C2.4 Solve one-dimensional problems with gravity, normal force, and friction

Ontario Grade 11 Physics

Forces

A Grade 11 method for drawing forces, choosing a direction, and solving motion problems

A force is a push or pull. Force is a vector, so it has both a size and a direction. Mass is a scalar: it has a size but no direction. In this lesson, the system is the object whose motion we are studying. We will use a straight-line coordinate axis and state which direction is positive before calculating. The main model is that the net force on an object equals its mass multiplied by its acceleration. We will apply that model to situations involving gravity, the support force from a surface, and friction.

What you will learn

Distinguish force vectors from scalar quantities such as mass and speed.
Identify gravity, normal force, and friction in a one-dimensional situation.
Choose a positive direction and use it consistently in a force calculation.
Solve problems using net force, mass, and acceleration, then check units and reasonableness.

1. Prerequisite bridge: forces and signs

A force is measured in newtons, symbol

\mathrm{N}

. Mass is measured in kilograms, symbol

\mathrm{kg}

. Acceleration is measured in metres per second squared, symbol

\mathrm{m/s^2}

. A positive or negative sign shows direction along the chosen axis; it does not mean that a force has a negative size.

The net force is the vector sum of all forces on the system. In one dimension, choose one direction as positive. Forces in that direction are positive, and forces in the opposite direction are negative. The same sign convention must be used for force and acceleration.

Newton’s second law connects the net force, mass, and acceleration. If the net force is zero, the acceleration is zero. The object may be at rest or moving at constant velocity; zero net force does not by itself mean zero motion.

\sum F = ma

System: the object being studied.
Reference frame: the observer’s chosen coordinate system; here, it is fixed to the ground.
A vector has magnitude and direction. A scalar has magnitude only.

2. The three forces and the free-body diagram

Gravity is the attractive force exerted by Earth on an object. Near Earth’s surface, its magnitude is the object’s weight,

F_g=mg

, where

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

downward. Weight is a force, so it is measured in newtons; it is not the same as mass.

The normal force is the contact force exerted by a surface. It acts perpendicular to the surface and pushes away from it. It is not always equal to weight. On a level surface, with no vertical acceleration and no other vertical forces, the normal force balances weight.

Friction is a contact force parallel to the surfaces touching. It acts opposite the sliding motion, or opposite the direction the surfaces would start to slide. Kinetic friction acts while surfaces slide. Static friction acts when they do not slide; its size adjusts as needed up to a maximum. The coefficient of friction, written

\mu

, is a unitless measure of how much friction a pair of surfaces can provide. In a simple model, kinetic friction has magnitude

F_f=\mu_k F_N

, and maximum static friction has magnitude

F_{f,\max}=\mu_s F_N

.

A free-body diagram shows only forces acting on the chosen object. Draw each force as an arrow from the object, label it, and make its direction agree with the situation. Do not draw a separate arrow for net force; net force is the combined effect of the arrows.

F_g=mg,\quad F_{f,k}=\mu_kF_N,\quad F_{f,s}\leq\mu_sF_N

Weight points vertically downward.
The normal force is perpendicular to the contact surface.
Friction is parallel to the surface and opposes sliding or the tendency to slide.

3. A consistent method for one-dimensional problems

First name the system and reference frame. Choose the positive direction along the line of motion. Then list known values and the unknown. Draw the forces and resolve them along the chosen axis. A force pointing opposite the positive direction receives a negative sign.

If an object stays in contact with a surface and has no acceleration perpendicular to that surface, the perpendicular forces balance. On level ground, this often gives a normal force equal to the weight. On a slope, weight has a component down the slope and a component into the slope. For a slope angle

\theta

measured above the horizontal, those components are

mg\sin\theta

down the slope and

mg\cos\theta

into the slope.

Add the signed forces along the motion axis, then use the net-force relationship to find acceleration or an unknown force. If the object is sliding, use the kinetic-friction model. If it is not sliding, do not automatically assume that static friction equals its maximum; use the force balance to find the amount needed, then check that it does not exceed the maximum.

Before finishing, report a direction as well as a magnitude. Check that the units work out, the sign matches the chosen axis, and the result makes sense. For example, friction should not speed up an object that is sliding in the opposite direction.

\sum F_{\parallel}=ma_{\parallel}

State the sign convention before substituting values.
Use the normal force in the friction model, not automatically the weight.
Keep units in each substitution and round sensibly.

4. Reading the result

The direction of acceleration follows the direction of the net force. This does not always match the direction of motion. For example, an object moving forward while slowing down has acceleration opposite its motion.

A negative acceleration is not an error if the chosen positive direction was forward. It means the acceleration points backward. When a problem asks for acceleration as a vector, include that direction in words.

A useful final check is to compare the forces before calculating. If opposing forces are nearly equal, acceleration should be small. If the net force points downhill on a slope, the acceleration should also point downhill.

A sign communicates direction relative to the selected axis.
Check both the numerical result and its physical direction.

Worked example

A crate pulled across a level floor

A

12.0\,\mathrm{kg}

crate is pulled right with a horizontal force of

55.0\,\mathrm{N}

. It slides on a floor with

\mu_k=0.25

. Find its acceleration.

Set the system and direction
The system is the crate, viewed from the ground. Take right as positive. The unknown is the crate’s horizontal acceleration. Its vertical acceleration is zero because it remains on the level floor.
Find the normal force
Vertically, the normal force balances the weight. The weight is the mass multiplied by gravitational field strength.
$F_N=F_g=mg=(12.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=117.6\,\mathrm{N}$
Find kinetic friction
The crate is sliding, so use the kinetic-friction model. Friction points left, opposite the sliding direction.
$F_{f,k}=\mu_kF_N=(0.25)(117.6\,\mathrm{N})=29.4\,\mathrm{N}$
Apply the net-force relationship
The pull is positive and friction is negative. Subtract the opposing force, then divide by the crate’s mass.
$a=\frac{55.0\,\mathrm{N}-29.4\,\mathrm{N}}{12.0\,\mathrm{kg}}=2.13\,\mathrm{m/s^2}$

Answer: The crate accelerates at

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

to the right.

Check: A newton divided by a kilogram is

\mathrm{m/s^2}

, so the units are correct. The pull is greater than friction, so rightward acceleration is reasonable. Three significant figures are retained.

Worked example

A box sliding down a slope

A

5.00\,\mathrm{kg}

box slides down a

30.0^\circ

slope. The kinetic-friction coefficient is

0.20

. Find its acceleration along the slope.

Set the system and direction
The system is the box, viewed from the ground. Choose down the slope as positive. The box slides, so kinetic friction points up the slope. The unknown is acceleration along the slope.
Find forces perpendicular to the slope
There is no acceleration through the slope. The normal force balances the component of weight into the slope, which is $mg\cos\theta$ .
$F_N=mg\cos\theta=(5.00\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos(30.0^\circ)=42.4\,\mathrm{N}$
Find friction and the downhill weight component
Kinetic friction acts uphill. The component of weight along the slope acts downhill, so it is positive with this sign convention.
$F_{f,k}=0.20(42.4\,\mathrm{N})=8.48\,\mathrm{N},\quad mg\sin\theta=(5.00\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\sin(30.0^\circ)=24.5\,\mathrm{N}$
Calculate acceleration along the slope
Subtract uphill friction from the downhill component of weight, then divide the net force by the mass.
$a=\frac{24.5\,\mathrm{N}-8.48\,\mathrm{N}}{5.00\,\mathrm{kg}}=3.20\,\mathrm{m/s^2}$

Answer: The box accelerates at

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

down the slope.

Check: The units reduce to

\mathrm{m/s^2}

. The downhill component of weight exceeds friction, so downhill acceleration is reasonable. The result is smaller than the acceleration from the downhill component alone because friction opposes the motion.

Worked example

Testing whether a box stays at rest

A

10.0\,\mathrm{kg}

box rests on a level floor. A horizontal force of

18.0\,\mathrm{N}

pushes it right. The coefficient of static friction is

0.25

. Does it remain at rest?

Set the system and direction
The system is the box, viewed from the ground. Choose right as positive. If the box remains at rest, its acceleration is zero and static friction must balance the push.
Find the normal force
There is no vertical acceleration or other vertical force, so the floor’s normal force balances the box’s weight.
$F_N=mg=(10.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=98.0\,\mathrm{N}$
Find maximum static friction
Static friction can adjust up to a maximum. Compare that maximum with the force needed to balance the push.
$F_{f,s,\max}=\mu_sF_N=(0.25)(98.0\,\mathrm{N})=24.5\,\mathrm{N}$
Compare the required and available friction
Balancing the $18.0\,\mathrm{N}$ push requires $18.0\,\mathrm{N}$ of friction to the left. This is below the available maximum, so static friction can prevent sliding.
$18.0\,\mathrm{N}<24.5\,\mathrm{N}$

Answer: Yes. The box remains at rest, and static friction is

18.0\,\mathrm{N}

to the left.

Check: The friction needed is less than its maximum, so the conclusion is reasonable. The horizontal forces cancel, giving zero net force and zero acceleration. All force values are in newtons.

Common mistakes and how to avoid them

Assuming the normal force always equals the object’s weight.

Correction: They are equal only when the perpendicular force balance makes them equal. On a slope, the normal force balances the perpendicular component of weight.

Using kinetic friction for an object that is not sliding.

Correction: Use static friction when the surfaces are not sliding. Its value is whatever is needed for the force balance, up to its maximum.

Treating friction as always opposite the direction the object is moving.

Correction: Friction opposes sliding between the surfaces or the tendency to slide. State the contact situation and draw its direction carefully.

Ignoring signs or reporting acceleration without direction.

Correction: Choose a positive direction first. Use signed forces consistently, then state the acceleration direction in words.

Lesson summary

Choose the system, reference frame, and positive direction before calculating.
Draw gravity, normal force, and friction in directions that match the physical situation.
Use the net force along the motion line to determine acceleration.
Use kinetic friction for sliding and check whether static friction can balance forces when an object is at rest.
Check units, significant figures, direction, and physical reasonableness.

Check your understanding

Question 1

A

4.0\,\mathrm{kg}

object rests on a level floor. What is its weight near Earth’s surface?

$4.0\,\mathrm{N}$ downward
$39\,\mathrm{N}$ downward
$39\,\mathrm{N}$ upward
$0\,\mathrm{N}$

Show answer and explanation

39\,\mathrm{N}

downward

Weight is

mg=(4.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=39.2\,\mathrm{N}

downward, which rounds to

39\,\mathrm{N}

.

Question 2

A block slides right on a horizontal surface. Which way does kinetic friction on the block point?

Right
Left
Up
Down

Show answer and explanation

Left

Kinetic friction acts opposite the sliding direction along the surface. The block slides right, so friction points left.

Question 3

A stationary box needs

12\,\mathrm{N}

of static friction to balance a push. Its maximum static friction is

17\,\mathrm{N}

. What happens?

It remains at rest; friction is $12\,\mathrm{N}$ opposite the push.
It remains at rest; friction must be $17\,\mathrm{N}$ .
It slides because static friction is zero.
It accelerates in the direction of friction.

Show answer and explanation

It remains at rest; friction is

12\,\mathrm{N}

opposite the push.

The needed friction is below the maximum, so static friction can balance the push with a magnitude of

12\,\mathrm{N}

. The net force and acceleration are zero.

Key terms

Acceleration: The rate at which velocity changes, including a change in speed or direction.
Friction: A contact force parallel to touching surfaces that opposes sliding or the tendency to slide.
Free-body diagram: A drawing that shows the forces acting on one chosen object.
Net force: The vector sum of all forces acting on an object.
Normal force: The contact force from a surface, directed perpendicular to that surface.
System: The object or group of objects chosen for the motion problem.
Weight: The force of gravity acting on an object.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation C2.4. 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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C2.4 · Solve one-dimensional problems with gravity, normal force, and friction

Ontario Grade 11 Physics

A Grade 11 method for drawing forces, choosing a direction, and solving motion problems

What you will learn

1. Prerequisite bridge: forces and signs

2. The three forces and the free-body diagram

3. A consistent method for one-dimensional problems

4. Reading the result

A crate pulled across a level floor

A box sliding down a slope

Testing whether a box stays at rest

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH3U

About this lesson and its review