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Simplifies math!
Forces on an Incline
Weight Components
Weight W=mg points straight down (not perpendicular to incline!).
Break weight into components:
Parallel to incline (down the slope):
Wx=mgsinθ
Perpendicular to incline (into the surface):
Wy=mgcosθ
Derivation from geometry:
Angle between weight and perpendicular to incline = θ
To push at constant velocity (a=0):
Fapp=mgsinθ+μkmgcosθFapp=mg(sinθ+μk
Problem-Solving Strategy
Draw the incline and identify angleθ
Rotate coordinate system (x parallel, y perpendicular to incline)
Draw free body diagram with rotated axes
Break weight into components:
Parallel: mgsinθ
Perpendicular: mgcosθ
Find normal force from y-direction equilibrium: N=mgcosθ
Determine friction (static or kinetic)
Apply Newton's Second Law in x-direction
Solve for unknowns
Common Angles and Trig Values
θ
sinθ
cosθ
tanθ
0°
0
1
0
30°
0.5
0.866
0.577
37°
0.6
0.8
0.75
45°
0.707
0.707
1
53°
0.8
0.6
1.33
90°
1
0
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Real-World Applications
Ramps: Reduce force needed to lift heavy objects
Roads on hills: Banking prevents cars from sliding
Ski slopes: Steeper = faster (larger gsinθ)
Conveyor belts: Angle and friction determine maximum load
💡 Show Solution
Given:
Mass: m=5 kg
Incline angle: θ=30°
Frictionless (no friction)
g=10 m/s²
sin30°=0.5
Find: Acceleration down the incline
Free Body Diagram (rotated axes):
Weight component parallel (down incline): mgsinθ
Weight component perpendicular: mgcosθ
Normal force: N (perpendicular, out from surface)
Apply Newton's Second Law (parallel to incline):
∑Fx=mamgsinθ=ma
Solve for acceleration:a=gsinθa=(10)(0.5)a=
Direction: Down the incline
Answer: The acceleration is 5 m/s² down the incline.
Key insight: Acceleration is independent of mass! A 5 kg block and a 50 kg block both accelerate at the same rate on a frictionless incline.
Check: At θ=90°: a=gsin90°=g (free fall) ✓
2Problem 2medium
❓ Question:
A box sits on an incline at 37°. The coefficient of static friction is μs=0.75. Does the box slide down? (Use sin37°=0.6, cos37°=0.8, tan37°=0.75)
💡 Show Solution
Given:
Incline angle: θ=37°
Coefficient of static friction: μs=0.75
Box initially at rest
Does it slide?
3Problem 3hard
❓ Question:
A 10 kg sled slides down a 30° incline with coefficient of kinetic friction μk=0.2. Find: (a) the acceleration of the sled, (b) the speed after sliding 20 m from rest. (Use g=10 m/s², sin30°=0.5, cos30°=0.866)
💡 Show Solution
Given:
Mass: m=10 kg
Angle: θ=30°
μ
▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
cos
θ
)
5
m/s2
Find:
Method 1: Compare forces
Step 1: Force trying to pull box down incline:
Fdown=mgsin37°=0.6mg
Step 2: Maximum friction force holding box:
fs,max=μsN=μsmgcos37°fs,max=(0.75)(mg)(0.8)=0.6mg
Step 3: Compare:
Fdown=0.6mg=fs,max
Conclusion: The box is on the verge of sliding (just barely held in place).
Method 2: Critical angle
Critical angle: θc=tan−1(μs)
tanθc=0.75θc=tan−1(0.75)=37°
Since θ=θc, the box is on the verge of sliding.
Answer: The box is on the verge of sliding but does not slide (at the critical angle). Any slight increase in angle or decrease in friction would cause sliding.
Key insight: When tanθ=μs, the object is at the critical angle. Notice that tan37°=0.75=μs!
k
=
0.2
Distance: d=20 m
Initial velocity: v0=0 (starts from rest)
g=10 m/s²
Part (a): Find acceleration
Step 1: Find normal forceN=mgcosθ=(10)(10)(0.866)=86.6 N
Step 2: Find kinetic frictionfk=μkN=(0.2)(86.6)=17.32 N
Step 3: Find weight component down inclineFparallel=mgsinθ=(10)(10)(0.5)=50 N
Step 4: Apply Newton's Second Law (down incline is positive)