Friction

Dynamics · FE Reference Handbook section

Dynamics

6 formulas

10 exam-style examples

~57 min

All Dynamics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Friction within Dynamics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what friction describes physically and when it applies.
State every one of the 6 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.

Lecture

Why this section exists. Friction is the part of Dynamics that lets you connect a particle or rigid body moving under known forces to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as kinematics first, then a work-energy or impulse-momentum shortcut. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: friction.
Capstone Studio instructional photograph

Dynamics — Friction: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a particle or rigid body moving under known forces. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 6 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

F ≤ µN	Quantity produced by "F ≤ µN" — read its definition and unit from the handbook line directly above the equation.
where µ	Quantity produced by "where µ = the coefficient of friction." — read its definition and unit from the handbook line directly above the equation.
F	Quantity produced by "F = µs N, at the point of impending slip" — read its definition and unit from the handbook line directly above the equation.
µs	Quantity produced by "µs = coefficient of static friction" — read its definition and unit from the handbook line directly above the equation.
µk	Quantity produced by "µk = coefficient of kinetic friction" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The Laws of Friction are
1. The total friction force F that can be developed is independent of the magnitude of the area of contact.
2. The total friction force F that can be developed is proportional to the normal force N.
3. For low velocities of sliding, the total frictional force that can be developed is practically independent of the sliding
velocity, although experiments show that the force F necessary to initiate slip is greater than that necessary to maintain
the motion.
The formula expressing the Laws of Friction is
In general
F < µs N, no slip occurring
Here,

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Weight, normal force and sliding friction — Friction

A 327 kg crate rests on a level floor with a friction coefficient of 0.45. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 988 N is applied.

Given

m = 327 kg
μ = 0.45
P = 988 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 327 (9.81) = 3, 208 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.45 (3, 208) = 1, 444 N$
Compare
$P = 988 N < F = 1, 444 N, so t h ecr a t es t a y s a t r es t$
Formula
$a = (P - F) / m$
Substituting
$a = (988 - 1, 444) /327 = - 1.393 m / s^{2}$

Answer: W = 3,208 N, F = 1,444 N, a = 0.00 m/s²

Why the other options are there

W = 327 N (mass reported as weight)
F = 147.2 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Example 2

Weight, normal force and sliding friction — Friction (2)

A 168 kg crate rests on a level floor with a friction coefficient of 0.30. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 1623 N is applied.

Given

m = 168 kg
μ = 0.30
P = 1623 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 168 (9.81) = 1, 648 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.30 (1, 648) = 494.4 N$
Compare
$P = 1623 N > F = 494.4 N, so t h ecr a t es l i d es$
Formula
$a = (P - F) / m$
Substituting
$a = (1623 - 494.4) /168 = 6.718 m / s^{2}$

Answer: W = 1,648 N, F = 494.4 N, a = 6.72 m/s²

Why the other options are there

W = 168 N (mass reported as weight)
F = 50.4 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Example 3

Weight, normal force and sliding friction — Friction (3)

A 228 kg crate rests on a level floor with a friction coefficient of 0.45. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 1323 N is applied.

Given

m = 228 kg
μ = 0.45
P = 1323 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 228 (9.81) = 2, 237 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.45 (2, 237) = 1, 007 N$
Compare
$P = 1323 N > F = 1, 007 N, so t h ecr a t es l i d es$
Formula
$a = (P - F) / m$
Substituting
$a = (1323 - 1, 007) /228 = 1.388 m / s^{2}$

Answer: W = 2,237 N, F = 1,007 N, a = 1.39 m/s²

Why the other options are there

W = 228 N (mass reported as weight)
F = 102.6 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Example 4

Weight, normal force and sliding friction — Friction (4)

A 69 kg crate rests on a level floor with a friction coefficient of 0.45. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 201 N is applied.

Given

m = 69 kg
μ = 0.45
P = 201 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 69 (9.81) = 676.9 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.45 (676.9) = 304.6 N$
Compare
$P = 201 N < F = 304.6 N, so t h ecr a t es t a y s a t r es t$
Formula
$a = (P - F) / m$
Substituting
$a = (201 - 304.6) /69 = - 1.501 m / s^{2}$

Answer: W = 676.9 N, F = 304.6 N, a = 0.00 m/s²

Why the other options are there

W = 69 N (mass reported as weight)
F = 31.1 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Example 5

Weight, normal force and sliding friction — Friction (5)

A 353 kg crate rests on a level floor with a friction coefficient of 0.25. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 1257 N is applied.

Given

m = 353 kg
μ = 0.25
P = 1257 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 353 (9.81) = 3, 463 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.25 (3, 463) = 865.7 N$
Compare
$P = 1257 N > F = 865.7 N, so t h ecr a t es l i d es$
Formula
$a = (P - F) / m$
Substituting
$a = (1257 - 865.7) /353 = 1.108 m / s^{2}$

Answer: W = 3,463 N, F = 865.7 N, a = 1.11 m/s²

Why the other options are there

W = 353 N (mass reported as weight)
F = 88.3 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Example 6

Weight, normal force and sliding friction — Friction (6)

A 212 kg crate rests on a level floor with a friction coefficient of 0.45. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 904 N is applied.

Given

m = 212 kg
μ = 0.45
P = 904 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 212 (9.81) = 2, 080 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.45 (2, 080) = 935.9 N$
Compare
$P = 904 N < F = 935.9 N, so t h ecr a t es t a y s a t r es t$
Formula
$a = (P - F) / m$
Substituting
$a = (904 - 935.9) /212 = - 0.150 m / s^{2}$

Answer: W = 2,080 N, F = 935.9 N, a = 0.00 m/s²

Why the other options are there

W = 212 N (mass reported as weight)
F = 95.4 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Example 7

Weight, normal force and sliding friction — Friction (7)

A 326 kg crate rests on a level floor with a friction coefficient of 0.40. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 1603 N is applied.

Given

m = 326 kg
μ = 0.40
P = 1603 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 326 (9.81) = 3, 198 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.40 (3, 198) = 1, 279 N$
Compare
$P = 1603 N > F = 1, 279 N, so t h ecr a t es l i d es$
Formula
$a = (P - F) / m$
Substituting
$a = (1603 - 1, 279) /326 = 0.993 m / s^{2}$

Answer: W = 3,198 N, F = 1,279 N, a = 0.99 m/s²

Why the other options are there

W = 326 N (mass reported as weight)
F = 130.4 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Example 8

Weight, normal force and sliding friction — Friction (8)

A 197 kg crate rests on a level floor with a friction coefficient of 0.50. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 1101 N is applied.

Given

m = 197 kg
μ = 0.50
P = 1101 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 197 (9.81) = 1, 933 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.50 (1, 933) = 966.3 N$
Compare
$P = 1101 N > F = 966.3 N, so t h ecr a t es l i d es$
Formula
$a = (P - F) / m$
Substituting
$a = (1101 - 966.3) /197 = 0.684 m / s^{2}$

Answer: W = 1,933 N, F = 966.3 N, a = 0.68 m/s²

Why the other options are there

W = 197 N (mass reported as weight)
F = 98.5 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Example 9

Weight, normal force and sliding friction — Friction (9)

A 127 kg crate rests on a level floor with a friction coefficient of 0.40. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 562 N is applied.

Given

m = 127 kg
μ = 0.40
P = 562 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 127 (9.81) = 1, 246 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.40 (1, 246) = 498.3 N$
Compare
$P = 562 N > F = 498.3 N, so t h ecr a t es l i d es$
Formula
$a = (P - F) / m$
Substituting
$a = (562 - 498.3) /127 = 0.501 m / s^{2}$

Answer: W = 1,246 N, F = 498.3 N, a = 0.50 m/s²

Why the other options are there

W = 127 N (mass reported as weight)
F = 50.8 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Example 10

Weight, normal force and sliding friction — Friction (10)

A 177 kg crate rests on a level floor with a friction coefficient of 0.50. Compute the weight, the friction force at impending motion, and the acceleration when a horizontal push of 602 N is applied.

Given

m = 177 kg
μ = 0.50
P = 602 N
g = 9.81 m/s²

Find

Weight W, friction force F, and resulting acceleration

Start with the thinking

Weight is a force, mass is not: W = m g. On a level surface the normal force equals the weight.
Motion begins only when the applied push exceeds the maximum static friction.

Step-by-step solution

Formula
$W = m g$
Substituting
$W = 177 (9.81) = 1, 736 N$
Formula
$F = μ N = μ W$
Substituting
$F = 0.50 (1, 736) = 868.2 N$
Compare
$P = 602 N < F = 868.2 N, so t h ecr a t es t a y s a t r es t$
Formula
$a = (P - F) / m$
Substituting
$a = (602 - 868.2) /177 = - 1.504 m / s^{2}$

Answer: W = 1,736 N, F = 868.2 N, a = 0.00 m/s²

Why the other options are there

W = 177 N (mass reported as weight)
F = 88.5 N (g omitted)

Reference: FE Reference Handbook — Dynamics → Friction

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a particle or rigid body moving under known forces, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Friction contains 6 relations; you must be able to find this page in under 15 seconds.
Exam style: kinematics first, then a work-energy or impulse-momentum shortcut.
Unit rule: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Friction

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

F≤µN

whereµ=thecoefficientoffriction.

F=µsN,atthepointofimpendingslip

F=µkN,whenslipisoccurring

µs=coefficientofstaticfriction

µk=coefficientofkineticfriction

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section