B2.5 Relate system motion to its forces using free-body diagrams

Ontario Grade 12 Physics

Dynamics

Ontario Grade 12 Physics — study topic B2.5

A free-body diagram helps connect forces to motion. It shows forces acting on a chosen object or system, but it does not show the system’s path or velocity. First choose the system and reference frame. A reference frame is the viewpoint and coordinate directions used to describe motion. Then choose a positive direction and keep it consistent. Add the forces as vectors and use their net effect to determine acceleration.

What you will learn

Identify the system, reference frame, and positive direction before analyzing motion.
Draw a free-body diagram showing forces acting on one system.
Use net force to relate a system’s acceleration to its motion.
Explain what a free-body diagram shows and what it does not show.

1. From motion to a force model

In SPH3U, you described motion using position, velocity, and acceleration. Velocity and acceleration are vectors: they have both magnitude and direction. Mass and time are scalars: they have magnitude but no direction. A force is also a vector. Force can change velocity by changing its magnitude, its direction, or both.

A system is the object or group of objects selected for analysis. In a free-body diagram, show the system as a dot or simple shape. Include only forces acting on that system. A force the system exerts on another object does not belong on its diagram.

Force is measured in newtons, written as N. Common forces include gravitational force, a surface’s support force, friction, and tension in a rope. Near Earth, gravitational force on an object is often called weight. Mass is measured in kilograms and is not the same as weight.

\vec{F}_{\mathrm{net}}=m\vec{a}

State the system, reference frame, and positive direction before solving.
A free-body diagram shows forces on the system, not its velocity or path.
Use arrows to show each force’s direction. Label each arrow.

2. Drawing and reading a free-body diagram

Isolate the system and represent it with a dot or box. Draw an arrow for each external force acting on it. The arrow points in the force’s direction. Arrow lengths can suggest relative sizes, but use given magnitudes or calculations for exact comparisons.

For an object on a horizontal surface, weight points down and the normal force points up. The normal force is the support force exerted by a surface, perpendicular to that surface. These forces are not automatically equal. They are equal only when the vertical net force is zero.

Tension acts along a taut rope or cable, pulling away from the system. Friction acts along the contact surface and opposes sliding or the tendency to slide. Identify the situation before deciding its direction.

The net force is the vector sum of all forces on the system. In one dimension, choose a positive direction. Forces in that direction have positive signs; forces in the opposite direction have negative signs. Zero net force means zero acceleration, not necessarily zero velocity.

\sum F_x=ma_x

Draw forces acting on the chosen system, not forces it exerts on other objects.
Opposite forces cancel only when their magnitudes are equal.
Zero acceleration means constant velocity; the system may be at rest or moving.

3. Relating net force to motion

Newton’s second law states that the net force equals mass multiplied by acceleration. Add signed force components along the chosen direction to find net force. Then divide by mass to find acceleration. Acceleration points in the direction of the net force.

This relationship describes a change in velocity, not velocity itself. In one-dimensional or collinear motion, a net force in the direction of motion makes an object speed up in that direction. In that same type of motion, a net force opposite the motion makes it slow down. In two-dimensional motion, a force can also change the direction of velocity. If an object moves while net force is zero, its velocity remains constant in the chosen frame.

Use SI units throughout. Force is measured in newtons, mass in kilograms, and acceleration in metres per second squared. One newton is one kilogram metre per second squared, so the units agree with Newton’s second law.

1\,\mathrm{N}=1\,\mathrm{kg\,m/s^2}

Find the net force before finding acceleration.
Acceleration points in the net-force direction; it does not automatically point in the velocity direction.
Check the sign, direction, and units of the calculated acceleration.

4. A diagram-to-motion method

Start by stating the system and reference frame. Choose a positive direction that makes force signs clear. Draw the system, then add the forces acting on it. Do not confuse individual forces with net force: net force is their combined effect.

Write the force relationship along the direction needed to analyze motion. If a force is angled, resolve it into components along the chosen axes. A component is the part of a vector in one direction. If the net force along an axis is zero, acceleration along that axis is zero.

Finally, describe the result relative to your chosen axes. A positive or negative acceleration sign indicates direction. Compare the result with the force diagram. If acceleration appears opposite the net force, recheck the signs and arrows.

a_x=\frac{\sum F_x}{m}

System → reference frame and positive direction → free-body diagram → net force → acceleration.
Use force components when a force is not aligned with an axis.
A force diagram supports a motion conclusion when axes and signs are clear.

Common forces in a free-body diagram

Force	Symbol	Direction in the situation
Weight	$F_g$	Toward Earth; near its surface, downward
Normal force	$F_N$	Perpendicular to the surface, away from it
Tension	$T$	Along a taut rope or cable, pulling away from the system
Friction	$F_f$	Along a contact surface, opposing sliding or its tendency

Worked example

A box pulled across a floor

A

6.0\,\mathrm{kg}

box is pulled horizontally to the right with a force of

28\,\mathrm{N}

. Friction on the box is

10\,\mathrm{N}

to the left. Find its horizontal acceleration.

Set the system and direction
The system is the box. Use the floor as the reference frame and choose right as positive. The unknown is the box’s horizontal acceleration.
Find the net horizontal force
The pull is positive and friction is negative because it points left. Add the signed forces to find the net force.
$\sum F_x=(+28\,\mathrm{N})+(-10\,\mathrm{N})=+18\,\mathrm{N}$
Relate net force to acceleration
The net force points right. Newton’s second law gives acceleration when the net force is divided by the box’s mass.
$a_x=\frac{+18\,\mathrm{N}}{6.0\,\mathrm{kg}}=+3.0\,\mathrm{m/s^2}$

Answer: The box’s acceleration is

3.0\,\mathrm{m/s^2}

to the right.

Check: The units reduce to

\mathrm{m/s^2}

. The positive sign matches the net force to the right. Rightward acceleration is reasonable because the pull exceeds friction.

Worked example

A load lifted by a cable

A

12\,\mathrm{kg}

load accelerates upward at

1.5\,\mathrm{m/s^2}

. Use

g=9.8\,\mathrm{m/s^2}

. Find the cable tension.

Set the system and direction
The system is the load, viewed from the ground. Choose upward as positive. Tension acts upward, weight acts downward, and tension is unknown.
Calculate the weight
Weight is the gravitational force on the load. Its magnitude is mass multiplied by gravitational field strength, and it points downward.
$F_g=mg=(12\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=1.2\times10^2\,\mathrm{N}$
Relate the forces to acceleration
With upward positive, tension is positive and weight is negative. The net force must point upward because the acceleration is upward.
$T-F_g=ma_y$
Solve for tension
Substitute the mass, acceleration, and weight. Round to two significant figures, consistent with the given values.
$T=(12\,\mathrm{kg})(1.5\,\mathrm{m/s^2})+1.2\times10^2\,\mathrm{N}=1.4\times10^2\,\mathrm{N}$

Answer: The cable tension is

1.4\times10^2\,\mathrm{N}

upward.

Check: Each term in the force equation is in newtons. Tension exceeds weight, so the net force and acceleration are upward, as stated.

Worked example

A box pulled at an angle

A

5.0\,\mathrm{kg}

box is on a level floor. A rope pulls with a force of

20\,\mathrm{N}

at

30^\circ

above the horizontal. Friction is

8.0\,\mathrm{N}

to the left. Find the horizontal acceleration. Assume no vertical acceleration.

Set the system and axes
The system is the box in the floor’s reference frame. Choose right and up as positive. The rope force has horizontal and vertical components. The horizontal component contributes to horizontal acceleration.
Find the horizontal net force
Since the angle is measured above the horizontal, the horizontal component uses cosine. Subtract friction to find the net horizontal force.
$\sum F_x=(20\,\mathrm{N})\cos 30^\circ-8.0\,\mathrm{N}=9.3\,\mathrm{N}$
Find the acceleration
Apply Newton’s second law horizontally. The positive result means the acceleration is to the right.
$a_x=\frac{9.3\,\mathrm{N}}{5.0\,\mathrm{kg}}=1.9\,\mathrm{m/s^2}$

Answer: The horizontal acceleration is

1.9\,\mathrm{m/s^2}

to the right.

Check: The units are metres per second squared. The horizontal rope component is about

17\,\mathrm{N}

, greater than the

8.0\,\mathrm{N}

friction force, so rightward acceleration is reasonable.

Common mistakes and how to avoid them

Drawing velocity or acceleration as if it were a force.

Correction: A free-body diagram contains forces acting on the system. Use net force to determine acceleration; describe velocity separately.

Assuming zero net force means the object is at rest.

Correction: Zero net force means zero acceleration. The object may be at rest or moving at constant velocity.

Giving every force a positive sign.

Correction: Choose a positive direction. Forces pointing opposite to it have negative components.

Confusing mass with weight.

Correction: Mass is measured in kilograms. Weight is a gravitational force measured in newtons.

Assuming acceleration always points in the direction of motion.

Correction: Acceleration points in the direction of net force. For one-dimensional motion, it can point opposite velocity when the object slows down. In two-dimensional motion, acceleration can change the direction of velocity.

Lesson summary

Choose and state the system, reference frame, and positive direction.
Draw a free-body diagram with only forces acting on the system.
Add signed force components to find net force.
Use net force and mass to determine acceleration; acceleration points in the net-force direction.
Check units, signs, direction, and the motion interpretation.

Check your understanding

Question 1

A cart moves right at constant velocity. What is the net force on it?

A net force to the right
A net force to the left
Zero net force
A net force equal to its weight

Show answer and explanation

Zero net force

Constant velocity means zero acceleration. Newton’s second law then gives zero net force.

Question 2

A system has a net force of

12\,\mathrm{N}

left and a mass of

3.0\,\mathrm{kg}

. What is its acceleration?

$4.0\,\mathrm{m/s^2}$ left
$4.0\,\mathrm{m/s^2}$ right
$36\,\mathrm{m/s^2}$ left
Zero

Show answer and explanation

4.0\,\mathrm{m/s^2}

left

Acceleration is net force divided by mass:

12\,\mathrm{N}/3.0\,\mathrm{kg}=4.0\,\mathrm{m/s^2}

. It points left, in the direction of net force.

Question 3

Which arrow belongs on a free-body diagram for a book resting on a table?

An arrow showing the book’s velocity
An arrow showing the force the book exerts on Earth
An upward arrow for the table’s normal force on the book
An arrow showing the book’s path

Show answer and explanation

An upward arrow for the table’s normal force on the book

The diagram shows forces acting on the book. The table exerts an upward normal force on it. Velocity and path are not forces, and the book’s force on Earth acts on a different system.

Key terms

System: The object or group of objects selected for analysis.
Reference frame: The viewpoint and coordinate directions used to describe position and motion.
Free-body diagram: A simplified drawing showing forces acting on one chosen system.
Net force: The vector sum of all forces acting on a system.
Acceleration: The rate at which velocity changes; it has magnitude and direction.
Normal force: A support force from a surface, directed perpendicular to the surface.
Component: The part of a vector in a chosen direction.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation B2.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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B2.5 · Relate system motion to its forces using free-body diagrams

Ontario Grade 12 Physics

Ontario Grade 12 Physics — study topic B2.5

What you will learn

1. From motion to a force model

2. Drawing and reading a free-body diagram

3. Relating net force to motion

4. A diagram-to-motion method

Common forces in a free-body diagram

A box pulled across a floor

A load lifted by a cable

A box pulled at an angle

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review