DoAssignment.ca

C2.2 · Investigate forces with free-body diagrams and Newton’s laws

Learn to investigate forces with free-body diagrams and newton’s laws through clear examples and targeted practice.

Ontario Grade 11 Physics

Forces

A Grade 11 guide to drawing forces, choosing a direction, and explaining motion

A force is a push or pull on an object. Forces can change an object’s motion, but several forces may act at once. A free-body diagram helps organize those forces before using Newton’s laws. In every problem, first identify the system: the object or group of objects being studied. Then choose a reference frame, the viewpoint used to describe motion, and a positive direction. In this lesson, right and up are positive when they are the chosen directions. A vector has both magnitude and direction; force and acceleration are vectors. A scalar has magnitude only; mass is a scalar. Force is measured in newtons (N), mass in kilograms (kg), and acceleration in metres per second squared (m/s²).

What you will learn

1. From the physical situation to a force diagram

Before this lesson, you have used the idea that motion can be described by speed and direction. A change in velocity is acceleration. For this expectation, the key question is what forces act on an object and how those forces relate to its motion.
A free-body diagram (FBD) is a simple drawing of one chosen system with arrows for the external forces acting on it. Each arrow starts on the system and points in the direction of the force. Its length can suggest the force’s relative size, but the diagram is not a picture of the object’s path. Label each arrow with the force type or a clear symbol.
Common forces include weight, the gravitational force on an object; the normal force, a support force from a surface; tension, a pull through a rope or cord; an applied force, a push or pull from a person or object; and friction, a force between surfaces that opposes their sliding or tendency to slide. Only include forces that act on the chosen system. Do not draw the system’s own force on another object as a force acting on the system.

2. Newton’s laws connect forces and motion

Newton’s first law says that if the net force on an object is zero, its velocity does not change. The object may be at rest, or it may keep moving at constant velocity. Zero net force does not mean that no forces act; forces in opposite directions can balance.
The net force is the vector sum of all external forces on the system. In one direction, choose a positive direction and give forces in that direction positive signs. Forces opposite to it receive negative signs. Newton’s second law says that net force equals mass times acceleration. Here, mass is in kg and acceleration is in m/s², so the product has units of newtons. Acceleration points in the same direction as the net force.
Newton’s third law describes a pair of forces: if object A exerts a force on object B, object B exerts an equal-sized force in the opposite direction on object A. The pair acts on two different objects. Therefore, the two forces do not cancel on a single-object FBD.
The laws are useful in an investigation as well as in calculations. A proposed investigation might vary the applied force on a cart while keeping the cart’s mass the same, then record measured acceleration. The FBD helps identify other forces that could affect the motion. A simulation can help explore a prediction, but simulated results are not measurements from a physical experiment. Do not claim results until evidence has actually been collected.
∑F⃗ext=ma⃗\sum \vec{F}_{\text{ext}}=m\vec{a}

3. A careful method for diagrams and investigations

Start by stating the system and the reference frame. For a cart on a track, the system could be the cart, and the frame could be the classroom. Choose a positive direction, such as right along the track. List known quantities and the unknown before calculating.
Draw the FBD without adding motion arrows as if they were forces. Check each force by asking what object or surface exerts it. Then compare forces in each direction. If a problem has horizontal and vertical forces, treat each direction separately using course-level components where needed. A force arrow’s direction must agree with the sign convention.
For an investigation, state what quantity will be changed and what will be measured. Keep other relevant conditions as consistent as possible. Record actual observations with units, and distinguish those measurements from predicted values. Use the FBD to explain which forces were included and whether the measured motion is consistent with the net-force prediction. If measurements do not match a prediction exactly, report the evidence rather than changing it to fit the model.

Worked example

1. Net force and acceleration

A 4.0 kg box is on a level floor. A person pushes it 18 N to the right while friction acts with a force of 6.0 N to the left. Find the box’s acceleration.
  1. Set the system and direction
    The system is the box, viewed from the floor. Choose right as positive. The known mass is 4.0 kg, and the unknown is the box’s acceleration.
  2. Draw and read the FBD
    The horizontal arrows are an 18 N applied force to the right and a 6.0 N friction force to the left. Vertically, weight acts down and the floor’s normal force acts up. On a level floor with no vertical acceleration, those vertical forces balance; they do not affect the horizontal calculation.
  3. Apply Newton’s second law
    The net horizontal force is the signed sum of the horizontal forces. Substitute the values with their directions, then divide by the mass.
    ∑Fx=18 N−6.0 N=12 Nax=∑Fxm=12 N4.0 kg=3.0 m/s2\sum F_x=18\,\mathrm{N}-6.0\,\mathrm{N}=12\,\mathrm{N} a_x=\frac{\sum F_x}{m}=\frac{12\,\mathrm{N}}{4.0\,\mathrm{kg}}=3.0\,\mathrm{m/s^2}
Answer: The box accelerates at 3.0 m/s² to the right.
Check: The units are N/kg, equivalent to m/s². The net force is positive, so the acceleration is rightward. A 12 N net force on a 4.0 kg object gives a modest acceleration, which is reasonable.

Worked example

2. Balanced forces on a hanging object

A 2.5 kg lamp hangs motionless from a vertical cord. Find the tension in the cord. Use 9.8 m/s² for the gravitational field strength.
  1. Set the system and direction
    The system is the lamp, viewed from the room. Choose up as positive. The lamp is motionless, so its acceleration is zero. The unknown is the cord tension.
  2. Draw the FBD
    The cord pulls the lamp upward with tension. Earth pulls it downward with weight. Since the lamp has no acceleration, the forces balance.
  3. Use the force balance
    Newton’s second law gives zero net force for zero acceleration. Weight is mass multiplied by gravitational field strength. Solve for the upward tension.
    ∑Fy=T−mg=0T=mg=(2.5 kg)(9.8 m/s2)=24.5 N≈25 N\sum F_y=T-mg=0 T=mg=(2.5\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=24.5\,\mathrm{N}\approx25\,\mathrm{N}
Answer: The tension is 25 N upward, to two significant figures.
Check: The product kg·m/s² is a newton. The tension is upward and balances the downward weight. Equal forces are reasonable because the lamp remains at rest.

Worked example

3. Find an applied force from the motion

A 6.0 kg cart accelerates at 1.5 m/s² to the right on a track. A 4.0 N resistive force acts to the left. Find the applied force to the right.
  1. Set the system and direction
    The system is the cart, viewed from the track. Choose right as positive. The known values are mass, acceleration, and the leftward resistive force. The unknown is the applied force.
  2. Draw the FBD
    Horizontally, the applied force points right and the resistive force points left. The net force must point right because the acceleration is rightward. The vertical support and weight balance on the level track.
  3. Solve for the applied force
    The signed horizontal net force is applied force minus resistance. Newton’s second law gives the net force from the cart’s mass and acceleration. Add the resistance to find the applied force.
    ∑Fx=ma=(6.0 kg)(1.5 m/s2)=9.0 NFapplied−4.0 N=9.0 NFapplied=13.0 N≈13 N\sum F_x=ma=(6.0\,\mathrm{kg})(1.5\,\mathrm{m/s^2})=9.0\,\mathrm{N} F_{\text{applied}}-4.0\,\mathrm{N}=9.0\,\mathrm{N} F_{\text{applied}}=13.0\,\mathrm{N}\approx13\,\mathrm{N}
Answer: The applied force is 13 N to the right, to two significant figures.
Check: The result is in newtons. Subtracting 4.0 N of resistance leaves 9.0 N to the right, which produces the stated rightward acceleration. The direction and size are consistent.

Common mistakes and how to avoid them

Treating zero net force as proof that no forces act.
Correction: Several forces can act and balance. Draw each force, then find their vector sum.
Putting velocity or acceleration arrows on the FBD as forces.
Correction: An FBD shows forces only. Describe motion separately.
Giving every force a positive sign, even when it points opposite the chosen direction.
Correction: Choose a positive direction first. Use negative signs for force components pointing the other way.
Cancelling a Newton’s third-law pair on one FBD.
Correction: The paired forces act on different objects. Draw each object as its own system to show the pair.

Lesson summary

Check your understanding

Question 1

A 3.0 kg object has a net force of 6.0 N to the left. What is its acceleration?
  1. 2.0 m/s² to the left
  2. 2.0 m/s² to the right
  3. 18 m/s² to the left
  4. Zero, because force does not determine direction
Show answer and explanation
2.0 m/s² to the left
Newton’s second law gives acceleration magnitude 6.0 N ÷ 3.0 kg = 2.0 m/s². Acceleration points in the direction of the net force, so it is leftward.

Question 2

A book rests on a table. Which statement is correct for the book’s FBD?
  1. The table’s upward normal force balances the book’s downward weight.
  2. There are no forces because the book is at rest.
  3. The book’s weight and the table’s force form a Newton’s third-law pair on the book.
  4. The book has a forward force because it could slide.
Show answer and explanation
The table’s upward normal force balances the book’s downward weight.
The book is at rest, so its net force is zero. The normal force and weight act on the book in opposite directions and balance. They are not a third-law pair because both act on the same system.

Key terms

System
The object or group of objects selected for study.
Reference frame
The viewpoint used to describe an object’s position and motion.
Vector
A quantity with both magnitude and direction, such as force or acceleration.
Scalar
A quantity with magnitude only, such as mass.
Free-body diagram
A drawing of one system with arrows showing the external forces acting on it.
Net force
The vector sum of the forces acting on a system.
Acceleration
A change in velocity over time, including a change in speed, direction, or both.

Continue through SPH3U

View the complete SPH3U Ontario Grade 11 Physics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation C2.2. 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.

Official curriculum reference

Report a correction or ask a question