Friction is the force that resists relative motion (or attempted motion) between surfaces in contact. It's what lets you walk, drive a car, and hold objects. Without friction, life would be... slippery.
Understanding the two types of friction — static and kinetic — is essential for AP Physics 1 dynamics problems.
What Is Friction?
Friction is a contact force that acts parallel to the contact surface and opposes relative motion (or the tendency of relative motion) between two surfaces.
Two Types of Friction
Type
Symbol
When It Acts
Direction
Static friction (fs)
fs
Object is NOT sliding
Opposes the tendency to slide
Kinetic friction (fk)
fk
Object IS sliding
Opposes the actual sliding motion
Key Differences
Property
Static Friction
Kinetic Friction
Object moving?
No
Yes (sliding)
Magnitude
Variable: 0≤fs≤μs
Static Friction in Detail
Static friction keeps an object from starting to move. It's a self-adjusting force:
Push a box with 5 N → fs=5 N (box stays still)
Push with 10 N → fs= N (box stays still)
Kinetic Friction in Detail
Once an object starts sliding, kinetic friction takes over.
fk=μkN
Key Properties
is (doesn't depend on speed in the AP model)
Static vs. Kinetic Friction Concepts 🎯
Friction Calculations 🧮
A 20 kg box sits on a horizontal surface. μs=0.50, μk=0.40, m/s².
Classify the Friction 🔍
Exit Quiz — Static vs. Kinetic Friction ✅
Part 2: Coefficient of Friction
📐 Friction Equations
Part 2 of 7 — Friction
Now that we understand the two types of friction, let's master the mathematical relationships. These equations are used in nearly every dynamics problem on the AP exam.
fs≤μsNf
Part 3: Friction on Flat Surfaces
📋 Free Body Diagrams with Friction
Part 3 of 7 — Friction
Adding friction to FBDs requires care. You must determine the direction of friction (it depends on the situation!) and correctly write Newton's Second Law equations for both axes.
Drawing Friction on FBDs
Step 1: Determine the Type
Is the object sliding? → Kinetic friction (fk=μk)
Part 4: Friction & Applied Forces
📐 Friction with Angled Forces
Part 4 of 7 — Friction
When you pull or push an object at an angle, the vertical component of your force changes the normal force, which in turn changes the friction force. This is a very common AP Physics 1 scenario.
Pulling at an Angle Above Horizontal
A force F is applied at angle θ above horizontal to a block on a rough surface.
Force Components
Fx (horizontal, in direction of motion)
Part 5: Air Resistance & Drag
🔬 Finding μ Experimentally
Part 5 of 7 — Friction
How do physicists and engineers actually measure the coefficient of friction? This part covers the classic experimental methods — which are also common AP lab questions.
Method 1: Horizontal Pull
Setup
Place a block on a horizontal surface. Attach a spring scale and pull horizontally, gradually increasing the force.
Finding μs
Record the force when the block just starts to move. That's .
Part 6: Problem-Solving Workshop
🛠️ Problem-Solving Workshop
Part 6 of 7 — Friction
This workshop brings together all friction concepts: static vs. kinetic, FBDs with friction, angled forces, and experimental methods. Work through these problems systematically using the FBD → equations → solve approach.
Friction Problem Strategy
Step 1: Determine the Type of Friction
Is the object sliding? → Kinetic (fk=μkN)
Part 7: Synthesis & AP Review
🎯 Synthesis & AP Review
Part 7 of 7 — Friction
Congratulations on completing the Friction unit! This final lesson reviews every key idea, connects friction to the broader dynamics framework, and tests you with AP-style questions.
Concept Map: Friction in Dynamics
Force→FBD→Fnet=ma
Key Friction Equations
N
Constant: fk=μkN
Adjusts to match applied force?
Yes (up to a maximum)
No (constant value)
Coefficient
μs (usually larger)
μk (usually smaller)
10
Push with 15 N → fs=15 N (box stays still)
Push with 20 N → fs,max is exceeded → box starts moving!
The Inequality
fs≤μsN
fs matches the applied force up to a maximum value
The maximum is fs,max=μsN
Once the applied force exceeds fs,max, the object starts to move
Static Friction Can Point in Any Direction
Common misconception: friction always opposes motion. Static friction opposes the tendency of motion:
A box on a truck accelerating forward: static friction points forward (prevents the box from sliding backward relative to the truck)
fk opposes the direction of sliding (velocity relative to the surface)
μk<μs for the same surfaces → it's easier to keep something moving than to start it moving
Why μk<μs?
When surfaces are stationary relative to each other, microscopic bonds form between surface irregularities. Once sliding begins, these bonds are continuously broken before they can fully form, reducing the friction force.
Typical Coefficients
Surfaces
μs
μk
Rubber on dry concrete
1.0
0.8
Wood on wood
0.5
0.3
Ice on ice
0.1
0.03
Steel on steel (dry)
0.6
0.4
Teflon on steel
0.04
0.04
g=10
What is the maximum static friction force (in N)?
If the box is sliding, what is the kinetic friction force (in N)?
You push with 60 N horizontally. Does the box move? What is the friction force (in N)?
k
=
μkN
The Static Friction Inequality
fs≤μsN
What This Means
fs = actual static friction force (what it is right now)
μs = coefficient of static friction (dimensionless, depends on surfaces)
N = normal force (perpendicular contact force)
μsN = maximum static friction force
The Inequality Is Important!
The ≤ sign means static friction can be anywhere from zero up toμsN:
0≤fs≤μsN
Applied Force F
Static Friction fs
Motion?
F<μ
When to Use fs=μsN
Only when the problem says:
"On the verge of sliding"
"About to move"
"Maximum static friction"
"What is the largest force before the object moves?"
The Kinetic Friction Equation
fk=μkN
What This Means
fk = kinetic friction force (constant while sliding)
μk = coefficient of kinetic friction
N = normal force
Key Properties
fk is constant — it doesn't depend on speed (AP model)
fk is always less than — since
Normal Force Affects Friction
Since f=μN, anything that changes the normal force changes friction:
Situation
Normal Force
Friction
Flat surface, no extra forces
N=mg
f=μmg
Pushing down at angle θ
Using Friction in F=ma Problems
Horizontal Surface — Constant Velocity
If an object slides at constant velocity (a=0):
Fapplied−fk=0Fapplied=μkN=μk
Horizontal Surface — Accelerating
If an object slides with acceleration a:
Fapplied−fk=ma
Example
A 10 kg box is pushed with 60 N on a surface where μk=0.30. (g=10 m/s²)
fk=μkmg=0.30×10
Friction Equation Concepts 🎯
Friction Equation Practice 🧮
Use g=10 m/s² for all problems.
A 15 kg block on a surface (μk=0.20) is pushed with 50 N horizontally. What is the acceleration (in m/s²)?
A 10 kg block slides on a surface (μk=0.50). What horizontal force is needed for the block to slide at constant velocity (in N)?
A 25 kg box sits on a surface with μs=0.60. What is the maximum horizontal push before it starts moving (in N)?
Round all answers to 3 significant figures.
Equation Selection 🔍
Exit Quiz — Friction Equations ✅
N
Is the object stationary? → Static friction (fs≤μsN)
Step 2: Determine the Direction
Kinetic friction: Opposes the direction of sliding (velocity relative to surface)
Static friction: Opposes the tendency to slide
Step 3: Draw the Arrow
Friction is parallel to the surface
At the contact point, pointing in the direction determined in Step 2
Complete FBD for a Block Pushed Across a Rough Floor
Forces:
W=mg downward
N upward (perpendicular to surface)
Fapplied in the direction of push
fk opposite to the direction of motion (along the surface)
FBD Examples with Friction
Example 1: Block Pushed Right, Sliding Right
Weight: down
Normal: up
Applied force: right
Kinetic friction: left (opposes rightward sliding)
Example 2: Block Sliding Right with NO Applied Force (Slowing Down)
Weight: down
Normal: up
Kinetic friction: left (opposes rightward motion)
No applied force!
Net force is to the left → block decelerates
Example 3: Box on a Truck Accelerating Forward (Box Not Sliding)
Weight: down
Normal: up
Static friction: forward (prevents box from sliding backward relative to the truck)
Key insight: Static friction can point in ANY direction along the surface. It points in whatever direction is needed to prevent relative motion.
Example 4: Block on a Ramp (Not Sliding)
Weight: straight down
Normal: perpendicular to ramp surface (tilted)
Static friction: up the ramp (prevents sliding down)
Writing Newton's Second Law with Friction
Horizontal Surface, Object Sliding Right
x-direction (horizontal):∑Fx=Fapp−fk=max
y-direction (vertical):∑Fy=N−mg=0⇒N
Combining:
Fapp−μkmg=ma
What If No Applied Force? (Object Sliding and Slowing)
∑Fx=−fk=ma
This is powerful: A sliding object on a horizontal surface decelerates at μkg, regardless of mass!
Example
A box slides across a surface with μk=0.30 at initial speed 12 m/s. How far does it slide?
a=−μkg=−0.30×9.8=−2.94 m/s
FBD with Friction 🎯
FBD with Friction — Calculations 🧮
A 5 kg block slides to the right on a surface with μk=0.40. No other horizontal force acts. What is the block's acceleration (magnitude, in m/s²)? Use g=10 m/s².
The block from #1 has an initial speed of 10 m/s. How far does it slide before stopping (in m)?
A 12 kg box is pushed with 80 N across a rough floor (μk=0.30). What is the acceleration (in m/s²)?
Round all answers to 3 significant figures.
Friction Direction Practice 🔍
Exit Quiz — FBDs with Friction ✅
=
Fcosθ
Fy=Fsinθ (vertical, upward)
y-direction (no vertical acceleration):
N+Fsinθ−mg=0N=mg−Fsinθ
Friction Force
fk=μkN=μk(mg−Fsinθ)
Pulling up reduces the normal force, which reduces friction. This is why it's easier to pull a suitcase at an angle than to push it!
x-direction:
Fcosθ−fk=maFcosθ−μk(mg−Fsinθ)=ma
Example
Pull a 20 kg box with 80 N at 30° (μk=0.25, g=10 m/s²):
N=200−80sin30°=200−40=160 N
fk=0.25×160=40 N
Fx=80cos30°=69.3 N
a=(69.3−40)/20=29.3/20=1.47 m/s²
Pushing at an Angle Below Horizontal
A force F is applied at angle θ below horizontal (pushing downward and forward).
Force Components
Fx=Fcosθ (horizontal, forward)
Fy=−Fsinθ (vertical, downward)
y-direction:
N−mg−Fsinθ=0N=mg+
Friction Force
fk=μkN=μ
Pushing down increases the normal force, which increases friction. More effort wasted fighting friction!
Comparing Pull vs. Push
For the same force magnitude and angle, pulling upward is always more efficient:
Method
Normal Force
Friction
Horizontal Acceleration
Pull up at θ
mg−Fsinθ
Lower
Higher
Push down at θ
The Optimal Pulling Angle
There's actually an optimal angle that maximizes acceleration (or minimizes the force needed to move at constant velocity).
For Constant Velocity (a=0):
Fcosθ=μk(mg−Fsinθ)
Solving for F:
F=cosθ+μk
To minimize F, take dF/dθ=0:
θoptimal=tan−1(μk)
Example
For μk=0.40: θopt
This means pulling at about 22° above horizontal requires the least force to keep the object moving at constant velocity.
This explains why suitcase handles are angled — it's not just for comfort; it's physics!
Angled Force with Friction 🎯
Angled Force Calculations 🧮
A 10 kg block on a rough surface (μk=0.40) is pulled with 50 N. Use g=10 m/s².
If pulled horizontally, what is the friction force (in N)?
If pulled at 37° above horizontal (sin37°=0.60, cos37°=0.80), what is the normal force (in N)?
For the 37° pull, what is the acceleration (in m/s²)?
Round all answers to 3 significant figures.
Pull vs. Push Comparison 🔍
Exit Quiz — Friction with Angled Forces ✅
fs,max
μs=Nfs,max=mgfs,max
Finding μk
Pull the block at constant velocity. The applied force equals kinetic friction.
μk=Nfk=mgFconstant velocity
Example
A 2 kg block requires 9.8 N to just start moving and 7.8 N to slide at constant speed.
μs=2(9.8)9.8=19.69.8=0.50
μk=19.67.8=0.40
Method 2: The Incline Method
Finding μs with a Ramp
Place a block on a ramp. Slowly increase the angle until the block just begins to slide. Record this critical angleθc.
At the critical angle, the block is on the verge of sliding:
fs,max=mgsinθcN=mgcosθc
μs=Nf
μs=tanθc
Beautiful result:μs equals the tangent of the critical angle. No need to know the mass!
Finding μk with a Ramp
Give the block a push and find the angle where it slides at constant velocity:
μk=tanθconstant velocity
Why This Method Is Elegant
Mass cancels out — you don't need to measure it
Simple equipment: just a ramp and a protractor
Highly reproducible
Method 3: Graphical Analysis
Using an f vs. N Graph
Vary the normal force (by stacking masses) and measure friction at each value. Plot f vs. N.
f=μN
This is a linear relationship passing through the origin:
Slope = μ
No y-intercept (friction is zero when N=0)
Using an Acceleration Method
Push a block with known force F and measure acceleration a:
F−μkmg=maa=
Plot a vs. F/m:
Slope = 1
y-intercept = −μkg
So μk=−(y-intercept)/g
Sources of Error
Source
Effect
Surface not perfectly flat
Inconsistent results
Measuring force while accelerating
Overestimates static friction
Surface wears down during experiment
μ changes over time
Vibration during measurement
Premature sliding
Experimental Methods Quiz 🎯
Experimental Calculations 🧮
A 3 kg block on a horizontal surface requires 12 N to just start sliding. What is μs? (g=10 m/s²)
A block slides at constant velocity down a ramp inclined at 20°. What is μk? (tan20°=0.364, round to 3 significant figures)
In an experiment, doubling the mass of a block doubles the measured friction force. This confirms that friction is proportional to what variable?
Lab Reasoning 🔍
Exit Quiz — Measuring Friction ✅
Is the object stationary? → Static (fs≤μsN)
"On the verge"? → Static at maximum (fs=μsN)
Step 2: Find the Normal Force First!
The normal force is NOT always mg. Set up the y-direction equation:
N=mg±FyorN=mgcosθ (on incline)
Step 3: Calculate Friction
f=μN
Step 4: Apply F=ma in the Direction of Motion
∑Falong motion=ma
Worked Example
A 10 kg block is pulled at 30° above horizontal with 60 N across a rough surface (μk=0.20, g=10 m/s²).
Normal force:N=mg−Fsin30°=100−30=70 N
Friction:fk=0.20×70=14 N
Horizontal:Fcos30°−fk=ma52−14=10a⇒a=3.8 m/s2
Worked Example 2: Will It Slide?
A 5 kg block sits on a horizontal surface (μs=0.50, μk=0.30). You push horizontally with 20 N.
Check:fs,max=μsmg=0.50× N
Since 20<25: The block does NOT move. fs=20 N.
Now push with 30 N:
Since 30>25: The block starts moving!
Once moving: fk=μkmg=0.30×50= N
a=530−15=5
Worked Example 3: Constant Velocity Pulling
What force is needed to pull a 20 kg box at constant velocity at 25° above horizontal? (μk=0.35, g=10 m/s²)
At constant velocity (a=0):
Fcos25°=μk(mg−
Workshop Multiple Choice 🎯
Workshop Calculations 🧮
A 8 kg block is pushed with 60 N horizontally across a rough floor (μk=0.25, g=10 m/s²). What is the acceleration (in m/s²)?
A block slides to a stop in 4 seconds from an initial speed of 12 m/s on a horizontal surface. What is μk? (g=10 m/s²)
What angle must a ramp be tilted to for a block to be on the verge of sliding if μs=0.577? (in degrees, given tan−1(0.577)=)
Round all answers to 3 significant figures.
Quick Reasoning Checks 🔍
Exit Quiz — Friction Workshop ✅
Concept
Equation
When to Use
Static friction (max)
fs=μsN
Object on the verge of moving
Static friction (general)
fs≤μsN
Object not yet sliding
Kinetic friction
fk=μkN
Object is sliding
Normal force (flat)
N=mg
Horizontal surface, no vertical components
Normal force (pulled up)
N=mg−Fsinθ
Pulling at angle above horizontal
Normal force (pushed down)
N=mg+Fsinθ
Pushing at angle below horizontal
Stopping distance
d=2μkgv
Critical angle
tanθc=μs
Object on verge of sliding on an incline
Key Principles
μs>μk always — it's harder to start motion than maintain it
Friction is independent of contact area and independent of speed
Friction always opposes relative motion (or tendency of motion)
The normal force determines friction — anything that changes N changes friction
Conceptual Review 🧠
AP-Style Calculations 📝
A 6 kg block slides with μk=0.30 on a horizontal surface. What is the magnitude of its deceleration (m/s²)? (g=10 m/s²)
A block on a flat surface requires 24 N to start moving and 18 N to keep it moving at constant velocity. The block weighs 60 N. What is μs?
A 50 N horizontal force pushes a 10 kg block at constant velocity. What is μk? (g=10 m/s²)
Round all answers to 3 significant figures.
AP Reasoning Questions 🎯
Final Exit Quiz — Friction Unit ✅
s
N
fs=F (matches)
No
F=μsN
fs=μsN (at max)
On the verge
F>μsN
Static friction fails
Object begins sliding
fs,max
μk<μs
N is NOT always mg — it depends on other vertical forces