CLASS 9 SCIENCE · CHAPTER 6 · STUDY NOTES
How Forces Affect Motion
Understand how forces affect motion through short explanations, labelled models and original practice. Read one topic, explain it in your own words, then check.
EXPLORE
A push or pull has a direction
Describe both how strong a force is and which way it acts.One object can have several interactions
- Force
- An interaction that can change motion or shape; its SI unit is the newton (N).
- Contact / non-contact force
- An interaction through touching / without direct contact between the objects.
A hand can push a trolley; gravity pulls a falling ball without touching it. Contact forces include a surface’s support, friction and a rope’s tension. Gravity, electric and magnetic forces act without direct contact.
A spring balance measures force through the extension of its spring, after calibration. A reading in newtons is a force, not a mass. Weight is the gravitational force on an object; near Earth, W=mg.
Take a moment. What do you think?
PAUSE & TRY · think first, then check with a free account
Question
What information is missing from “a force of 5 N”?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 1, 2.
EXPLORE
Find the resultant force
For a straight line, choose a positive direction and add with signs.Opposite directions subtract
- Net or resultant force
- The vector sum of all forces acting on one object.
- Balanced forces
- Forces whose vector sum is zero.
Two rightward pushes of 4 N and 7 N give 11 N rightward. A 7 N rightward push opposed by 4 N leftward gives 3 N rightward.
Equal opposite forces on the same object can balance. That means no change in its velocity in the simple particle model. The object might be resting, or might already be moving steadily.
Keep the signs
Fnet=+7+(−4)=+3 NPositive was chosen to the right. The result says both size and direction.
PAUSE & TRY · think first, then check with a free account
Question
If 9 N right and 6 N left act on one cart, what is the net force?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 2, 3.
EXPLORE
A force can also turn or deform
Changing straight-line motion is only one effect.Separated opposite forces can make a turn
- Turning effect
- The tendency of a force to rotate an object about a pivot.
- Deformation
- A change in an object’s shape or size.
A push farther from a door’s hinge generally turns it more easily than the same suitably directed push close to the hinge. A spring stretches when pulled.
Two opposite forces at different places can turn an extended object even when their net force is zero. So “net force zero” alone does not rule out rotation or deformation.
PAUSE & TRY · think first, then check with a free account
Question
Why is a door handle placed far from its hinge?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 3, 4.
EXPLORE
Friction responds at the contact
Ask how the two surfaces would slide relative to each other.Static friction adjusts before sliding
Opposing friction magnitude (N)
Applied horizontal push (N)
- Static region · Friction matches the push while the box remains at rest.
- Sliding model · An illustrative constant 4 N after motion begins.
Assumptions: A teaching model with limiting static friction 6 N and sliding friction 4 N. These numbers are illustrative, not a rule for every surface.
Read the plotted values
Static region
| Applied horizontal push (N) | Opposing friction magnitude (N) |
|---|---|
| 0 | 0 |
| 6 | 6 |
Sliding model
| Applied horizontal push (N) | Opposing friction magnitude (N) |
|---|---|
| 6 | 4 |
| 10 | 4 |
A teaching model with limiting static friction 6 N and sliding friction 4 N. These numbers are illustrative, not a rule for every surface.
- Static friction
- Contact friction before relative sliding; it adjusts up to a limiting value.
- Sliding friction
- Contact friction while surfaces slide relative to each other.
A small push on a box may be balanced by static friction. Increasing the push can reach the limit, after which the box slides. Once sliding, the friction need not equal that earlier limit.
Friction opposes relative sliding or its tendency at the contact. It does not always point opposite to an object’s overall motion: during walking, the ground can push your foot forward. A spring-balance pull equals opposing friction only when other horizontal forces and acceleration are properly accounted for.
PAUSE & TRY · think first, then check with a free account
Question
Before a pushed box starts moving, must static friction already have its maximum value?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 4, 5, 6, 7.
EXPLORE
Steady motion does not need a net push
Forces are needed to change velocity.Equal position changes continue
- Inertia
- The tendency to resist a change in velocity.
- Newton’s first law
- In an inertial reference frame, an object maintains rest or constant straight-line velocity when net force is zero.
A sliding object on an ordinary floor slows because friction acts. A nearly frictionless model helps us see that motion itself would continue without a net force.
Mass measures inertia: a larger mass is harder to accelerate with the same net force. A car moving steadily can still have engine force and resistance; these can balance. Say “no net force”, rather than “no forces”.
PAUSE & TRY · think first, then check with a free account
Question
Why does a puck slow on a rough floor?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 7, 8.
EXPLORE
Change one factor at a time
Compare acceleration while keeping the relevant other factors controlled.Same mass: acceleration follows net force
Acceleration (m/s²)
Net force (N)
- Mass 2 kg · a=F/2.
Assumptions: Ideal model with fixed mass 2 kg. Use net force, not just one applied force.
Read the plotted values
Mass 2 kg
| Net force (N) | Acceleration (m/s²) |
|---|---|
| 0 | 0 |
| 2 | 1 |
| 4 | 2 |
| 6 | 3 |
| 8 | 4 |
| 10 | 5 |
| 12 | 6 |
Ideal model with fixed mass 2 kg. Use net force, not just one applied force.
- Controlled comparison
- A comparison where the tested factor changes while important other factors are held fixed.
With the same mass, a larger measured net force gives larger acceleration. With the same net force, adding mass reduces acceleration.
A cart pulled by a hanging cup can investigate this. But the rope tension is not simply the cup’s weight when the cup itself accelerates. Measure or model the actual force, account for friction, and do not call approximate apparatus data an exact law.
PAUSE & TRY · think first, then check with a free account
Question
Why is doubling a hanging mass not automatically an exact doubling of the cart’s net force?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 9, 10, 11.
EXPLORE
Divide net force by mass
The rule connects an interaction to the change in motion.Same force: more mass means less acceleration
Acceleration (m/s²)
Mass (kg)
- Net force 12 N · a=12/m.
Assumptions: Original constant-force model. The curve is not a straight line because acceleration is inversely proportional to mass.
Read the plotted values
Net force 12 N
| Mass (kg) | Acceleration (m/s²) |
|---|---|
| 1 | 12 |
| 2 | 6 |
| 3 | 4 |
| 4 | 3 |
| 5 | 2.4 |
| 6 | 2 |
| 7 | 1.7142857142857142 |
| 8 | 1.5 |
Original constant-force model. The curve is not a straight line because acceleration is inversely proportional to mass.
- Newton’s second law
- For constant mass in an inertial frame, Fnet=ma.
One newton gives a 1 kg mass an acceleration of 1 m/s². Force, mass and acceleration units must be consistent.
For an original 3 kg trolley, a 12 N pull opposed by 3 N friction gives Fnet=9 N and a=3 m/s² in the pull’s direction. Using 12/3 would forget friction.
Include the opposing force
a=(12−3)/3=3 m/s²Mass 3 kg; rightward pull 12 N; leftward friction 3 N. The acceleration is rightward.
- 1
Draw forces on the chosen object.
- 2
Add them with directions to find the net force.
- 3
Use a=Fnet/m in SI units.
- 4
State the acceleration’s direction.
PAUSE & TRY · think first, then check with a free account
Question
With 12 N net force, what happens to acceleration when mass changes from 2 to 4 kg?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 11, 12, 13.