Dynamics in a plane

Dynamics connects the forces acting on an object to its acceleration. In a plane, both force and acceleration are vectors, so Newton’s second law must hold in two independent directions:

$$\boxed{\sum \mathbf F=m\mathbf a}.$$

If horizontal and vertical unit vectors are $\mathbf i$ and $\mathbf j$, write

$$\sum \mathbf F=F_x\mathbf i+F_y\mathbf j, \qquad \mathbf a=a_x\mathbf i+a_y\mathbf j.$$

Equality of vectors means equality of their components. Therefore

$$\boxed{F_x=ma_x,\qquad F_y=ma_y.}$$

These are not two different laws. They are the two component equations of one vector law.

Prerequisites

You should be able to:

• draw and label a free body diagram;
• use Newton’s laws, including $F=ma$ in one dimension;
• resolve forces using sine and cosine;
• find the magnitude and direction of a vector;
• distinguish velocity from acceleration in two dimensional motion.

Unless stated otherwise, this lesson models bodies as particles, uses constant mass, and works relative to an inertial frame.

The central idea: one vector equation, two scalar equations

Suppose the resultant force is

$$\sum\mathbf F=(12\mathbf i-5\mathbf j)\ \mathrm N$$

and the mass is $4\ \mathrm{kg}$. Then

$$\mathbf a=\frac{\sum\mathbf F}{m} =\left(3\mathbf i-\frac54\mathbf j\right)\ \mathrm{m\,s^{-2}}.$$

The $\mathbf i$ component and $\mathbf j$ component are divided by the same scalar mass. The acceleration points in the direction of the resultant force, not necessarily in the direction of motion.

For any vector $\mathbf q=q_x\mathbf i+q_y\mathbf j$,

$$|\mathbf q|=\sqrt{q_x^2+q_y^2}.$$

Its direction must be found with attention to the signs of both components. A calculator value of $\tan^{-1}(q_y/q_x)$ gives a reference angle, but may not identify the correct quadrant.

Use the resultant force

In $\sum\mathbf F=m\mathbf a$, $\sum\mathbf F$ is the vector sum of all external forces. Do not set one selected force equal to $m\mathbf a$ unless every other force has already cancelled.

A reliable solution method

For most problems in a plane:

1. Isolate the body and draw every external force acting on it.
2. Choose two perpendicular positive directions and mark them clearly.
3. Resolve every force into those directions.
4. Write $\sum F_x=ma_x$ and $\sum F_y=ma_y$ with signs.
5. Solve the simultaneous equations.
6. Recombine components only if a magnitude or direction is required.
7. Check units, signs, and whether the answer fits the diagram.

Horizontal and vertical axes are often convenient, but not compulsory. On a slope, axes parallel and perpendicular to the slope usually produce simpler equations.

Finding acceleration from several forces

Worked example 1: resultant force and acceleration

A particle of mass $5\ \mathrm{kg}$ is acted on by the forces

$$(18\mathbf i+7\mathbf j)\ \mathrm N, \qquad (-3\mathbf i+9\mathbf j)\ \mathrm N, \qquad (-5\mathbf i-6\mathbf j)\ \mathrm N.$$

Find its acceleration, its acceleration magnitude, and its direction measured anticlockwise from the positive $\mathbf i$ direction.

First add corresponding components:

$$\begin{aligned} \sum\mathbf F &=(18-3-5)\mathbf i+(7+9-6)\mathbf j\\ &=10\mathbf i+10\mathbf j. \end{aligned}$$

Apply $\sum\mathbf F=m\mathbf a$:

$$\mathbf a=\frac{10\mathbf i+10\mathbf j}{5} =2\mathbf i+2\mathbf j\ \mathrm{m\,s^{-2}}.$$

Hence

$$|\mathbf a|=\sqrt{2^2+2^2}=2\sqrt2\ \mathrm{m\,s^{-2}}.$$

Both components are positive, so the direction is in the first quadrant. If $\theta$ is measured from positive $\mathbf i$,

$$\tan\theta=\frac{2}{2}=1,$$

giving

$$\boxed{\theta=45^\circ}.$$

Self-check 1

A $3\ \mathrm{kg}$ particle is acted on by forces $(8\mathbf i-2\mathbf j)\ \mathrm N$ and $(-2\mathbf i-7\mathbf j)\ \mathrm N$. Find its acceleration and acceleration magnitude.

Answer

$$\sum\mathbf F=(6\mathbf i-9\mathbf j)\ \mathrm N,$$

so

$$\boxed{\mathbf a=(2\mathbf i-3\mathbf j)\ \mathrm{m\,s^{-2}}}.$$

Its magnitude is

$$\boxed{|\mathbf a|=\sqrt{13}\ \mathrm{m\,s^{-2}}}.$$

Resolving forces given by magnitude and direction

A force of magnitude $P$ acting at angle $\alpha$ above the positive horizontal has components

$$P\cos\alpha\ \mathbf i+P\sin\alpha\ \mathbf j.$$

This is a geometric statement, not a rule that cosine is always horizontal. The component adjacent to the marked angle uses cosine; the opposite component uses sine. Signs come from direction.

Worked example 2: a pull at an angle

A $6\ \mathrm{kg}$ particle moves on a smooth horizontal plane. It is pulled by a force of $20\ \mathrm N$ at $30^\circ$ above the horizontal. A horizontal resistance of $5\ \mathrm N$ opposes the motion. Find:

1. the normal reaction from the plane;
2. the horizontal acceleration.

The forces are:

• weight $6g$ vertically downwards;
• normal reaction $R$ vertically upwards;
• pull with components $20\cos30^\circ$ horizontally and $20\sin30^\circ$ vertically upwards;
• resistance $5\ \mathrm N$ horizontally backwards.

There is no vertical acceleration because the particle remains on the horizontal plane. Vertically,

$$R+20\sin30^\circ-6g=0.$$

Using $g=9.8\ \mathrm{m\,s^{-2}}$,

$$R+10-58.8=0,$$

so

$$\boxed{R=48.8\ \mathrm N}.$$

Horizontally,

$$20\cos30^\circ-5=6a.$$

Therefore

$$a=\frac{10\sqrt3-5}{6} =2.05\ \mathrm{m\,s^{-2}}\quad\text{to 3 significant figures}.$$

Notice that $R\ne6g$. The upward component of the pull reduces the contact force.

Zero acceleration in one direction only

The particle accelerates horizontally but not vertically. It is correct to write $\sum F_y=0$ and $\sum F_x=ma$ in the same problem.

Self-check 2

A $4\ \mathrm{kg}$ particle lies on a smooth horizontal plane. A force of $25\ \mathrm N$ acts at $40^\circ$ below the horizontal. Find the normal reaction and horizontal acceleration, taking $g=9.8\ \mathrm{m\,s^{-2}}$.

Answer

The force has a downward component, so

$$R-4g-25\sin40^\circ=0.$$

Thus

$$\boxed{R=4g+25\sin40^\circ=55.3\ \mathrm N}$$

to $3$ significant figures. Horizontally,

$$25\cos40^\circ=4a,$$

so

$$\boxed{a=4.79\ \mathrm{m\,s^{-2}}}.$$

Finding an unknown force

Newton’s second law works in reverse. If mass and acceleration are known, the required resultant force is $m\mathbf a$. Any missing applied force is then found by subtracting the known forces as vectors.

Worked example 3: determine a missing force

A particle of mass $2\ \mathrm{kg}$ has acceleration

$$\mathbf a=(4\mathbf i-3\mathbf j)\ \mathrm{m\,s^{-2}}.$$

Two forces acting on it are

$$\mathbf F_1=(5\mathbf i+7\mathbf j)\ \mathrm N, \qquad \mathbf F_2=(-3\mathbf i-2\mathbf j)\ \mathrm N.$$

Find the third force $\mathbf F_3$.

The required resultant is

$$m\mathbf a=2(4\mathbf i-3\mathbf j) =(8\mathbf i-6\mathbf j)\ \mathrm N.$$

Since

$$\mathbf F_1+\mathbf F_2+\mathbf F_3=m\mathbf a,$$ $$\begin{aligned} \mathbf F_3 &=m\mathbf a-\mathbf F_1-\mathbf F_2\\ &=(8\mathbf i-6\mathbf j)-(5\mathbf i+7\mathbf j)-(-3\mathbf i-2\mathbf j)\\ &=6\mathbf i-11\mathbf j. \end{aligned}$$

Therefore

$$\boxed{\mathbf F_3=(6\mathbf i-11\mathbf j)\ \mathrm N}.$$

A quick check is to add all three forces:

$$(5-3+6)\mathbf i+(7-2-11)\mathbf j =8\mathbf i-6\mathbf j=m\mathbf a.$$

Acceleration need not point along velocity

Velocity describes the current direction of motion. Acceleration describes the rate at which velocity changes. There is no general requirement that they be parallel.

Worked example 4: force, velocity and subsequent motion

At $t=0$, a $2\ \mathrm{kg}$ particle has velocity

$$\mathbf u=(5\mathbf i+2\mathbf j)\ \mathrm{m\,s^{-1}}.$$

A constant resultant force

$$\mathbf F=(-2\mathbf i+4\mathbf j)\ \mathrm N$$

then acts. Find its velocity after $3$ seconds and its displacement during those $3$ seconds.

First find acceleration:

$$\mathbf a=\frac{\mathbf F}{m} =(-\mathbf i+2\mathbf j)\ \mathrm{m\,s^{-2}}.$$

For constant acceleration, apply $\mathbf v=\mathbf u+t\mathbf a$ component by component:

$$\begin{aligned} \mathbf v &=(5\mathbf i+2\mathbf j)+3(-\mathbf i+2\mathbf j)\\ &=\boxed{(2\mathbf i+8\mathbf j)\ \mathrm{m\,s^{-1}}}. \end{aligned}$$

The displacement is

$$\mathbf s=t\mathbf u+\frac12t^2\mathbf a.$$

Therefore

$$\begin{aligned} \mathbf s &=3(5\mathbf i+2\mathbf j)+\frac12(3^2)(-\mathbf i+2\mathbf j)\\ &=(15\mathbf i+6\mathbf j)+(-4.5\mathbf i+9\mathbf j)\\ &=\boxed{(10.5\mathbf i+15\mathbf j)\ \mathrm m}. \end{aligned}$$

The initial velocity points into the first quadrant, while the acceleration points into the second. The particle initially moves right, but its rightward velocity decreases.

Self-check 3

A $5\ \mathrm{kg}$ particle initially has velocity $(-2\mathbf i+6\mathbf j)\ \mathrm{m\,s^{-1}}$. A constant resultant force $(10\mathbf i-5\mathbf j)\ \mathrm N$ acts for $4$ seconds. Find the final velocity.

Answer

$$\mathbf a=\frac{10\mathbf i-5\mathbf j}{5} =(2\mathbf i-\mathbf j)\ \mathrm{m\,s^{-2}}.$$

Hence

$$\mathbf v=\mathbf u+t\mathbf a =(-2\mathbf i+6\mathbf j)+4(2\mathbf i-\mathbf j) =\boxed{(6\mathbf i+2\mathbf j)\ \mathrm{m\,s^{-1}}}.$$

Choosing axes to fit the geometry

Vector equations do not depend on the orientation of the axes. Choose axes that reduce the number of components.

For a particle on a plane inclined at angle $\alpha$ to the horizontal, take one axis up the slope and one perpendicular out of the slope. Weight $mg$ then has components

$$mg\sin\alpha\quad\text{down the slope}, \qquad mg\cos\alpha\quad\text{into the slope}.$$

Worked example 5: motion on a smooth inclined plane

A $3\ \mathrm{kg}$ particle is pulled up a smooth plane inclined at $25^\circ$ to the horizontal by a force of $20\ \mathrm N$ parallel to the plane. Find its acceleration and the normal reaction. Take $g=9.8\ \mathrm{m\,s^{-2}}$.

Parallel to the plane, taking uphill as positive,

$$20-3g\sin25^\circ=3a.$$

Hence

$$a=\frac{20-3(9.8)\sin25^\circ}{3} =2.53\ \mathrm{m\,s^{-2}}$$

to $3$ significant figures, uphill.

Perpendicular to the plane there is no acceleration, so

$$R-3g\cos25^\circ=0.$$

Thus

$$\boxed{R=3g\cos25^\circ=26.6\ \mathrm N}.$$

If horizontal and vertical axes had been used, both $R$ and the pulling force would need resolving. The answer would be the same, but the algebra would be longer.

Equilibrium as a special case

If $\mathbf a=\mathbf 0$, then

$$\sum\mathbf F=\mathbf 0.$$

In components,

$$\sum F_x=0, \qquad \sum F_y=0.$$

This includes a particle at rest and a particle moving with constant velocity. Equilibrium does not mean that no forces act. It means their vector sum is zero.

Worked example 6: force required for constant velocity

Forces $(7\mathbf i-4\mathbf j)\ \mathrm N$ and $(-2\mathbf i+9\mathbf j)\ \mathrm N$ act on a particle. Find the third force needed for the particle to move with constant velocity.

Constant velocity means $\mathbf a=\mathbf0$, so the resultant force must be zero:

$$\begin{aligned} \mathbf F_3 &=-\left[(7\mathbf i-4\mathbf j)+(-2\mathbf i+9\mathbf j)\right]\\ &=\boxed{(-5\mathbf i-5\mathbf j)\ \mathrm N}. \end{aligned}$$

The mass is irrelevant because $m\mathbf a=\mathbf0$ for every positive mass.

Common misconceptions

Treating magnitudes as signed components

A magnitude is non-negative. Direction is represented by the signs of components. For example, a $6\ \mathrm N$ force to the left has horizontal component $-6\ \mathrm N$ if right is positive.

Using $F=ma$ separately for every force

Acceleration is caused by the resultant. If forces $P$ and $Q$ act in opposite directions, the equation is $P-Q=ma$, not $P=ma$ and $Q=ma$.

Assuming the normal reaction equals the weight

$R=mg$ only when vertical forces balance and no other force has a vertical component. Resolve perpendicular to the contact surface every time.

Confusing motion with acceleration

A particle can move east while accelerating north, or move upwards while accelerating downwards. Use velocity to describe motion and resultant force to determine acceleration.

Combining components too early

Keep component equations separate until unknowns have been found. Taking magnitudes too soon discards directional information and often creates unnecessary square roots.

Giving an ambiguous direction

An angle such as $30^\circ$ is incomplete unless its reference direction and sense are stated. Write, for example, ”$30^\circ$ above the positive horizontal” or give the vector components.

Mixed self-check

A particle of mass $4\ \mathrm{kg}$ is acted on by three forces:

$$12\mathbf i\ \mathrm N, \qquad (-4\mathbf i+6\mathbf j)\ \mathrm N, \qquad P(\cos\theta\,\mathbf i+\sin\theta\,\mathbf j)\ \mathrm N,$$

where $P>0$ and $0^\circ<\theta<90^\circ$. Its acceleration is $(3\mathbf i+2\mathbf j)\ \mathrm{m\,s^{-2}}$. Find $P$ and $\theta$.

Answer

The required resultant is

$$m\mathbf a=4(3\mathbf i+2\mathbf j) =12\mathbf i+8\mathbf j.$$

The known forces sum to

$$12\mathbf i+(-4\mathbf i+6\mathbf j) =8\mathbf i+6\mathbf j.$$

Therefore the third force is

$$(12\mathbf i+8\mathbf j)-(8\mathbf i+6\mathbf j) =4\mathbf i+2\mathbf j.$$

Comparing components,

$$P\cos\theta=4, \qquad P\sin\theta=2.$$

Hence

$$P=\sqrt{4^2+2^2}=\boxed{2\sqrt5\ \mathrm N},$$

and

$$\tan\theta=\frac24=\frac12, \qquad \boxed{\theta=26.6^\circ}$$

to $3$ significant figures.

Exam checklist

Before finishing a dynamics problem in a plane, ask:

• Have I included every external force exactly once?
• Are my positive directions clear?
• Have I resolved components with the correct signs?
• Does each equation use the resultant component, such as $\sum F_x=ma_x$?
• Have I used zero acceleration only in directions where the motion is constrained?
• Is the final direction unambiguous?
• Are force, mass and acceleration units consistent?

Where to go next

• Apply the same component method with a limiting or kinetic force in friction.
• Combine equations for several bodies in connected particles and pulleys.
• Use constant vertical acceleration and independent horizontal motion in projectiles.
• Study rotational effects of forces in moments.