3.2 Apply Newton’s second law in Cartesian components | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Particle Kinetics: Force and Acceleration

EN PH 131 Engineering Mechanics: Dynamics — Study topic 3.2

Newton’s second law connects the forces on a system to how its motion changes. For this topic, the system is a particle: an object whose size and rotation do not need to be considered. The observer measures its motion in a reference frame, usually one fixed to the ground and treated as non-accelerating. In a Cartesian frame, choose perpendicular

x

and

y

axes and state which direction is positive on each. The same choices must be used in the force components and the acceleration components. The central rule is vectorial: the net force equals mass times acceleration. Writing that rule once for each axis turns a vector equation into equations that can be solved with algebra.

What you will learn

Choose a system, observer, Cartesian axes, and positive directions for a particle problem.
Draw a free-body diagram and identify the forces acting on the system.
Apply Newton’s second law separately in the horizontal and vertical directions.
Solve for unknown accelerations or forces and check signs, units, and physical meaning.

1. Set up the system and coordinates

Start by identifying the system whose motion you want to predict. Include only the object or particle named in the problem. The free-body diagram shows external forces exerted on that system by other objects, such as gravity, a surface, a cable, or an applied push. Do not put the object’s acceleration on this diagram as though it were a force.

Choose an observer and reference frame. In the usual introductory problems, the observer is stationary relative to the ground, and the ground-fixed frame is treated as inertial. This means Newton’s second law can be applied in its familiar form. Then choose axes that make the force directions easy to describe. Horizontal and vertical axes are common, but another perpendicular pair is acceptable if it is clearly defined.

A Cartesian component is the signed amount of a vector along one selected axis. For a force of magnitude

F

at angle

\theta

counterclockwise from positive

x

, its components are

F_x=F\cos\theta

and

F_y=F\sin\theta

. The signs follow from the direction of the components, not from whether a force is considered helpful or opposing.

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

A free-body diagram includes forces acting on the system, not forces the system exerts on other objects.
State both positive axis directions before writing signed components.
Mass is measured in kilograms; weight is a force, commonly modeled near Earth as $mg$ downward.

2. Apply the law in each direction

Newton’s second law states that the vector sum of external forces equals mass times the particle’s acceleration. Because the axes are perpendicular, the vector equation is equivalent to two scalar equations: one along

x

and one along

y

. Each force contributes its signed component to the appropriate equation.

For a particle of constant mass, use

\sum F_x=ma_x

and

\sum F_y=ma_y

. These equations apply whether acceleration is zero or nonzero. If the particle stays at rest or moves at constant velocity in a direction, its acceleration in that direction is zero, so the net force component in that direction is also zero.

A useful solution order is to list known quantities and unknowns, resolve angled forces, write one equation per axis, and solve. If an unknown reaction force appears in the vertical equation while the particle remains on a horizontal surface, the vertical acceleration is zero. This does not mean the vertical forces are absent; it means their signed sum is zero.

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

Resolve angled forces before adding them.
Do not assume acceleration is zero just because velocity is constant at one instant; use the stated motion condition.
A negative acceleration component means acceleration points opposite the chosen positive direction.

3. Check the result

After solving, attach units and interpret the signs. A force divided by mass has units

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

, as acceleration should. If the answer is negative, report the magnitude and state that the direction is opposite the positive axis; do not silently change the sign during the calculation.

Check that the diagram and equations agree. A force pointing left must have a negative

x

component if positive

x

points right. If there is no vertical motion, verify that the vertical force components balance. These checks often reveal a reversed component or a missing force.

Finally, test simple limits where practical. For example, if the net force is zero, the calculated acceleration should be zero. If a force increases while mass stays fixed, the acceleration in its direction should increase. These checks do not replace the equations, but they help catch arithmetic and sign errors.

\mathbf{a}=\frac{\sum\mathbf{F}}{m}

Use signs to preserve direction throughout the calculation.
Check units and force balance in any direction known to have zero acceleration.
The net force, not an individual force, determines acceleration.

Worked example

A horizontal pull on a particle

A 4.0 kg particle is pulled horizontally to the right by a 18 N force. Its only other force is its 39.2 N weight downward, balanced by an upward support force. Find its acceleration. Treat the support as smooth, so there is no friction.

Horizontal pull

Forces on the particle; positive x is right and positive y is up.

Define the system and frame
The system is the particle, observed from a ground-fixed frame treated as inertial. Choose positive $x$ to the right and positive $y$ upward. The particle remains at the same height, so its vertical acceleration is zero.
Write the component equations
The vertical support force $N$ balances weight. Horizontally, the only force is the pull, so it produces the horizontal acceleration.
$\sum F_x=18=4.0a_x,\quad \sum F_y=N-39.2=4.0(0)$
Solve and interpret
Dividing the horizontal net force by mass gives the acceleration. The positive result means it points right. The vertical equation also gives $N=39.2\,\mathrm{N}$ .
$a_x=\frac{18}{4.0}=4.5\,\mathrm{m/s^2}$

Answer: The particle accelerates at

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

to the right, with zero vertical acceleration.

Check: The horizontal units are

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

. The vertical forces balance, consistent with no change in height.

Worked example

A pull angled above a rough horizontal surface

A 6.0 kg crate moves on a level floor. A 50 N force pulls it to the right at 30° above horizontal. The floor exerts 12 N of friction to the left. Find the crate’s acceleration and the normal force, assuming it has no vertical acceleration.

Angled pull

The angled pull has both horizontal and vertical components.

Resolve the pull
The system is the crate in a ground-fixed frame. Choose positive $x$ right and positive $y$ up. The pull’s horizontal component is rightward, and its vertical component is upward. The weight is $mg=6.0(9.81)=58.86\,\mathrm{N}$ downward.
$F_x=50\cos30^\circ=43.3\,\mathrm{N},\quad F_y=50\sin30^\circ=25.0\,\mathrm{N}$
Use the horizontal equation
The crate has no vertical motion, but it can accelerate horizontally. Subtract the leftward friction force from the rightward pull component, then divide the net force by the mass.
$a_x=\frac{43.3-12}{6.0}=5.22\,\mathrm{m/s^2}$
Use the vertical equation
Zero vertical acceleration means the vertical forces sum to zero. The normal force must balance the weight after accounting for the upward component of the pull.
$N+25.0-58.86=0,\quad N=33.86\,\mathrm{N}$

Answer: The crate accelerates at approximately

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

to the right. The normal force is approximately

33.9\,\mathrm{N}

upward.

Check: The normal force is less than the weight because the pull partly supports the crate. The horizontal net force is positive, so rightward acceleration is consistent.

Worked example

Two forces on a free particle

A 5.0 kg particle is acted on by two forces: 13 N directed with components 12 N right and 5 N up, and 5 N directed with components 4 N left and 3 N down. Find its acceleration in Cartesian components and its magnitude.

Two applied forces

The component descriptions define each force direction.

Set the frame and components
The system is the particle, observed in a ground-fixed frame. Let positive $x$ point right and positive $y$ point up. The problem gives the signed force components directly: $(12,5)\,\mathrm{N}$ and $(-4,-3)\,\mathrm{N}$ .
Add force components
Newton’s second law uses the net force. Add the components separately before dividing by the particle’s mass.
$\sum F_x=12-4=8\,\mathrm{N},\quad \sum F_y=5-3=2\,\mathrm{N}$
Find acceleration
Divide each net-force component by the same mass. Both components are positive, so acceleration points right and upward.
$a_x=\frac{8}{5.0}=1.6\,\mathrm{m/s^2},\quad a_y=\frac{2}{5.0}=0.40\,\mathrm{m/s^2}$
Find the magnitude
The acceleration magnitude follows from the perpendicular Cartesian components. Its direction is above the positive horizontal axis.
$|\mathbf{a}|=\sqrt{1.6^2+0.40^2}=1.65\,\mathrm{m/s^2}$

Answer: The acceleration is

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

right and

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

up, with magnitude approximately

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

.

Check: The net force magnitude is

\sqrt{8^2+2^2}=8.25\,\mathrm{N}

; dividing by

5.0\,\mathrm{kg}

gives

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

, matching the component result.

Common mistakes and how to avoid them

Using the full angled force in the horizontal equation.

Correction: Resolve the force into components first, then use only its horizontal component in the horizontal equation.

Treating a negative acceleration as an impossible answer.

Correction: A negative component simply indicates acceleration opposite the positive direction chosen for that axis.

Setting every force component to zero when the particle has no motion in that direction.

Correction: No motion or no change in velocity in a direction means acceleration is zero there, so the net force component is zero; individual forces may still be present.

Using weight as though it were mass.

Correction: Mass is measured in kilograms. Near Earth, weight is a downward force with magnitude

mg

in newtons.

Lesson summary

Define the particle system, observer, reference frame, axes, and positive directions.
Draw the forces acting on the system and resolve angled forces into signed Cartesian components.
Apply $\sum F_x=ma_x$ and $\sum F_y=ma_y$ independently.
Interpret signs and check units, force balance, and directions.

Check your understanding

Question 1

A 2.0 kg particle has a net horizontal force of 6.0 N to the left. If positive

x

is right, what is

a_x

?

$+3.0\,\mathrm{m/s^2}$
$-3.0\,\mathrm{m/s^2}$
$-12\,\mathrm{m/s^2}$
$+12\,\mathrm{m/s^2}$

Show answer and explanation

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

The force is negative in the chosen coordinate, so

a_x=(-6.0)/(2.0)=-3.0\,\mathrm{m/s^2}

.

Question 2

A particle has zero vertical acceleration. What must be true of its vertical forces?

There are no vertical forces.
The upward forces must each equal the particle’s mass.
Their signed sum is zero.
The horizontal net force must also be zero.

Show answer and explanation

Their signed sum is zero.

The vertical form of Newton’s second law is

\sum F_y=ma_y

. With

a_y=0

, the signed sum of vertical forces is zero.

Key terms

System: The object or particle selected for analysis.
Reference frame: The observer’s chosen viewpoint and coordinate system for describing motion.
Net force: The vector sum of all external forces acting on the system.
Cartesian component: The signed part of a vector along one of two perpendicular coordinate axes.

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3.2 · Apply Newton’s second law in Cartesian components

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 3.2

What you will learn

1. Set up the system and coordinates

2. Apply the law in each direction

3. Check the result

A horizontal pull on a particle

A pull angled above a rough horizontal surface

Two forces on a free particle

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through EN PH 131

About this lesson and its review