4.2 Free body diagrams and equilibrium

Achievable AP Physics 1

4. Laws of motion

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Free body diagrams and equilibrium

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This subchapter explains how to construct and use free-body diagrams (FBDs) to analyze forces and apply Newton’s laws of motion to solve problems.

In this section, we focus on:

Constructing free-body diagrams to visualize forces acting on an object.
Resolving forces into their horizontal and vertical components.
Applying the equilibrium conditions for static systems ( $\sum F_{x} = 0$ , $\sum F_{y} = 0$ , $\sum τ = 0$ ).
Analyzing dynamic equilibrium (objects moving at constant velocity with zero net force) as well as situations with acceleration.

In this subchapter, we do not introduce friction. Our focus is on ideal force conditions, tensions, and multiple-body configurations.

Introduction to free-body diagrams

A free-body diagram is a simplified sketch that shows all the forces acting on a single, isolated object. You use it to translate a physical situation into equations for motion or equilibrium.

The typical steps in drawing an FBD are:

Isolate the object: Imagine the object removed from its environment.
Identify all forces: List every force acting on the object. Common forces include:
- Weight, $m g$ , acting downward.
- Normal force, $N$ , acting perpendicular to the supporting surface.
- Applied forces, $F_{app}$ , from pushes or pulls.
- Tension, $T$ , in strings or ropes.
Choose a coordinate system: Pick axes that simplify the math (for example, one axis along an inclined surface).
Resolve forces into components: For forces not aligned with your axes, break them into components along the chosen directions.
Label and draw: Draw an arrow for each force, showing its direction and labeling its magnitude (or an unknown symbol).

Resolving forces into components

When a force acts at an angle, it’s usually easier to work with its components along your coordinate axes (often horizontal and vertical). Those components are what you plug into Newton’s second law in each direction.

Steps to resolve a force

Suppose a force $F$ acts on an object at an angle $θ$ from the horizontal. We can resolve $F$ into:

A horizontal component, $F_{x} = F cos θ,$
A vertical component, $F_{y} = F sin θ .$

Once you have $F_{x}$ and $F_{y}$ , you can write separate equations for the $x$ - and $y$ -directions.

Example: Resolving an applied force

Consider a force $F = 30 N$ applied at an angle $θ = 6 0^{\circ}$ above the horizontal. The components are:

$F_{x} = 30 cos 6 0^{\circ}, F_{y} = 30 sin 6 0^{\circ} .$

Since $cos 6 0^{\circ} = 0.5$ and $sin 6 0^{\circ} \approx 0.866$ , we get:

$F_{x} = 30 \times 0.5 = 15 N, F_{y} = 30 \times 0.866 \approx 25.98 N .$

You’d then use $F_{x}$ to analyze motion in the horizontal direction and $F_{y}$ to analyze motion in the vertical direction.

Resolving an applied force into components

Static equilibrium

An object is in static equilibrium when it is at rest and both the net force and the net torque on it are zero. In equation form:

$\sum F_{x} = 0, \sum F_{y} = 0, \sum τ = 0.$

These conditions let you solve for unknown forces (like tensions or normal forces) in a system.

Object on a horizontal surface

Consider an object of mass $m$ resting on a horizontal surface with an applied force that has both horizontal and vertical components. In equilibrium:

Vertical forces: $N + F_{y} = m g$ .
Horizontal forces: Any horizontal applied force must be balanced by an opposing force.

FBD of the object on the horizontal surface

Suspended object with two strings

An object of mass $m$ is suspended from the ceiling by two strings. The left string makes an angle $θ_{1}$ with the vertical, and the right string makes an angle $θ_{2}$ with the vertical. In static equilibrium, the forces satisfy:

Vertical balance: $T_{1} cos θ_{1} + T_{2} cos θ_{2} = m g$ .
Horizontal balance: $T_{1} sin θ_{1} = T_{2} sin θ_{2}$ .

From the horizontal equation,

$T_{1} = T_{2} \frac{sin θ _{2}}{sin θ _{1}} .$

Substitute this expression into the vertical balance equation to solve for the tensions.

FBD of an object suspended by two strings

Dynamic equilibrium

Dynamic equilibrium occurs when an object moves at a constant velocity (possibly non-zero), so the net force on it is zero:

$\sum F_{x} = 0 and \sum F_{y} = 0.$

Constant velocity means the acceleration is zero, so the force balance looks the same as in static equilibrium.

Constant velocity motion

Consider a sled moving across frictionless ice at a constant velocity while being pulled by a rope at an angle. Even though the sled is moving, the net force is zero.

Since the sled moves at constant velocity, the net force in all directions is zero:

$F_{net} = 0 ⟹ a = 0.$

Accelerated motion

When a box rests on a frictionless inclined plane, its weight can be resolved into two components:

A component perpendicular to the plane (balanced by the normal force)
A component parallel to the plane (which causes acceleration down the plane if nothing counters it)

Free-body diagram of a box on an inclined plane

Consider a box of mass $m$ on a frictionless incline that makes an angle $θ$ with the horizontal. The weight of the box is $m g$ , acting vertically downward. We resolve this weight into:

A component perpendicular to the incline: $m g cos θ$ ,
A component parallel to the incline: $m g sin θ$ .

Since the plane is frictionless, the only unbalanced force causing the box to accelerate is the parallel component.

Equilibrium in the perpendicular direction

For the box to remain in contact with the plane (that is, not accelerate through the surface), the normal force $N$ must balance the perpendicular component of the weight:

$N = m g cos θ .$

Acceleration along the incline

In the absence of friction, the net force acting along the plane is:

$F_{∥} = m g sin θ .$

Using Newton’s second law along the incline:

$m g sin θ = ma,$

which simplifies to:

$a = g sin θ .$

Example problem 1

Consider two blocks of masses $m_{1}$ and $m_{2}$ on a frictionless horizontal surface connected by a light string. A force $F$ is applied to $m_{1}$ , causing both blocks to accelerate together.

(a) Draw the free-body diagram for each block. (b) Derive the acceleration of the system and the tension in the string.

Solution:

(spoiler)

For Block 1 (mass $m_{1}$ ): The forces acting are the applied force $F$ to the right and the tension $T$ to the left.

Newton’s second law gives:

$F - T = m_{1} a .$

For Block 2 (mass $m_{2}$ ): The only horizontal force is the tension $T$ pulling it to the right:

$T = m_{2} a .$

Combine the equations: Substitute $T = m_{2} a$ into the first equation:

$F - m_{2} a = m_{1} a \Rightarrow F = (m_{1} + m_{2}) a .$

Thus, the acceleration is:

$a = \frac{F}{m _{1} + m _{2}} .$

And the tension is:

$T = m_{2} \cdot \frac{F}{m _{1} + m _{2}} .$

Example Problem 2

A $1.50$ kg puck glides on perfectly frictionless ice. It is simultaneously pulled by two massless ropes:

Rope 1 exerts a force of magnitude $F_{1} = 8.0$ N at an angle of $2 0^{\circ}$ north of east.
Rope 2 exerts a force of magnitude $F_{2} = 5.0$ N at an angle of $5 5^{\circ}$ south of east.

Draw a complete free-body diagram of the puck, labeling $F_{1}$ and $F_{2}$ .
Decompose each force into its components $F_{x}$ (eastward) and $F_{y}$ (northward). Write your results in terms of sines and cosines, then compute numerical values.
Using Newton’s second law, determine the puck’s acceleration $a$ . Give both the magnitude $a = ∣ a ∣$ and the direction $ϕ$ , measured as “degrees north of east.”
Suppose a third rope is tied to the puck so that it remains at rest (net acceleration zero). What must be the magnitude and direction of the third force $F_{3}$ to hold it in equilibrium?

Solution:

(spoiler)

Free-body diagram.

Free body diagram of the puck with F1 and F2

Resolve into components.

$F_{1} = F_{1 x} \hat{i} + F_{1 y} \hat{j}, F_{1 x} = F_{1} cos 2 0^{\circ}, F_{1 y} = F_{1} sin 2 0^{\circ},$

$F_{2} = F_{2 x} \hat{i} + F_{2 y} \hat{j}, F_{2 x} = F_{2} cos 5 5^{\circ}, F_{2 y} = - F_{2} sin 5 5^{\circ} .$

Numerically:
- $F_{1 x} = 8.0 cos 2 0^{\circ} \approx 7.52$ N
- $F_{1 y} = 8.0 sin 2 0^{\circ} \approx 2.74$ N
- $F_{2 x} = 5.0 cos 5 5^{\circ} \approx 2.87$ N
- $F_{2 y} = - 5.0 sin 5 5^{\circ} \approx - 4.10$ N
Net force and acceleration.

$F_{net, x} = F_{1 x} + F_{2 x} = 7.52 + 2.87 = 10.39 N,$

$F_{net, y} = F_{1 y} + F_{2 y} = 2.74 - 4.10 = - 1.36 N .$

With $m = 1.50$ kg,

$a_{x} = \frac{F _{net, x}}{m} = \frac{10.39}{1.50} \approx 6.93 m/ s^{2},$

$a_{y} = \frac{F _{net, y}}{m} = \frac{- 1.36}{1.50} \approx - 0.91 m/ s^{2} .$

Then

$a = a_{x}^{2} + a_{y}^{2} \approx 6.9 3^{2} + (- 0.91)^{2} \approx 6.98 m/ s^{2},$

$ϕ = arctan (\frac{a _{y}}{a _{x}}) \approx arctan (- 0.91/6.93) \approx - 7.4 5^{\circ} (or 7.4 5^{\circ} south of east) .$
Third force for equilibrium.

$F_{3} = - (F_{1} + F_{2}),$

so

$F_{3 x} = - 10.39 N, F_{3 y} = + 1.36 N .$

Magnitude and direction:

$F_{3} = (- 10.39)^{2} + (1.36)^{2} \approx 10.48 N,$

$θ_{3} = arctan (\frac{1.36}{10.39}) \approx 7.4 5^{\circ} (north of west) .$

Introduction to free-body diagrams

Visual tool: shows all forces on a single, isolated object
Steps:
- Isolate object, identify all forces (weight, normal, applied, tension)
- Choose coordinate system, resolve forces into components
- Draw and label all force vectors

Resolving forces into components

For force $F$ at angle $θ$ :
- Horizontal: $F_{x} = F cos θ$
- Vertical: $F_{y} = F sin θ$
Use components to write Newton’s laws in $x$ and $y$ directions

Static equilibrium

Conditions: $\sum F_{x} = 0$ , $\sum F_{y} = 0$ , $\sum τ = 0$
Object at rest: net force and net torque are zero
- Horizontal surface: $N + F_{y} = m g$ (vertical), horizontal forces must balance
- Two-string suspension:
  - Vertical: $T_{1} cos θ_{1} + T_{2} cos θ_{2} = m g$
  - Horizontal: $T_{1} sin θ_{1} = T_{2} sin θ_{2}$
  - $T_{1} = T_{2} \frac{s i n θ _{2}}{s i n θ _{1}}$

Dynamic equilibrium

Object moves at constant velocity: $\sum F_{x} = 0$ , $\sum F_{y} = 0$
Net force zero implies $a = 0$ (no acceleration)
Example: sled on frictionless ice, pulled at constant velocity

Accelerated motion on an incline

Box on frictionless incline:
- Resolve weight: $m g cos θ$ (perpendicular), $m g sin θ$ (parallel)
- Normal force: $N = m g cos θ$
- Net force down incline: $F_{∥} = m g sin θ$
- Acceleration: $a = g sin θ$

Multiple-body systems (example: two blocks with string)

For $m_{1}$ (pulled by $F$ ): $F - T = m_{1} a$
For $m_{2}$ : $T = m_{2} a$
System acceleration: $a = \frac{F}{m _{1} + m _{2}}$
String tension: $T = m_{2} \cdot \frac{F}{m _{1} + m _{2}}$

Vector addition and equilibrium (example: puck with two ropes)

Resolve each force into $x$ and $y$ components using sines and cosines
Net force: sum $x$ and $y$ components
Acceleration: $a_{x} = F_{net, x} / m$ , $a_{y} = F_{net, y} / m$
Magnitude: $a = a_{x}^{2} + a_{y}^{2}$ ; direction: $arctan (a_{y} / a_{x})$
For equilibrium: third force must exactly cancel net force (equal magnitude, opposite direction)

Key strategies

Always start with a clear free-body diagram
Resolve all forces into components along chosen axes
Apply equilibrium or Newton’s laws to relate forces and accelerations
Solve algebraically for unknowns

Next | 4.3 Friction and spring forces