When an object sits on a tilted surface, gravity doesn't just pull it "down" — it has components along and perpendicular to the slope. Decomposing weight into these components is the foundation of every inclined plane problem.
The Tilted Coordinate System
On an incline, we tilt our x- and y-axes:
x-axis: parallel to the surface (positive = down the slope)
y-axis: perpendicular to the surface (positive = away from surface)
Why Tilt the Axes?
The normal force and friction are already along these tilted axes. The only force we need to decompose is gravity.
Weight Components
For an object on an incline of angle θ:
Wx=mgsinθ(along the slope, downhill)
Wy=mgcosθ(into the slope)
Remembering Which Is Which
At θ=0°: No component along the slope (sin0°=0), full weight into surface (cos0°=1). ✅
At : Full weight along the "slope" (), nothing into surface (). ✅
The sin component drives the object down the slope.
The cos component is balanced by the normal force.
The Geometry Trick
Why is the angle of the incline equal to the angle between W and the perpendicular-to-surface direction?
Imagine the incline angle θ at the base. The surface is tilted θ from horizontal. The perpendicular-to-surface direction is tilted from vertical. Since gravity points straight down (vertical), the angle between gravity and the perpendicular direction is also .
Check Your Understanding 🧠
Calculate the Components 🧮
A 20 kg block sits on a 37° incline (g=10 m/s², sin37°=0.60, cos37°=0.80).
What is the component of weight parallel to the slope (in N)?
Conceptual Checks 🔍
Exit Quiz ✅
Part 2: Component Decomposition
🎿 Frictionless Inclines
Part 2 of 7 — Inclined Planes
Before adding friction, let's master the simpler case: objects sliding on perfectly smooth inclines. The only forces are gravity and the normal force — and since the normal force is perpendicular to motion, only gravity's component along the slope accelerates the object.
Acceleration on a Frictionless Incline
Free Body Diagram
On a frictionless incline of angle θ:
N=mgcosθ (perpendicular — balanced)
(along the slope — unbalanced!)
Part 3: Frictionless Inclines
🧱 Inclines with Friction
Part 3 of 7 — Inclined Planes
Most real inclines have friction. Combining the incline weight components with friction forces is one of the most common AP Physics 1 problems. Let's master the approach.
FBD on an Incline with Friction
Forces on an object on an incline with friction:
Perpendicular to surface (y-axis):
N (away from surface)
mgcosθ (into surface)
These balance: N=
Part 4: Inclines with Friction
🔗 Connected Systems on Inclines
Part 4 of 7 — Inclined Planes
Many AP Physics problems involve objects connected by strings over pulleys, with one or both on inclines. These are classic Atwood-on-incline problems. The key: all connected objects share the same magnitude of acceleration.
The Classic Setup
Object on Incline + Hanging Mass
A block of mass m1 sits on an incline of angle θ, connected by a string over a frictionless pulley to a hanging mass m.
Part 5: Connected Objects on Inclines
📐 Acceleration and Velocity on Inclines
Part 5 of 7 — Inclined Planes
This lesson focuses on using kinematics on inclines — calculating how fast objects move, how far they travel, and how long it takes. We'll combine the incline acceleration formula with the kinematic equations you already know.
Kinematics on Inclines — Review
Once you find a on an incline, the kinematics equations work exactly the same:
v=v0+a
Part 6: Problem-Solving Workshop
🛠️ Problem-Solving Workshop
Part 6 of 7 — Inclined Planes
This workshop combines all incline concepts: weight components, friction on inclines, connected systems, and kinematics. Practice the complete problem-solving approach for AP-level incline problems.
Incline Problem-Solving Strategy
Step 1: Draw the FBD with Tilted Axes
x-axis along the slope, y-axis perpendicular
Weight components: mgsinθ (along), mgcosθ (perp)
Step 2: Find the Normal Force
Part 7: Synthesis & AP Review
🎯 Synthesis & AP Review
Part 7 of 7 — Inclined Planes
Congratulations on completing the Inclined Planes unit — and the entire Dynamics section! This final lesson reviews all incline concepts and connects them to the broader AP Physics 1 framework.
Complete Inclined Planes Reference
Weight Decomposition
W∥=mgsinθ(drives motion along slope)
θ=90°
sin90°=1
cos90°=0
θ
θ
This means:
Component perpendicular to surface=mgcosθComponent parallel to surface=mgsinθ
The Normal Force
Since there's no acceleration perpendicular to the surface:
N=mgcosθ
This is less thanmg — the steeper the incline, the smaller the normal force.
What is the normal force (in N)?
What is the ratio W∥/N?
Round all answers to 3 significant figures.
mgsinθ
Newton's Second Law (Along the Slope)
mgsinθ=maa=gsinθ
Key Insight
The acceleration is independent of mass! A 1 kg block and a 100 kg block slide at the same rate on the same frictionless incline.
Example Values
Angle
sinθ
a (m/s²)
10°
0.174
1.7
30°
0.500
5.0
45°
0.707
7.1
60°
0.866
8.7
90°
1.000
10.0 (free fall!)
Notice: At 90° the "incline" is vertical, and the acceleration equals g.
Combining with Kinematics
Once you know a=gsinθ, use the standard kinematics equations:
v=v0+atΔx=v0t+21atv2=v02+2aΔx
Example: Sliding Down from Rest
A block starts from rest and slides 2 m down a 30° frictionless incline. How fast is it going?
a=gsin30°=10×0.50=5 m/s2
Example: Sliding Up
A block is launched UP a 30° frictionless incline at 6 m/s. How far up does it go?
The deceleration is a=−gsin30°=−5 m/s² (opposing motion up the slope).
0=36+2(−5)ΔxΔx=36/10=3.6 m
Check Your Understanding 🧠
Frictionless Incline Problems 🧮
A block starts from rest on a 37° frictionless incline and slides for 2 seconds. What is its speed (m/s)? (g=10 m/s², sin37°=0.60)
How far has it traveled in those 2 seconds (in m)?
A block is launched at 10 m/s up a 30° frictionless incline. How far up the slope does it travel before stopping (in m)? (g=10 m/s², sin30°=0.50)
Conceptual Reasoning 🔍
Exit Quiz ✅
mgcosθ
Parallel to surface (x-axis):
mgsinθ (down the slope)
f (friction — direction depends on motion!)
Friction Direction Rules
Sliding down: Friction acts up the slope
Sliding up: Friction acts down the slope
Stationary: Friction opposes the tendency of motion
Case 1: Sliding Down the Incline
ma=mgsinθ−fk=mgsinθ−μkmgcosθa=g(sinθ−μkcosθ)
Case 2: Sliding Up the Incline
Both gravity and friction oppose motion:
ma=mgsinθ+fk=mgsinθ+μkmgcosθadecel=g(sinθ+μk
Note: The deceleration going up is greater than the acceleration going down!
Worked Example
A 5 kg block slides down a 37° incline with μk=0.25 (g=10 m/s², sin37°=0.60, cos37°=0.80).
Step 1: Normal force: N=mgcos37°=50×0.80=40 N
Step 2: Friction: fk=μkN=0.25×40= N (up the slope)
Step 3: Net force along slope: Fnet=mgsin37°−fk N
Step 4: Acceleration: a=20/5=4 m/s²
Compare to frictionless: a=gsin37°=6 m/s²
Friction reduced the acceleration by μkgcos37°=2 m/s².
Check Your Understanding 🧠
Incline + Friction Problems 🧮
A 10 kg block slides down a 53° incline with μk=0.30 (g=10 m/s², sin53°=0.80, cos53°=0.60).
What is the normal force (in N)?
What is the kinetic friction force (in N)?
What is the acceleration down the slope (in m/s²)?
Round all answers to 3 significant figures.
Special Case: Constant Velocity on an Incline
If a block slides at constant velocity down an incline:
mgsinθ=μkmgcosθμk=tanθ
This is identical to the critical angle for static friction! Measuring the angle at which a block slides at constant velocity gives you μk directly.
Will It Slide?
A block is placed on a rough incline. It will slide if:
mgsinθ>fs,max=μ
If μs=0.577, the critical angle is tan−1(0.577)=30°. Any steeper and it slides.
Exit Quiz ✅
2
For m1 (on the incline, taking up-the-slope as positive):T−m1gsinθ=m1a
For m2 (hanging, taking downward as positive):m2g−T=m2a
Solving for Acceleration
Add the equations to eliminate T:
m2g−m1gsinθ=(m1+m2)a
a=m1+m2m2g−m1gsinθ
Solving for Tension
Substitute a back into either equation:
T=m2(g−a)=m1+m2m1m
Direction Check
If m2g>m1gsinθ: the hanging mass falls and m1 goes up the incline
If m2g<m1gsinθ: m slides down and rises
Worked Example
m1=4 kg on a frictionless 30° incline, connected to m2=3 kg hanging (g=10 m/s²).
Driving force:m2g−m1gsin30°=30− N
Total mass:m1+m2=7 kg
Acceleration:a=10/7≈1.43 m/s²
Tension:T=m2(g−a)=3(10−1.43)= N
Check: For m1: T−m1gsin30° N. N. ✅
With Friction
If the incline has friction (μk), add friction to m1's equation:
m1 sliding up:T−m1g
a=m1+m2
Check Your Understanding 🧠
Connected System Calculations 🧮
m1=6 kg on a frictionless 30° incline, connected over a pulley to m2=5 kg hanging freely (g=10 m/s², sin30°=0.50).
What is the net driving force of the system (in N)?
What is the acceleration (in m/s², to one decimal)?
What is the tension in the string (in N, to one decimal)?
Round all answers to 3 significant figures.
System Analysis 🔍
Exit Quiz ✅
t
Δx=v0t+21at2
v2=v02+2aΔx
Key Accelerations
Scenario
Acceleration
Frictionless, sliding down
a=gsinθ
With friction, sliding down
a=g(sinθ−μkcosθ)
With friction, sliding up
a=g(sinθ+μkcosθ) (deceleration)
Worked Example: Up and Back Down
A block is launched at v0=8 m/s up a 30° incline with μk=0.20 (g=10 m/s²).