Free-Body Diagrams and Friction: From Forces to Equations

1. Choose the system before the forces

The system is the object or group of objects you choose to study. Here, choose the box alone. Draw a boundary around it mentally: Earth, the floor and the rope are outside. Each can exert an external force on the box. Ask “What is pushing or pulling this object?” rather than “Which way is it moving?”

Let $m$ denote the box’s mass in kilograms and $g$ the magnitude of gravitational acceleration near Earth, in metres per second squared. A force is measured in newtons, abbreviated $\mathrm{N}$; the italic variable $N$ below names a particular force, not its unit. Weight $mg$ is Earth’s pull, vertically downward. The normal force $N$ is the surface’s push, perpendicular to that surface. Friction, labelled $f$ in the first diagram, acts along the contact surface. The rope’s pull $F$ follows the rope. “Normal” means perpendicular; it does not mean “equal to weight.”

A larger system is possible, but changes the force inventory. If you include both box and floor, their mutual contact forces become internal. You cannot switch systems halfway through an equation.

Use the same order each time

Choose the object → identify contacts and sliding → draw external forces → choose axes → write Newton’s law → test friction and contact → solve and check.

2. Separate force from motion

Use the stationary floor or room as an approximately inertial reference frame: a frame in which a force-free object moves at constant velocity. Newton’s second law connects the net external force to acceleration. The subscripts $x$ and $y$ identify the two chosen coordinate axes. The quantities $a_x$ and $a_y$ are the acceleration components; $F_x$ and $F_y$ are force components. The symbol $\sum$ means add the signed force components along the chosen axis:

$$\sum F_x=ma_x,\qquad \sum F_y=ma_y$$

Acceleration describes a change in velocity. A box moving right speeds up if acceleration points right, slows down if acceleration points left, and keeps constant velocity if acceleration is zero. Newton’s first law is the zero-net-force case; motion does not require a continuing net force.

Newton’s third law pairs two forces on different objects. The floor pushes up on the box while the box pushes down on the floor. Only the first belongs on the box’s diagram. The box’s weight and normal force are not a third-law pair: both act on the box.

A: do not invent a force to explain velocity. B: name the recipient of each force. C: resolve an angled force into components for calculation, but do not add the original force again. Components describe the same interaction.

Choose friction from the contact motion

The subscript $s$ means static and $k$ means kinetic; $\mu_s$ and $\mu_k$ are the corresponding friction coefficients for the pair of contacting surfaces. They are dimensionless numbers supplied by measurement or by the problem. The subscript $\max$ means the largest permitted value.

For surfaces sliding past one another, kinetic friction has magnitude $f_k=\mu_kN$ and opposes their relative sliding. For contact without sliding, static friction opposes the tendency to slip and takes the value needed to prevent it. Its magnitude $f_s$ satisfies:

$$0\le f_s\le\mu_sN,\qquad f_{s,\max}=\mu_sN$$

To test rest on a stationary floor, assume equilibrium, calculate the required friction, then compare it with the limit. A requirement above the limit disproves rest. Equality is limiting equilibrium; it does not by itself guarantee motion. Static friction can also accelerate an object without slipping, as the belt practice will show.

3. Worked example 1: A horizontal pull

A 5.0 kg box is already sliding right on a level, stationary floor. A horizontal pull of 18 N acts right. The kinetic coefficient is 0.20. Find the acceleration.

Throughout the numerical problems, use $g=9.8\ \mathrm{m/s^2}$. Model each box as non-tipping, ignore air resistance, and use constant friction coefficients. Keep unrounded values in calculations; final accelerations are shown to two decimals for comparison, not as a claim of measurement precision.

Model and solve

Choose right as positive $x$ and up as positive $y$. Contact with the level floor constrains vertical acceleration to zero. Start vertically because friction depends on the unknown normal force:

$$\begin{aligned}N-mg&=0\\\Rightarrow\quad N&=(5.0)(9.8)\\&=49\ \mathrm{N}\end{aligned}$$

“Already sliding” selects kinetic friction. Its direction is left because the box slips right relative to the floor:

$$f_k=\mu_kN=0.20(49)=9.8\ \mathrm{N}$$

$$\begin{aligned}F-f_k&=ma_x,\\a_x&=\frac{18-9.8}{5.0}\\&=1.64\ \mathrm{m/s^2}.\end{aligned}$$

Interpret and check

The positive acceleration is rightward, so this right-moving box speeds up. Units agree: $\mathrm{N/kg}=\mathrm{m/s^2}$. Without friction, acceleration would be $18/5.0=3.60\ \mathrm{m/s^2}$; a smaller result is sensible. Setting $\mu_k=0$ in the symbolic equation recovers that frictionless limit.

If the pull falls to 9.8 N while the box is still sliding, acceleration becomes zero and the box continues at constant velocity. Equal opposing forces do not imply that an object is at rest; they imply that its velocity is not changing.

4. Worked example 2: Will it move

A 6.0 kg box begins at rest on a level floor. Here $\mu_s=0.35$ and $\mu_k=0.25$. Compare two separate trials, each starting from rest: a horizontal pull of 16 N, then a pull of 25 N.

The vertical equation gives $N=mg=58.8\ \mathrm{N}$. Before choosing an acceleration, calculate the largest available static friction:

$$f_{s,\max}=0.35(58.8)=20.58\ \mathrm{N}$$

Trial A: Test the assumption of rest

Horizontal equilibrium requires $16-f_s=0$, hence $f_s=16\ \mathrm{N}$ left. That lies below 20.58 N, so static friction can supply it. The box stays at rest and acceleration is zero.

Using $f_s=\mu_sN=20.58\ \mathrm{N}$ here would predict a net force left even though the modest pull cannot make the box slide. The mistake is replacing an adjustable force with its maximum possible value.

Trial B: Reject rest and change the model

Rest would require 25 N of static friction, exceeding the limit. The box begins slipping right. Now use kinetic friction, which is a different contact condition:

$$f_k=0.25(58.8)=14.7\ \mathrm{N}$$

$$a_x=\frac{25-14.7}{6.0}\approx1.72\ \mathrm{m/s^2}$$

Subtracting the static limit from 25 N would give the wrong sliding acceleration. Once sliding begins, the kinetic model determines friction. At the threshold itself, the ideal static model still permits equilibrium.

What the graph does and does not say

The graph describes gradually increasing pull from an initially resting box. It is not a universal friction-versus-force curve independent of history. An already sliding box can have kinetic friction even when the current pull is below the static threshold. These friction laws approximate dry contact; real surfaces, rolling resistance and fluid drag may need different models.

5. Worked example 3: Does lifting help

Return to the already sliding 5.0 kg box, 18 N pull and $\mu_k=0.20$. Tilt the pull $30^\circ$ above horizontal. Predict the result before calculating: lifting reduces friction, but it also reduces the horizontal part of the pull.

With angle $\theta$ above horizontal, resolve the pull as $F_x=F\cos\theta$ and $F_y=F\sin\theta$. The box still has zero vertical acceleration while it remains on the floor:

$$\begin{aligned}&N+F\sin\theta-mg=0,\\&N=mg-F\sin\theta.\end{aligned}$$

$$\begin{aligned}N&=49-18\sin30^\circ\\&=40\ \mathrm{N},\\f_k&=0.20(40)\\&=8.0\ \mathrm{N}.\end{aligned}$$

$$a_x=\frac{F\cos\theta-\mu_k(mg-F\sin\theta)}{m}$$

$$a_x=\frac{18\cos30^\circ-8.0}{5.0}\approx1.52\ \mathrm{m/s^2}$$

The acceleration is slightly smaller than $1.64\ \mathrm{m/s^2}$ for the horizontal pull. Friction decreased by 1.8 N, but horizontal pull decreased by about 2.41 N. The competing changes explain the result; “less friction” alone cannot settle the question.

For an 18 N push 30° downward, the vertical component adds to the floor’s load: $N=49+9=58\ \mathrm{N}$. Friction becomes 11.6 N and acceleration is about $0.80\ \mathrm{m/s^2}$ rightward.

Check the contact assumption

A floor can push upward but cannot pull the box downward. Require $N\ge0$. If a calculation gives $F\sin\theta>mg$, the negative normal force signals loss of contact; the zero-vertical-acceleration model must be replaced. Here 18 N is smaller than the 49 N weight, so no upward angle can lift the box off the floor.

As $\theta\to0$, the formula returns the horizontal-pull result. As $\mu_k\to0$, it becomes $a_x=F\cos\theta/m$, which is the horizontal acceleration with no friction.

6. Practice the decisions

Draw a diagram and write symbolic equations before inserting numbers. For every answer, state the force directions and the contact condition that justifies your friction choice. Try each problem before revealing its explained answer below.

Guided practice: A pull that disappears

A 4.0 kg box slides right on a level stationary floor, pulled horizontally right by 9.0 N. Let $\mu_k=0.15$.

1. List the four external forces. Which fact sets $a_y=0$? Use it to find $N$.

2. Find kinetic friction, then write the signed horizontal force equation.

3. Find the acceleration and explain whether the box speeds up.

4. Remove the pull while the box is still sliding right. Find its new acceleration. What changes when the box reaches rest?

Independent 1: A box on a slope

A 3.0 kg box is released from rest on a stationary 25° incline. Let $\mu_s=0.40$ and $\mu_k=0.15$. Does it stay at rest? If it slides, find its acceleration. Choose axes along and perpendicular to the slope. The weight components are $mg\sin\theta$ down the slope and $mg\cos\theta$ into it; these replace the weight in component equations.

Independent 2: A moving belt

A 2.0 kg box initially shares a level conveyor belt’s velocity. The belt then accelerates right at $1.2\ \mathrm{m/s^2}$; the box does not slip. No other horizontal forces act and $\mu_s=0.30$. Which way is friction on the box? Find its magnitude, check that no slipping is consistent with the friction limit, and determine the maximum belt acceleration compatible with no slip.

Before you look at the answers

For the slope, compare required static friction with available static friction. For the belt, ask which force can produce the box’s rightward acceleration. An object can move through the room while its contact remains static.

7. Explained answers and a deeper connection

Quick check: What changes while the box slides uphill?

A 3.0 kg box now slides uphill on the same stationary 25° ramp after a brief shove has ended. Only gravity and the ramp’s contact forces act. Use $\mu_s=0.40$, $\mu_k=0.15$ and downhill as positive. Which way does kinetic friction act? Is the downhill acceleration still $2.81\ \mathrm{m/s^2}$?

Optional extension: Choose the pulling angle

Here $a(\theta)$ is the horizontal acceleration with the pull at angle $\theta$, and $a(0)$ is the horizontal-pull result. At fixed force magnitude and continued rightward sliding, compare them:

$$\begin{aligned}a(\theta)-a(0)&=\frac{F}{m}\bigl(\cos\theta\\&\qquad+\mu_k\sin\theta-1\bigr)\end{aligned}$$

Lifting helps when their difference is positive. Rearrange the bracket as $\mu_k\sin\theta>1-\cos\theta$. For $0<\theta<90^\circ$, divide by the positive $\sin\theta$, then use $(1-\cos\theta)/\sin\theta=\tan(\theta/2)$. Thus, with maintained contact, lifting helps exactly when $\tan(\theta/2)<\mu_k$. With $\mu_k=0.20$, the benefit ends at about $22.62^\circ$. The function $\arctan$ returns the angle whose tangent is the given number. Writing $\cos\theta+\mu_k\sin\theta=\sqrt{1+\mu_k^2}\cos(\theta-\arctan\mu_k)$ shows the best angle is about $11.31^\circ$, giving $a\approx1.71\ \mathrm{m/s^2}$. Check contact again before using that optimum in another problem.

Keep this test for your next problem

Can you name the agent behind each force, justify the friction model, and explain why the direction and size of your acceleration make sense? If one answer is unclear, revisit the setup before doing more algebra.

Further reading

• OpenStax 5.3 Newton’s second law
• OpenStax 5.5 Newton’s third law
• OpenStax 5.7 Drawing free body diagrams
• OpenStax 6.2 Friction

In a Nablio mechanics tutoring session, we can identify the forces, draw the diagram and write the equations together, or check a step you have already tried.