-time graph: Since the position does not change, the graph is a horizontal line. (ii) Velocity-time graph: Since the velocity is always zero, the graph is a horizontal line on the time axis.

Case 2: Object is moving with a constant velocity (i) Position-time graph: Since the object covers equal distances in equal time intervals, the graph is a straight line with a constant slope. (ii) Velocity-time graph: Since the velocity does not change, the graph is a horizontal line above the time axis.

Newton's Second Law of Motion

Newton's first law describes what happens when the net force is zero. But what happens when there is a net force? A net force causes an object to accelerate (change its velocity).

The relationship between force, mass, and acceleration is described by Newton's second law of motion:

When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is proportional to the magnitude of the net force and is inversely proportional to the mass of the object.

This gives us two key relationships:

For the same mass, a larger net force produces a larger acceleration.
For the same net force, a larger mass experiences a smaller acceleration.

This relationship is mathematically expressed as: $a = \frac{F}{m} \quad \text{or} \quad F = ma$ Where:

$F$ is the net force in newtons (N)
$m$ is the mass in kilograms (kg)
$a$ is the acceleration in $\text{m s}^{-2}$

Example

To understand how Newton's second law applies in real life, consider a cricket fielder catching a fast-moving ball. By pulling their hands backward with the ball, the fielder increases the time it takes for the ball's velocity to become zero. Since the change in velocity is the same, increasing the time (

t

) reduces the acceleration (

a

). A smaller acceleration requires a smaller force (

F=ma

) from the fielder's hands, which prevents injury. The same principle is used in vehicle airbags and high-jump landing mats.

Example

Example A weight lifter is holding a barbell with a mass of 10 kg fixed on each side of the bar. The mass of the bar itself is 10 kg. How much force is she applying to keep the barbell steady? (Use

g = 9.8 \text{ m s}^{-2}

)

Given

Mass of weights on each side = $10 \text{ kg}$
Mass of the bar = $10 \text{ kg}$
Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$

To Find

The upward force applied by the weightlifter.

Formula

$F = mg$

Solution

First, calculate the total mass of the barbell. Total mass, $m = 10 \text{ kg} + 10 \text{ kg} + 10 \text{ kg} = 30 \text{ kg}$

Next, calculate the gravitational force (weight) acting downwards on the barbell. $F = mg = 30 \text{ kg} \times 9.8 \text{ m s}^{-2} = 294 \text{ N}$

To keep the barbell steady, the forces must be balanced. This means the weightlifter must apply an equal and opposite (upward) force.

Final Answer She is applying a force of $294 \text{ N}$ in the upward direction.

Example

Example A student is trying to push a stationary block of 25 kg on a horizontal floor. The maximum force of friction opposing this motion is 50 N. Determine the displacement of the block in 2 seconds if the student pushes it with a constant force of (i) 50 N and (ii) 55 N in the forward direction.

Given

Mass of the block, $m = 25 \text{ kg}$
Maximum force of friction, $F_{friction} = 50 \text{ N}$
Time, $t = 2 \text{ s}$
Initial velocity, $u = 0 \text{ m s}^{-1}$ (since it's stationary)

To Find

The displacement of the block, $s$ , for two different applied forces.

Formula

$a = \frac{F_{net}}{m}$ $s = ut + \frac{1}{2}at^2$

Solution

(i) Applied force = 50 N

The forward force applied by the student (50 N) is equal to the opposing force of friction (50 N). Net force, $F_{net} = 50 \text{ N} - 50 \text{ N} = 0 \text{ N}$ Since the net force is zero, the block will not accelerate and will remain stationary.

Answer for part (i) = The displacement is $0 \text{ m}$ .

(ii) Applied force = 55 N

The forward force (55 N) is greater than the force of friction (50 N). Net force, $F_{net} = 55 \text{ N} - 50 \text{ N} = 5 \text{ N}$

Now, calculate the acceleration of the block using Newton's second law. $a = \frac{F_{net}}{m} = \frac{5 \text{ N}}{25 \text{ kg}} = 0.2 \text{ m s}^{-2}$

Finally, calculate the displacement in 2 seconds. $s = ut + \frac{1}{2}at^2 = (0 \text{ m s}^{-1} \times 2 \text{ s}) + \frac{1}{2} \times (0.2 \text{ m s}^{-2}) \times (2 \text{ s})^2$ $s = 0 + \frac{1}{2} \times 0.2 \times 4 = 0.4 \text{ m}$

Answer for part (ii) = The displacement is $0.4 \text{ m}$ in the forward direction.

Example

Example A sports car of mass 1500 kg is moving towards the east and its velocity-time graph is shown. Calculate the force acting on the car during (i) 0 s to 5 s, (ii) 5 s to 10 s, and (iii) 10 s to 15 s.

Given

Mass of the car, $m = 1500 \text{ kg}$
From the graph:
- At $t=0$ , $u=0 \text{ m s}^{-1}$
- At $t=5$ , $v=10 \text{ m s}^{-1}$
- At $t=10$ , $v=10 \text{ m s}^{-1}$
- At $t=15$ , $v=0 \text{ m s}^{-1}$

To Find

The force acting on the car during three time intervals.

Formula

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

Solution

(i) During 0 s to 5 s

First, calculate the acceleration. Initial velocity, $u = 0 \text{ m s}^{-1}$ . Final velocity, $v = 10 \text{ m s}^{-1}$ . Time interval, $t = 5 \text{ s}$ . $a = \frac{10 \text{ m s}^{-1} - 0 \text{ m s}^{-1}}{5 \text{ s}} = 2 \text{ m s}^{-2}$

Now, calculate the force. $F = ma = 1500 \text{ kg} \times 2 \text{ m s}^{-2} = 3000 \text{ N}$

Answer for part (i) = $3000 \text{ N}$ acting towards the east.

(ii) During 5 s to 10 s

The velocity-time graph is a horizontal line, which means the velocity is constant. If velocity is constant, the acceleration is zero. $a = 0 \text{ m s}^{-2}$ Therefore, the net force is also zero. $F = ma = 1500 \text{ kg} \times 0 \text{ m s}^{-2} = 0 \text{ N}$

Answer for part (ii) = $0 \text{ N}$ .

(iii) During 10 s to 15 s

First, calculate the acceleration. Initial velocity, $u = 10 \text{ m s}^{-1}$ . Final velocity, $v = 0 \text{ m s}^{-1}$ . Time interval, $t = 15 \text{ s} - 10 \text{ s} = 5 \text{ s}$ . $a = \frac{0 \text{ m s}^{-1} - 10 \text{ m s}^{-1}}{5 \text{ s}} = -2 \text{ m s}^{-2}$

Now, calculate the force. $F = ma = 1500 \text{ kg} \times (-2 \text{ m s}^{-2}) = -3000 \text{ N}$ The negative sign indicates the force is in the opposite direction of motion (west).

Answer for part (iii) = $-3000 \text{ N}$ , or $3000 \text{ N}$ acting towards the west.

Newton's Third Law of Motion

Forces always come in pairs. When you kick a ball, the ball also pushes back on your foot. This interaction is described by Newton's third law of motion:

Whenever one object is exerting a force on a second object, the second object is simultaneously exerting an equal and opposite force on the first object.

These two forces are often called the action-reaction pair.

Example

Walking: You push the ground backward with your foot (action). The ground's friction pushes your foot forward (reaction), propelling you.
Rocket Launch: The rocket engine pushes hot gas downward (action). The gas pushes the rocket upward (reaction).
Rowing a Canoe: The paddle pushes the water backward (action). The water pushes the paddle and canoe forward (reaction).

Note

The action and reaction forces in a pair always act on two different objects. Because they act on different objects, they do not cancel each other out.

Note

Even though the action and reaction forces are always equal in magnitude, they do not necessarily produce accelerations of equal magnitude. According to Newton's second law (

a = F/m

), if the masses of the two interacting objects are different, their accelerations will also be different.

Example

Example The Earth and a falling fruit apply equal and opposite gravitational forces on each other. Then why does the fruit move towards the Earth while the Earth doesn't seem to move towards the fruit?

Solution

According to Newton's third law, the force the Earth exerts on the fruit is equal in magnitude to the force the fruit exerts on the Earth. However, their masses are vastly different.

From Newton's second law, acceleration is $a = F/m$ .

For the fruit, the force acts on a very small mass, causing a large, noticeable acceleration.
For the Earth, the same amount of force acts on an enormous mass, causing an infinitesimally small and unnoticeable acceleration.

Final Answer The Earth does accelerate towards the fruit, but its acceleration is too small to be detected because its mass is so large.

Example

Example When a 0.1 kg bullet is fired from a 5 kg gun with a force of 2 N, the gun recoils. What are the magnitudes of the initial accelerations of the bullet and the gun?

Given

Mass of the bullet, $m_{bullet} = 0.1 \text{ kg}$
Mass of the gun, $m_{gun} = 5 \text{ kg}$
Force on the bullet, $F_{bullet} = 2 \text{ N}$

To Find

Initial acceleration of the bullet, $a_{bullet}$
Initial acceleration of the gun, $a_{gun}$

Formula

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

Solution

According to Newton's third law, the force exerted by the gun on the bullet is equal and opposite to the recoil force exerted by the bullet on the gun. Therefore, the recoil force on the gun is also $2 \text{ N}$ .

Acceleration of the bullet: $a_{bullet} = \frac{\text{force}}{\text{mass of bullet}} = \frac{2 \text{ N}}{0.1 \text{ kg}} = 20 \text{ m s}^{-2}$

Acceleration of the gun: $a_{gun} = \frac{\text{force}}{\text{mass of gun}} = \frac{2 \text{ N}}{5 \text{ kg}} = 0.4 \text{ m s}^{-2}$

Final Answer The initial acceleration of the bullet is $20 \text{ m s}^{-2}$ , and the initial acceleration of the gun is $0.4 \text{ m s}^{-2}$ .

Forces Acting on a System of Objects

Newton's laws can also be applied to a system of two or more objects connected together. When analyzing such a system, it's helpful to distinguish between internal and external forces.

Internal forces are forces that objects within the system exert on each other (e.g., the tension in a string connecting two boxes).
External forces are forces exerted on the system by objects outside of it (e.g., a person pulling on one of the boxes).

To find the acceleration of the entire system, we can treat the connected objects as a single object with a total mass equal to the sum of the individual masses. The acceleration of the system depends only on the net external force.

For two boxes of masses $m_1$ and $m_2$ connected by a string and pulled by an external force $F$ on a frictionless surface, the acceleration of the system is: $a = \frac{F}{\text{mass of the system}} = \frac{F}{(m_1 + m_2)}$

At a Glance

Friction is a force that acts in the direction opposite to an object's motion.
Newton's First Law (Law of Inertia): An object's state of motion (at rest or constant velocity) does not change unless a net force acts on it.
Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass ( $F=ma$ ).
Newton's Third Law: For every action, there is an equal and opposite reaction. These forces act on different objects.

How Forces Affect MotionClass 9 Science Notes

How Forces Affect Motion

The Concept of Force

Measuring the magnitude of a force

Balanced and Unbalanced Forces

Given

To Find

Solution

The Force of Friction: Often Overlooked but Always Present

Newton's First Law of Motion

Solution

Solution

Newton's Second Law of Motion

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Newton's Third Law of Motion

Solution

Given

To Find

Formula

Solution

Forces Acting on a System of Objects

At a Glance