Particle kinetics in rectangular coordinates

Kinematics tells you how a particle moves; kinetics tells you why. Once you allow forces to be unbalanced, Newton's second law $\sum \vec F = m\vec a$ stops being a statics equality and becomes the engine that drives motion. Rectangular coordinates are the simplest arena for this: fix an $x$ - $y$ - $z$ frame, resolve every force in the free-body diagram (FBD) along those axes, and Newton's law splits cleanly into three independent scalar equations, one per direction. This article picks up right after particle kinematics in the "Kinematics and Kinetics of Particles" unit. The new ingredient is mass: given the forces, you now solve for acceleration (or vice versa), then hand that acceleration to the kinematics toolkit ( $a=\dot v$ , $v=\dot s$ , or the constant-acceleration set) to get velocity and position. The habit to build here is choosing axes that match the constraint — often rotated to lie along and perpendicular to a slope — so each direction decouples as much as possible.

The core insight

Resolving Newton's second law $\sum \vec F = m\vec a$ along fixed perpendicular axes turns one vector equation into independent scalar equations $\sum F_x = m\ddot x$ , $\sum F_y = m\ddot y$ , $\sum F_z = m\ddot z$ , which you solve for acceleration and then feed into kinematics for velocity and position.

The core idea

Newton's second law, componentwise. For a particle of mass $m$ under a net force $\sum \vec F$ ,

\sum \vec F = m\vec a.

where: $\sum \vec F$ is the vector sum of every real force on the particle (weight, normal reaction, tension, applied load, friction), $m$ is its mass, and $\vec a$ is the resulting acceleration. Along fixed rectangular axes this becomes three scalar equations that can be solved one at a time:

\sum F_x = ma_x = m\ddot x, \qquad \sum F_y = ma_y = m\ddot y, \qquad \sum F_z = ma_z = m\ddot z.

where: $a_x, a_y, a_z$ are the acceleration components, and $\ddot x, \ddot y, \ddot z$ are the second time-derivatives of position — the same quantities you already computed in kinematics, now caused by forces rather than merely described.

The standard procedure. Draw the FBD showing every force acting on the particle, exactly as in statics — but now the sums need not be zero. Choose axes that suit the geometry (horizontal/vertical is natural on flat ground; along-slope/perpendicular-to-slope is natural on an incline — both still count as "rectangular," just rotated). Write $\sum F_x = ma_x$ and $\sum F_y = ma_y$ for each direction in turn. Finally, use kinematics to convert the acceleration you found into the velocity or displacement the problem actually asks for.

The constraint subtlety. Acceleration components must match the particle's actual path. If the particle is confined to a straight line — a block sliding on a level or inclined surface — the acceleration perpendicular to that line is zero, even though the force balance in that direction is still very much needed: it is exactly what fixes the normal reaction $N$ , which in turn may feed into a friction force in the along-motion direction.

Figure

A block on a frictionless incline with weight resolved into along slope and perpendicular to slope components and rotated x y axes drawn along and normal to the slope

Derivation AAcceleration of a block sliding down a frictionless incline

Find the acceleration $a_x$ of a block of mass $m$ sliding down a frictionless incline of angle $\theta$ , by resolving Newton's second law along rotated rectangular axes chosen to lie along and perpendicular to the slope, and show that the mass cancels out of the final result.

Step 1. Draw the FBD: the block experiences its weight $mg$ acting vertically downward, and the normal reaction $N$ acting perpendicular to the slope surface (no friction is present by assumption).

Step 2. Choose rotated rectangular axes with $x$ pointing down the slope and $y$ perpendicular to it, since the block's motion is confined to the slope direction. Resolve the weight into these axes: the component along the slope is $mg\sin\theta$ , and the component perpendicular to the slope is $mg\cos\theta$ .

Step 3. Apply Newton's second law perpendicular to the slope. Because the block cannot move off the surface, $a_y = 0$ , so $\sum F_y = ma_y$ gives $N - mg\cos\theta = 0$ , hence $N = mg\cos\theta$ . This equation fixes the reaction force; it does not by itself tell you how the block accelerates along the slope.

Step 4. Apply Newton's second law along the slope. The only force with a component in this direction is the resolved weight, so $\sum F_x = ma_x$ becomes

mg\sin\theta = ma_x.

Step 5. Divide both sides by $m$ to isolate the acceleration:

a_x = g\sin\theta.

where: $g$ is the acceleration due to gravity and $\theta$ is the incline angle measured from the horizontal. The mass $m$ has cancelled entirely — a frictionless incline accelerates a block at a rate that depends only on the angle, exactly as free fall resolved along the slope direction would, and this is the key structural result the worked example below uses directly.

Worked example

A crate of mass $m = 10\ \text{kg}$ slides down a frictionless incline of angle $25^\circ$ to the horizontal, starting from rest. Take $g = 9.8\ \text{m/s}^2$ . Find the acceleration down the slope, the speed after $4\ \text{s}$ , and the distance travelled in that time.

Step 1. Set up rotated rectangular axes along and perpendicular to the slope, exactly as in the derivation above, and draw the FBD: weight $mg$ resolved into $mg\sin25^\circ$ along the slope and $mg\cos25^\circ$ perpendicular to it, plus the normal reaction $N$ .

Step 2. Apply $\sum F_y = ma_y$ perpendicular to the slope with $a_y = 0$ : $N = mg\cos25^\circ$ . This value is not needed further here since the surface is frictionless, but it would be essential the moment a friction force $\mu N$ entered the problem.

Step 3. Apply $\sum F_x = ma_x$ along the slope using the derived result $a_x = g\sin\theta$ :

a_x = g\sin25^\circ = 9.8 \times 0.4226 = 4.14\ \text{m/s}^2.

Step 4. Use kinematics to find the speed after $4\ \text{s}$ , starting from rest ( $v_0 = 0$ ):

v = v_0 + a_x t = 0 + 4.14 \times 4 = 16.6\ \text{m/s}.

Step 5. Use kinematics again to find the distance travelled in the same $4\ \text{s}$ :

x = v_0 t + \tfrac12 a_x t^2 = 0 + 0.5 \times 4.14 \times 4^2 = 33.1\ \text{m}.

The crate reaches 16.6 m/s after 4 s, having travelled 33.1 m down the slope, with acceleration a constant 4.14 m/s^2 independent of its mass.

Competitive stretch

If the same incline instead had friction with coefficient $\mu_k = 0.2$ , Step 2's normal reaction $N = mg\cos25^\circ$ would feed a friction force $f = \mu_k N$ opposing the motion (up the slope, since the crate slides down), and Step 3 would become $mg\sin25^\circ - \mu_k mg\cos25^\circ = ma_x$ , giving $a_x = g(\sin25^\circ - \mu_k\cos25^\circ) = 9.8(0.4226 - 0.2 \times 0.9063) = 2.36\ \text{m/s}^2$ — noticeably smaller, and this time the mass still cancels because friction is also proportional to $N$ , which is proportional to $m$ .

Common traps

Solving only the along-motion equation and skipping the perpendicular one — but the perpendicular equation is exactly what determines $N$ , and $N$ is often needed to compute a friction force in the other direction.
Assuming acceleration is zero whenever velocity is momentarily zero (e.g., at the top of a vertical throw) — gravity still supplies $a = -g$ at that instant, since $\sum F = ma$ depends on force, not on velocity.
Confusing mass and weight: $m$ in $\sum F = ma$ is in kilograms, while $mg$ is a force in newtons — a very common unit slip when plugging numbers in.
Keeping horizontal/vertical axes on an incline problem out of habit, which couples the two equations together, instead of rotating axes to lie along and perpendicular to the slope so each equation decouples.

Ask the tutor: Want to see how connected blocks over a pulley, or a particle under a time-varying force, extend this same rectangular-coordinate approach — ask the AI tutor for a guided walkthrough.

Why do we bother writing the perpendicular equation if the block cannot accelerate off the surface anyway

Because $a_y = 0$ is a statement about acceleration, not about the forces themselves. Setting $\sum F_y = m(0) = 0$ is precisely how you solve for the normal reaction $N$ , which is a genuine unknown force. If friction is present, $N$ directly sets the friction magnitude $\mu N$ , so skipping the perpendicular equation leaves you unable to complete the along-motion equation at all.

Why does mass cancel out of a x = g sin theta but not always cancel in kinetics problems generally

It cancels here because every force resolved along the slope — just the weight component $mg\sin\theta$ — is itself proportional to $m$ , so dividing $\sum F_x = ma_x$ by $m$ removes it completely. This is special to gravity-driven problems. The moment a force that does not scale with $m$ appears, such as a fixed applied force $F$ in newtons, the mass no longer cancels, because $a_x = F/m$ depends explicitly on the particular block's mass.

How do I decide which axes to use, horizontal and vertical or rotated along the incline

Pick axes so that one axis lies along the direction the particle is actually free to accelerate in, and the other lies along the direction where the constraint forces you into zero acceleration. On a flat floor that is simply horizontal and vertical. On an incline, rotating the axes to lie along and perpendicular to the slope makes the perpendicular equation trivial ( $a_y=0$ ) and isolates the along-slope equation to one unknown, $a_x$ — far less algebra than resolving everything on unrotated axes.

Particle kinetics in rectangular coordinates

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads