Forces and free-body diagrams

A force is a push or pull acting on an object. A free-body diagram isolates one object and shows every external force acting on it. Getting this diagram right is usually the most important step in a mechanics problem: the equations can only describe the forces you have modelled.

Prerequisites

You should be able to:

• distinguish scalars from vectors;
• use basic trigonometry in right-angled triangles;
• work with SI units from quantities and units in mechanics;
• understand the role of assumptions in mechanical models.

Force is a vector

A force has both magnitude and direction. Its SI unit is the newton, $mathrm N$. A force of $8 mathrm N$ to the right is not the same as a force of $8 mathrm N$ to the left.

Several forces acting together have a resultant force. This is their vector sum:

$$\mathbf R=\sum \mathbf F.$$

In one dimension, choose a positive direction and attach signs accordingly. If right is positive, forces of $11 mathrm N$ right and $4 mathrm N$ left give

$$R=11-4=7 mathrm N.$$

The resultant is therefore $7 mathrm N$ to the right. A negative answer would mean that the resultant points opposite to the chosen positive direction.

Resultant force is not an extra force

The resultant is a single force with the same overall effect as all the actual forces combined. Do not draw the actual forces and then add the resultant as another arrow.

Common forces

The force name should describe the physical interaction, not merely the arrow’s direction.

Force	Symbol often used	Direction and meaning
Weight	$W$ or $mg$	Vertically downwards, towards the centre of the Earth
Normal reaction	$R$ or $N$	Perpendicular to a contact surface, away from it
Tension	$T$	Along a taut string or cable, pulling away from the object
Friction	$F$	Along a rough contact surface, opposing actual or impending relative motion
Thrust or driving force	$P$, $D$ or $F$	In the direction in which an engine, propeller or person pushes
Resistance or drag	$D$ or $F$	Opposite to motion through air, water or another medium

Symbols vary between questions. Always define yours clearly.

Weight and mass are different

Mass measures the amount of matter and is measured in kilograms. Weight is the gravitational force on that mass:

$$\boxed{W=mg}.$$

Here $g$ is the magnitude of gravitational acceleration. At the Earth’s surface, questions commonly use $g=9.8 mathrm{m,s^{-2}}$, unless another value is given. Since

$$1 mathrm N=1 mathrm{kg,m,s^{-2}},$$

a mass of $6 mathrm{kg}$ has weight

$$W=6(9.8)=58.8 mathrm N.$$

The mass is still $6 mathrm{kg}$. Writing its weight as $6 mathrm N$ or $6g mathrm{kg}$ mixes up different quantities and units.

The normal reaction is not always equal to weight

The word normal means perpendicular. A table exerts a normal reaction perpendicular to its surface. On a horizontal table this is vertical, but on a slope it is not.

For a stationary object on a horizontal table with no other vertical forces, vertical balance gives $R=mg$. That equality is a consequence of this particular situation, not a definition. An additional vertical pull, a vertical acceleration, or an inclined surface can make $R\ne mg$.

Tension pulls

An ideal light, inextensible string under tension pulls an attached object along the string. It cannot push. At an object, the tension arrow therefore points away from the object and towards the rest of the string.

In the usual ideal model, tension has the same magnitude throughout a light string passing over a smooth pulley. The assumptions matter, and are used later in connected particles and pulleys.

Friction opposes relative motion

Friction does not necessarily point opposite to the velocity of the object’s centre. It opposes sliding, or the tendency to slide, between the contacting surfaces.

For example, if a block on a rough slope would otherwise slide down the slope, friction acts up the slope. Its magnitude is not automatically $\mu R$. The equation $F=\mu R$ applies only in limiting equilibrium; kinetic models depend on the question’s assumptions. See friction.

What a free-body diagram shows

A free-body diagram, often abbreviated to FBD, should:

1. isolate one chosen object;
2. replace the object by a dot or simple box;
3. show every external force acting on that object as an arrow;
4. place each arrow in its correct direction and label it;
5. omit forces exerted by the object on other bodies.

Arrow lengths may indicate relative magnitudes if these are known, but a diagram need not be drawn to scale.

For a book resting on a horizontal table, the forces are

$$\begin{array}{c} \uparrow\ R\\[-2pt] \boxed{\text{book}}\\[-2pt] \downarrow\ mg \end{array}$$

The Earth pulls the book down, giving its weight. The table pushes the book up, giving the normal reaction. There is no force labelled “stationary” and no upward force supplied by the book itself.

A reliable question

For every arrow, complete the sentence: “The ___ exerts this force on the object.” If you cannot name the agent producing the force, the arrow may not belong on the diagram.

A systematic drawing method

Use the same process every time.

Step 1: choose the body

State exactly what you are isolating. A block and the table beneath it are different bodies. Connected particles usually need separate diagrams.

Step 2: add non-contact forces

At this level, the main non-contact force is weight. Draw $mg$ vertically downwards from the body’s centre of mass.

Step 3: inspect every contact

Each contact can produce forces:

• a surface may produce a normal reaction and friction;
• a string or cable may produce tension;
• a person, engine or connector may produce an applied force or thrust;
• a fluid may produce resistance.

Step 4: check directions

Weight is vertical, reaction is perpendicular to the surface, tension follows the string, and friction follows the surface.

Step 5: choose axes

Choose axes that make the equations simple. On a slope, axes parallel and perpendicular to the slope are usually best. Do not resolve forces until the complete diagram is visible.

Step 6: check completeness

Count physical interactions, not directions. Ask whether any surface, string, engine, fluid or gravitational field has been missed.

Worked example 1: a pulled crate

A crate of mass $12 mathrm{kg}$ rests on a rough horizontal floor. A horizontal force of $45 mathrm N$ pulls it to the right. Friction of magnitude $17 mathrm N$ acts on the crate. Draw its free-body diagram and find the horizontal resultant.

The four forces acting on the crate are:

$$\begin{array}{c} \uparrow\ R\\[-2pt] 17\ \mathrm N\ \leftarrow\ \boxed{12\ \mathrm{kg}}\ \rightarrow\ 45\ \mathrm N\\[-2pt] \downarrow\ 12g \end{array}$$

The force $12g$ is weight, not mass. Taking right as positive,

$$R_x=45-17=28 mathrm N.$$

There is no vertical acceleration, so the vertical forces balance:

$$R-12g=0,$$

and therefore $R=12g=117.6 mathrm N$. The resultant force is oxed{28 mathrm N} to the right.

Notice that the symbol $R$ is being used for the normal reaction, while $R_x$ denotes a resultant component. To avoid ambiguity, you could call the normal reaction $N$ instead.

Self-check 1

A $5 mathrm{kg}$ box rests on a horizontal floor. A horizontal force of $32 mathrm N$ acts left and friction of $11 mathrm N$ acts right. State all four forces and find the resultant.

Answer

The forces are weight $5g$ down, normal reaction $R$ up, applied force $32 mathrm N$ left, and friction $11 mathrm N$ right.

The vertical forces balance. Horizontally, the resultant has magnitude

$$32-11=21 mathrm N$$

and acts to the left.

Worked example 2: a pull at an angle

A suitcase of mass $20 mathrm{kg}$ is pulled along a horizontal floor by a force of $80 mathrm N$ at $30^\circ$ above the horizontal. It has no vertical acceleration. Resistance has magnitude $25 mathrm N$. Find the normal reaction and the horizontal resultant.

The free-body diagram contains:

• weight $20g$ vertically downwards;
• normal reaction $R$ vertically upwards;
• the $80 mathrm N$ pull at $30^\circ$ above the horizontal;
• resistance $25 mathrm N$ horizontally backwards.

The pull has components

$$80\cos 30^\circ \quad\text{horizontally}$$

and

$$80\sin 30^\circ \quad\text{vertically upwards}.$$

Since there is no vertical acceleration, upward forces equal downward forces:

$$R+80\sin 30^\circ=20g.$$

Thus

$$R=196-40=156 mathrm N.$$

The upward component of the pull reduces the contact force. Horizontally,

$$\begin{aligned} R_x&=80\cos 30^\circ-25\\ &=44.3 mathrm N \quad\text{to 3 significant figures}. \end{aligned}$$

The horizontal resultant is oxed{44.3 mathrm N} forwards.

Do not rotate weight

Weight remains vertically downwards, even when the object or surface is tilted. Axes may be rotated; gravity is not.

Self-check 2

A $10 mathrm{kg}$ sled is pulled by a $50 mathrm N$ force at $20^\circ$ above the horizontal. It remains in contact with horizontal ground and has no vertical acceleration. Find the normal reaction, using $g=9.8 mathrm{m,s^{-2}}$.

Answer

Resolve vertically:

$$R+50\sin20^\circ-10g=0.$$

Therefore

$$R=98-50\sin20^\circ=80.9 mathrm N$$

to $3$ significant figures.

Worked example 3: an object on a slope

A block of mass $4 mathrm{kg}$ rests on a rough plane inclined at $25^\circ$ to the horizontal. Draw its free-body diagram and determine the normal reaction and the friction required for equilibrium.

The actual forces are:

• weight $4g$, vertically downwards;
• normal reaction $R$, perpendicular to the plane;
• friction $F$, up the plane because the block would otherwise slide down.

The quantities $4g\sin25^\circ$ and $4g\cos25^\circ$ are components of weight, not additional forces. They may be drawn in a separate resolving triangle, but should not be added to a diagram that already contains $4g$.

Resolve perpendicular to the plane:

$$R-4g\cos25^\circ=0,$$

so

$$R=4(9.8)\cos25^\circ=35.5 mathrm N$$

to $3$ significant figures.

Resolve parallel to the plane:

$$F-4g\sin25^\circ=0,$$

so

$$F=4(9.8)\sin25^\circ=16.6 mathrm N.$$

The result $F>0$ confirms that friction acts up the plane. This is an equilibrium calculation, not evidence that friction always equals $mg\sin\theta$.

Self-check 3

A particle of mass $7 mathrm{kg}$ is held at rest on a smooth plane inclined at $35^\circ$ by a force $P$ acting up and parallel to the plane. Find $P$ and the normal reaction.

Answer

Because the plane is smooth, there is no friction. Parallel to the plane,

$$P=7g\sin35^\circ=39.4 mathrm N.$$

Perpendicular to the plane,

$$R=7g\cos35^\circ=56.2 mathrm N.$$

Both values are to $3$ significant figures, using $g=9.8 mathrm{m,s^{-2}}$.

Worked example 4: tension and separate bodies

Two particles $A$ and $B$ hang at rest, one below the other. Particle $A$ has mass $3 mathrm{kg}$ and is attached to a ceiling by an upper string. Particle $B$ has mass $2 mathrm{kg}$ and hangs from $A$ by a lower string. Find the tension in each string.

Draw a separate diagram for each particle.

For $B$, the lower string pulls upwards with tension $T_2$, and weight $2g$ acts downwards:

$$T_2-2g=0 \quad\Longrightarrow\quad T_2=19.6 mathrm N.$$

For $A$, the upper string pulls upwards with tension $T_1$. Weight $3g$ and the lower-string tension $T_2$ both act downwards:

$$T_1-3g-T_2=0.$$

Hence

$$T_1=3g+2g=5g=49.0 mathrm N.$$

The two tensions are different because these are two different strings. The upper string supports both masses, while the lower string supports only $B$.

Alternatively, treating $A$ and $B$ as one system makes the lower tension internal, so it does not appear:

$$T_1-(3+2)g=0.$$

This gives the same $T_1$. Choosing a larger system can remove unknown internal forces, but it cannot give the internal tension $T_2$ directly.

Self-check 4

A lamp of mass $8 mathrm{kg}$ hangs at rest from a vertical cable. Find the tension. If a second $3 mathrm{kg}$ lamp is attached below it by a separate cable, find the tension in the upper cable.

Answer

For the single lamp,

$$T=8g=78.4 mathrm N.$$

With both lamps, the upper cable supports total mass $11 mathrm{kg}$, so

$$T_{\text{upper}}=11g=107.8 mathrm N.$$

Equilibrium and resultant force

An object is in translational equilibrium when its resultant force is zero:

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

In two dimensions this means

$$\sum F_x=0 \qquad\text{and}\qquad \sum F_y=0.$$

Equilibrium does not mean that no forces act. It means that the forces balance. Nor does it necessarily mean that the object is stationary: an object moving with constant velocity also has zero resultant force. The connection with acceleration is formalised in Newton’s laws.

Worked example 5: three forces in equilibrium

A ring is held in equilibrium by a horizontal force of $12 mathrm N$ to the right, a vertical force of $5 mathrm N$ upwards, and a third force $\mathbf F$. Find the magnitude and direction of $\mathbf F$.

The first two forces have combined vector

$$\begin{pmatrix}12\\5\end{pmatrix} mathrm N.$$

For equilibrium, the third force must be the opposite vector:

$$\mathbf F= \begin{pmatrix}-12\\-5\end{pmatrix} mathrm N.$$

Its magnitude is

$$|\mathbf F|=\sqrt{12^2+5^2}=13 mathrm N.$$

The reference angle is

$$\tan^{-1}\left(\frac{5}{12}\right)=22.6^\circ.$$

Therefore $\mathbf F$ acts oxed{22.6^\circ} below the horizontal towards the left, with magnitude oxed{13 mathrm N}.

Newton’s third-law pairs

If body $A$ exerts a force on body $B$, then $B$ exerts an equal and opposite force on $A$. These two forces:

• are the same type of interaction;
• have equal magnitudes and opposite directions;
• act on different bodies.

They therefore never both appear on the same free-body diagram.

For a book on a table:

• the table’s upward force on the book appears on the book’s diagram;
• the book’s downward force on the table appears on the table’s diagram.

The book’s weight and the table’s reaction on the book are not a third-law pair because both act on the book. They may balance, but balancing forces and third-law partners are different ideas.

Common mistakes

Drawing motion as a force

Velocity and acceleration are not forces. A moving object does not need a forward force if resistance is absent. Draw only physical interactions.

Assuming every object has a reaction force

A normal reaction exists only when there is contact with a surface. A falling object that is no longer touching a platform has no reaction from it.

Assuming $R=mg$

This is true only when the other vertical forces and vertical acceleration make it true. Write the vertical force equation first.

Mixing forces and components

If $mg$ is drawn, do not also count $mg\sin\theta$ and $mg\cos\theta$ as extra forces. They are two descriptions of the same weight vector.

Putting both sides of an interaction on one diagram

An FBD shows forces on the chosen body. Forces that it exerts on other bodies belong on those bodies’ diagrams.

Deciding friction from the direction an object faces

Friction is determined by relative sliding or impending sliding, not by the way a diagram is facing. First ask which way the surfaces would slip relative to each other.

Mixed self-check

Question 1

A lift of mass $600 mathrm{kg}$ is pulled vertically upwards by a cable of tension $6500 mathrm N$. Ignore resistance and use $g=9.8 mathrm{m,s^{-2}}$. Find the resultant force, including direction.

Answer

The forces are tension upwards and weight downwards. Taking upwards as positive,

$$R=6500-600(9.8)=620 mathrm N.$$

The resultant is oxed{620 mathrm N} upwards.

Question 2

A block is pressed against a vertical wall by a horizontal force $P$. It remains at rest. State the direction of the wall’s normal reaction and of friction on the block.

Answer

The wall’s normal reaction is horizontal, away from the wall. The block would otherwise slide down under its weight, so friction on the block acts vertically upwards.

Question 3

A parachutist is falling vertically at constant speed. Name the forces and state their relationship. The parachute then opens further, increasing air resistance immediately. What is the initial direction of the resultant?

Answer

At constant speed, weight acts downwards and air resistance acts upwards with equal magnitude, so the resultant is zero.

Immediately after the resistance increases, it exceeds weight. The resultant is upwards, even though the parachutist is still moving downwards. The upward resultant causes the downward speed to decrease.

Question 4

A $15 mathrm{kg}$ crate is pulled by a force of $100 mathrm N$ at $40^\circ$ below the horizontal. It has no vertical acceleration. Find the normal reaction.

Answer

The pull has a downward vertical component $100\sin40^\circ$. Therefore

$$R-15g-100\sin40^\circ=0,$$

so

$$R=15(9.8)+100\sin40^\circ=211 mathrm N$$

to $3$ significant figures. A downward angled pull increases the normal reaction.

Final checklist

Before writing force equations, check that you have:

• isolated one clearly defined body or system;
• drawn weight as $mg$ vertically downwards;
• included every relevant contact force;
• made reactions perpendicular and friction parallel to surfaces;
• drawn tension along strings and away from the body;
• excluded velocity, acceleration, components and the resultant as extra forces;
• chosen convenient positive directions;
• kept units in newtons.

Next steps

Continue with:

• Newton’s laws to connect resultant force with acceleration;
• resolving forces and equilibrium for systematic two-dimensional calculations;
• friction for limiting equilibrium and rough surfaces;
• connected particles and pulleys for linked free-body diagrams;
• dynamics in a plane for motion under forces in two dimensions.