B2.3 Solve two-dimensional force and friction problems

Ontario Grade 12 Physics

Dynamics

Resolve forces into components, then apply Newton’s second law

In SPH3U, you used force diagrams and Newton’s laws to describe how forces affect motion. Here, you will apply those ideas when forces act in more than one direction. A force is a vector: it has magnitude and direction. Mass and time are scalars: they have magnitude but no direction. The main strategy is to choose axes, resolve angled forces into components, and apply Newton’s second law separately along each axis. Friction is a contact force that acts parallel to a surface and opposes sliding or the tendency to slide.

What you will learn

Choose a system, reference frame, and positive directions for a force problem.
Resolve angled forces into horizontal and vertical components.
Use a free-body diagram and Newton’s second law to find forces or acceleration.
Determine friction and its direction in two-dimensional situations.

1. Set up the system and axes

The system is the object, or group of objects, whose motion you are studying. The reference frame is the viewpoint used to describe its position and motion. For these problems, use a frame fixed to the ground. State which direction is positive before writing equations. On a level surface, a useful choice is positive

x

to the right and positive

y

upward. On a ramp, choose one axis parallel to the ramp and another perpendicular to it.

A free-body diagram shows the forces acting on one chosen system. Draw each force as an arrow starting on the object. Label the arrow and show its direction. Include only forces acting on the chosen system. Common forces are weight, the normal force, an applied force, and friction. Weight is the gravitational force on an object. The normal force is the contact force perpendicular to a surface.

Resolve a force into components when it points at an angle to your axes. Components are the parts of a vector along the chosen axes. If a force of magnitude

F

is at angle

\theta

above the positive horizontal direction, its horizontal component is

F\cos\theta

and its vertical component is

F\sin\theta

. The angle in this case is measured from the horizontal. Signs depend on the chosen positive directions.

F_x=F\cos\theta,\quad F_y=F\sin\theta

Name the system and reference frame before solving.
Choose axes that simplify the surface or motion.
Show vector directions and signs on the free-body diagram.

2. Apply Newton’s second law on each axis

Newton’s second law relates the net force on a system to its mass and acceleration. Net force means the vector sum of all forces acting on that system. In two dimensions, write one equation for each axis. The symbol

\sum

means to add the forces on that axis. Use newtons for force, kilograms for mass, and metres per second squared for acceleration.

An object may accelerate horizontally while having no vertical acceleration. For example, an object that stays on a level floor has

a_y=0

. Its vertical forces balance, even if its horizontal forces do not. Do not assume that the normal force always equals the weight. An angled applied force can partly lift or press down on an object, changing the normal force.

On an incline, weight still points vertically downward. Resolve it into components parallel and perpendicular to the ramp. For a ramp at angle

\theta

above the horizontal, the component down the ramp has magnitude

mg\sin\theta

, and the component into the ramp has magnitude

mg\cos\theta

. Use the ramp axes to write the force equations.

\sum F_x=ma_x,\quad \sum F_y=ma_y

Write a separate net-force equation for each axis.
Use zero acceleration on an axis only when the situation supports it.
Resolve forces to match the axes you selected.

3. Include friction in the force model

Friction acts parallel to the surfaces in contact. Kinetic friction acts when the surfaces slide relative to each other. Its magnitude is

f_k=\mu_kN

, where

\mu_k

is the coefficient of kinetic friction and

N

is the normal-force magnitude. A coefficient of friction has no units.

Static friction acts when the surfaces are not sliding relative to each other. It adjusts as needed, up to a maximum value. The coefficient of static friction is

\mu_s

. Static friction is not automatically at its maximum; its actual value depends on the force balance.

Friction opposes sliding or the tendency of the surfaces to slide relative to each other. It does not always point opposite an object’s motion. For example, friction on a box pulled right across a floor points left. On a ramp, friction on a block sliding down the ramp points up the ramp.

Find the normal force from the force balance perpendicular to the surface before calculating friction. Use kinetic friction when the object is sliding. For a stationary object, find the friction needed to balance the other forces, then check whether it is no larger than the maximum static friction. If the required value is larger, static friction cannot keep the object at rest.

f_k=\mu_kN,\quad f_s\leq\mu_sN

Kinetic friction has magnitude $\mu_kN$ while surfaces slide.
Static friction varies as needed, up to its maximum.
Friction is parallel to the contact surface.

4. Solve and check

A reliable solution follows a consistent order. Identify the system and frame. Choose positive axes. Draw and label a free-body diagram. Resolve angled forces. Write Newton’s second-law equations for both axes. Find the normal force if friction depends on it. Then solve for the requested quantity.

Keep units in substitutions. Check that a force result is in newtons and an acceleration result is in metres per second squared. Check the sign: a negative component or acceleration points opposite to the direction labelled positive. Finally, ask whether the result makes sense. For example, an upward pull on a box reduces its normal force, so it also reduces the kinetic friction when the coefficient stays the same.

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

Use the diagram to decide signs and friction direction.
Keep units through the calculation.
Check direction, magnitude, and physical reasonableness.

Worked example

An angled pull across a level floor

A

20.0\,\mathrm{kg}

crate is pulled to the right by a

100\,\mathrm{N}

force at

30.0^\circ

above the horizontal. It slides on a level floor with

\mu_k=0.200

. Find its horizontal acceleration. Use

g=9.80\,\mathrm{m/s^2}

.

Choose the system and axes
The system is the crate, viewed from a ground-fixed frame. Choose positive $x$ to the right and positive $y$ upward. The crate remains on the floor, so its vertical acceleration is zero.
Resolve the pull
The angle is measured above the horizontal. The upward component reduces the normal force. The rightward component contributes to the horizontal net force.
$F_x=(100\,\mathrm{N})\cos30.0^\circ=86.6\,\mathrm{N},\quad F_y=(100\,\mathrm{N})\sin30.0^\circ=50.0\,\mathrm{N}$
Find the normal force and kinetic friction
Vertically, the normal force and upward pull balance the crate’s weight. Then use the kinetic-friction relationship.
$N=mg-F_y=(20.0)(9.80)-50.0=146\,\mathrm{N},\quad f_k=\mu_kN=(0.200)(146)=29.2\,\mathrm{N}$
Apply Newton’s second law horizontally
Friction points left, opposite the sliding direction. Subtract it from the rightward pull component, then divide the net force by the mass.
$a_x=\frac{F_x-f_k}{m}=\frac{86.6-29.2}{20.0}=2.87\,\mathrm{m/s^2}$

Answer: The crate’s acceleration is

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

to the right.

Check: The units are newtons divided by kilograms, which gives

\mathrm{m/s^2}

. The positive result agrees with the chosen rightward direction. The upward pull lowers the normal force and therefore lowers friction.

Worked example

A block sliding down a ramp

An

8.00\,\mathrm{kg}

block slides down a ramp inclined at

25.0^\circ

to the horizontal. The coefficient of kinetic friction is

0.150

. Find the block’s acceleration. Use

g=9.80\,\mathrm{m/s^2}

.

Choose the system and axes
The system is the block in a ground-fixed frame. Choose positive $x$ down the ramp and positive $y$ perpendicular to the ramp, away from its surface. The block does not accelerate away from or into the ramp.
Resolve the weight
Weight is vertical. Relative to the ramp axes, its parallel component points down the ramp and its perpendicular component points into the ramp.
$mg\sin25.0^\circ=33.1\,\mathrm{N},\quad mg\cos25.0^\circ=71.1\,\mathrm{N}$
Find the friction force
The perpendicular force balance gives the normal force. Because the block slides down, kinetic friction points up the ramp.
$N=mg\cos25.0^\circ=71.1\,\mathrm{N},\quad f_k=(0.150)(mg\cos25.0^\circ)=10.7\,\mathrm{N}$
Find the acceleration
Along the ramp, gravity’s component is positive and friction is negative. Keep the unrounded values in the calculation and round the final acceleration to three significant figures.
$a_x=9.80\left(\sin25.0^\circ-0.150\cos25.0^\circ\right)=2.81\,\mathrm{m/s^2}$

Answer: The block accelerates at

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

down the ramp.

Check: The units are

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

. The positive result is in the chosen down-ramp direction. Friction reduces the acceleration compared with the value without friction.

Worked example

Testing whether static friction prevents motion

A

12.0\,\mathrm{kg}

box rests on a level floor. A

40.0\,\mathrm{N}

force pulls it to the right at

35.0^\circ

above the horizontal. The coefficient of static friction is

0.500

. Determine whether the box stays at rest and, if it does, find the static-friction force. Use

g=9.80\,\mathrm{m/s^2}

.

Choose the system and axes
The system is the box in a ground-fixed frame. Choose positive $x$ to the right and positive $y$ upward. First test whether static friction can balance the horizontal pull.
Resolve the applied force
The upward component reduces the normal force. If the box is at rest, vertical forces balance.
$F_x=(40.0)\cos35.0^\circ=32.8\,\mathrm{N},\quad F_y=(40.0)\sin35.0^\circ=22.9\,\mathrm{N}$
Find the maximum static friction
Use the vertical balance to find the normal force. Keep its unrounded value when calculating the maximum static friction. This maximum is a limit, not necessarily the friction present.
$N=(12.0)(9.80)-(40.0)\sin35.0^\circ=94.657\ldots\,\mathrm{N},\quad f_{s,\max}=(0.500)\left[(12.0)(9.80)-(40.0)\sin35.0^\circ\right]=47.3\,\mathrm{N}$
Compare the required and maximum friction
For the box to remain at rest, friction must balance the rightward component. The required friction is smaller than the available maximum, so static friction can hold the box.
$f_s=32.8\,\mathrm{N}<47.3\,\mathrm{N}$

Answer: The box stays at rest. Static friction has magnitude

32.8\,\mathrm{N}

and points left.

Check: The required and maximum friction values are in newtons. The required static friction is below its maximum, so the conclusion is physically consistent. Friction balances the horizontal pull rather than automatically taking its maximum value.

Common mistakes and how to avoid them

Using the full angled force in the horizontal equation.

Correction: Use only the force component parallel to the horizontal axis.

Setting the normal force equal to the weight in every situation.

Correction: Use the perpendicular force balance. An angled force can change the normal force.

Treating static friction as always equal to its maximum.

Correction: Find the friction needed for the force balance, then compare it with the maximum.

Choosing friction opposite the object’s velocity in every case.

Correction: Friction opposes sliding or the tendency to slide between the contact surfaces.

Using ramp components with the wrong signs.

Correction: Choose ramp axes first, then assign signs from each force’s arrow direction.

Lesson summary

Choose a system, reference frame, and positive axes before solving.
Use a free-body diagram and resolve angled forces into components.
Apply Newton’s second law separately along each axis.
Find the normal force before calculating friction.
Use kinetic friction for sliding and test whether static friction is within its maximum.

Check your understanding

Question 1

A box slides to the right across a level floor. Which way does the kinetic-friction force on the box point?

Right
Left
Upward
Downward

Show answer and explanation

Left

Kinetic friction acts parallel to the floor and opposes the sliding of the box relative to the floor.

Question 2

A

10.0\,\mathrm{kg}

object rests on a horizontal surface. A

20.0\,\mathrm{N}

horizontal force acts on it, and the maximum static friction is

30.0\,\mathrm{N}

. What is the actual static-friction magnitude if the object remains at rest?

$0\,\mathrm{N}$
$10.0\,\mathrm{N}$
$20.0\,\mathrm{N}$
$30.0\,\mathrm{N}$

Show answer and explanation

20.0\,\mathrm{N}

For zero horizontal acceleration, friction balances the applied force. The required

20.0\,\mathrm{N}

is below the

30.0\,\mathrm{N}

maximum.

Question 3

A force of

50.0\,\mathrm{N}

acts at

60.0^\circ

above the horizontal. What is its horizontal component?

$25.0\,\mathrm{N}$
$43.3\,\mathrm{N}$
$50.0\,\mathrm{N}$
$86.6\,\mathrm{N}$

Show answer and explanation

25.0\,\mathrm{N}

The horizontal component is

F\cos\theta=(50.0)\cos60.0^\circ=25.0\,\mathrm{N}

.

Key terms

System: The object or group of objects selected for analysis.
Reference frame: The viewpoint used to describe position and motion.
Component: The part of a vector that acts along one chosen axis.
Free-body diagram: A diagram showing the external forces acting on one chosen system.
Normal force: The contact force perpendicular to the surfaces in contact.
Static friction: Friction between surfaces that are not sliding relative to each other.
Kinetic friction: Friction between surfaces that are sliding relative to each other.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation B2.3. 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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B2.3 · Solve two-dimensional force and friction problems

Ontario Grade 12 Physics

Resolve forces into components, then apply Newton’s second law

What you will learn

1. Set up the system and axes

2. Apply Newton’s second law on each axis

3. Include friction in the force model

4. Solve and check

An angled pull across a level floor

A block sliding down a ramp

Testing whether static friction prevents motion

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