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Unit 2.3 | Drawing FBD (Force Diagrams) Quickly and Accurately

Vector graphic of a car in motion with force arrows showing direction and magnitude, accompanied by equations representing Newton's three laws
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Jason Kuma

Writer | Coach | Builder | Fremont, CA

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This will be a quick and fun lesson. You will learn how to draw FBDs or force diagrams. An FBD is incredibly important tool to solving hard force problems, as you will see in the next lesson. Before moving on to the next lesson Make sure to use all the resources below to master drawing FBDs.

This course article contains videos that can only be accessed once enrolled in the Learn AP Physics from Scratch Course.

Unit 2 Breakdown

You are on Lesson 3 of 8:

  1. Unit 2.1 | Understanding and Applying Newton’s Law in Depth
  2. Unit 2.2 | Common Linear Forces, Equations, and Misconceptions
  3. Unit 2.3 | Drawing and Understanding Force Body Diagrams [Current Lesson]
  4. Unit 2.4 | Deriving Equations to Solve Linear Force Problems
  5. Unit 2.5 | Advanced Force Problems – Tension and Elevators
  6. Unit 2.6 | Advanced Force Problems – Ramps (Inclines)
  7. Unit 2.7 | Advanced Force Problems – Pulley System
  8. Unit 2.8 | Advanced Force Problems – Multi-Body System

In this lesson you will learn: 

  • What are force body diagrams (FBD)
  • Steps to draw an FBD
  • Direction of net force and acceleration
  • Practice drawing FBDs of complex situations
  • Common Misconceptions

What are FBDs?

FBD is short for “Force Body Diagram.”

It is a drawing, like the one below, showing all forces acting ON a object.

An FBD of two tensions forces and a weight force
FBD of two tensions forces and a weight force.

We use FBDs to visualize and simplify complex problems.

While this may seem simple, there’s a lot of important rules and concepts to apply on FBDs.

The best way to learn is to read the steps below, then do a lot of practice.

Steps for Drawing FBDs

As we go through the steps, keep the following example situation in mind. We will refer back to this and see how the rules apply: A box being pushed to the right at a constant speed across the floor. 

  1. Draw a DOT to represent an object.
    • For the situation above you would draw a small dot to represent the box.
  2. Identify, then draw all forces acting on the object according to the force rules (cover in 2.2)
    • In this case, there is a very clear pushing force. On your dot, you would draw an arrow to the right and label it Fpush or Fapplied.
  3. Check for hidden forces.
    • In this case the box is moving at constant speed. This means that there should be zero acceleration and zero net force. So another force is needed to cancel out Fpush. This would be the force of friction (f) acting equal but opposite to Fpush.

Let’s try some harder problems together. Be sure to read the important rules first.

5 Important Rules

  1. Generally all objects will have a weight force. Weight ALWAYS points straight down.
  2. The direction of the net force is the direction of your acceleration
  3. Constant speed means 0 acceleration and thus 0 net force
  4. Kinetic friction always acts OPPOSITE to the object’s direction of motion.
  5. Normal force is drawn perpendicular (90°) to the surface.

PS – FDBs

Below are some common situations you will encounter. Try drawing an FBD for these 8 problems. Then watch the short video for the explanations for each. 

  1. A basket is suspended by a rope from the ceiling.
  2. A rightward force is applied to a book to accelerate it across a rough desk.
  3. A car applies its brakes to slow down.
  4. A skydiver descends with constant velocity.
  5. A box rests on a sloped incline.
  6. An elevator is traveling up and slowing down 
  7. Five seconds after you kick a soccer ball in the air (air resistance is negligible)
  8. Two cables suspend a heavy neon sign at 45° angles

How many did you get right? If you got 7 or more right, you are ready to move on to the next lesson. 

If you need more practice, here are some FBD worksheets: [Enroll For Material Access].

Lesson 2.3 Recap

In this lesson we covered how to quickly and accurately draw an FBD.

You got to apply your knowledge to challenging practice questions then watched me solve them.

Lesson 2.4 Preview

In the next lesson we will use FBDs to to derive equations and solve interesting, real world problems. This is generally what most students find hard in Physics. But we will break it down and make it super easy to understand. 

Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA

Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy 

Unit 5 – Momentum 

Unit 6 – Torque 

Unit 7 – Oscillations 

Unit 8 – Fluids

Reading Key

LRN
RE
PS
PQ
Black
White
Blue
Orange

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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