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): $W_x = mg\sin\theta$

Perpendicular to incline (into the surface): $W_y = mg\cos\theta$

Derivation from geometry:

• Angle between weight and perpendicular to incline = $\theta$
• Therefore: parallel component uses $\sin\theta$, perpendicular uses $\cos\theta$

Normal Force

Normal force $N$ acts perpendicular to the surface (along y-axis).

In y-direction (no acceleration perpendicular to incline): $N - mg\cos\theta = 0$ $N = mg\cos\theta$

Key point: Normal force is NOT equal to $mg$ on an incline!

Friction

Friction acts parallel to incline, opposing motion or attempted motion.

• If object slides down: friction points up the incline
• If object is pushed up: friction points down the incline

Maximum static friction: $f_{s,max} = \mu_s N = \mu_s mg\cos\theta$

Kinetic friction: $f_k = \mu_k N = \mu_k mg\cos\theta$

Motion Down a Frictionless Incline

Free body diagram:

• Weight component parallel: $mg\sin\theta$ (down incline)
• Weight component perpendicular: $mg\cos\theta$ (into surface)
• Normal force: $N = mg\cos\theta$ (out from surface)
• No friction

Apply Newton's Second Law (x-direction, parallel to incline):

$\sum F_x = ma$ $mg\sin\theta = ma$ $a = g\sin\theta$

Key results:

• Acceleration is independent of mass!
• Acceleration increases with steeper incline ($\theta$ larger)
• At $\theta = 90°$: $a = g$ (free fall)
• At $\theta = 0°$: $a = 0$ (horizontal surface)

Motion with Friction

Object at Rest on Incline

Question: What coefficient of friction is needed to prevent sliding?

Analysis:

Parallel to incline (equilibrium): $f_s = mg\sin\theta$

Maximum static friction: $f_{s,max} = \mu_s N = \mu_s mg\cos\theta$

Condition for no sliding: $mg\sin\theta \leq \mu_s mg\cos\theta$ $\sin\theta \leq \mu_s \cos\theta$ $\tan\theta \leq \mu_s$

Critical angle: $\theta_c = \tan^{-1}(\mu_s)$

• If $\theta < \theta_c$: object stays at rest
• If $\theta = \theta_c$: object on verge of sliding
• If $\theta > \theta_c$: object slides down

Object Sliding Down Incline

Free body diagram:

• Weight component parallel: $mg\sin\theta$ (down)
• Kinetic friction: $f_k = \mu_k mg\cos\theta$ (up)
• Normal force: $N = mg\cos\theta$

Apply Newton's Second Law:

$\sum F_x = ma$ $mg\sin\theta - \mu_k mg\cos\theta = ma$ $a = g(\sin\theta - \mu_k\cos\theta)$

Special cases:

• If $\sin\theta = \mu_k\cos\theta$: $a = 0$ (slides at constant velocity)
• If $\sin\theta > \mu_k\cos\theta$: $a > 0$ (speeds up)
• If $\sin\theta < \mu_k\cos\theta$: Can't slide down (would need initial motion)

Object Pushed Up Incline

Forces parallel to incline:

• Applied force $F_{app}$ (up incline)
• Weight component: $mg\sin\theta$ (down incline)
• Friction: $f_k = \mu_k mg\cos\theta$ (down incline, opposes motion)

Newton's Second Law: $F_{app} - mg\sin\theta - \mu_k mg\cos\theta = ma$

To push at constant velocity ($a = 0$): $F_{app} = mg\sin\theta + \mu_k mg\cos\theta$ $F_{app} = mg(\sin\theta + \mu_k\cos\theta)$

Problem-Solving Strategy

1. Draw the incline and identify angle $\theta$
2. Rotate coordinate system (x parallel, y perpendicular to incline)
3. Draw free body diagram with rotated axes
4. Break weight into components:
   • Parallel: $mg\sin\theta$
   • Perpendicular: $mg\cos\theta$
5. Find normal force from y-direction equilibrium: $N = mg\cos\theta$
6. Determine friction (static or kinetic)
7. Apply Newton's Second Law in x-direction
8. Solve for unknowns

Common Angles and Trig Values

Real-World Applications

• Ramps: Reduce force needed to lift heavy objects
• Roads on hills: Banking prevents cars from sliding
• Ski slopes: Steeper = faster (larger $g\sin\theta$)
• Conveyor belts: Angle and friction determine maximum load

Method 1: Compare forces

Step 1: Force trying to pull box down incline: $F_{down} = mg\sin 37° = 0.6mg$

Step 2: Maximum friction force holding box: $f_{s,max} = \mu_s N = \mu_s mg\cos 37°$ $f_{s,max} = (0.75)(mg)(0.8) = 0.6mg$

Step 3: Compare: $F_{down} = 0.6mg = f_{s,max}$

Conclusion: The box is on the verge of sliding (just barely held in place).

Method 2: Critical angle

Critical angle: $\theta_c = \tan^{-1}(\mu_s)$

$\tan\theta_c = 0.75$ $\theta_c = \tan^{-1}(0.75) = 37°$

Since $\theta = \theta_c$, 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 $\tan\theta = \mu_s$, the object is at the critical angle. Notice that $\tan 37° = 0.75 = \mu_s$!

Part (a): Find acceleration

Step 1: Find normal force $N = mg\cos\theta = (10)(10)(0.866) = 86.6 \text{ N}$

Step 2: Find kinetic friction $f_k = \mu_k N = (0.2)(86.6) = 17.32 \text{ N}$

Step 3: Find weight component down incline $F_{parallel} = mg\sin\theta = (10)(10)(0.5) = 50 \text{ N}$

Step 4: Apply Newton's Second Law (down incline is positive)

$\sum F_x = ma$ $mg\sin\theta - f_k = ma$ $50 - 17.32 = 10a$ $32.68 = 10a$ $a = 3.268 \text{ m/s}^2 \approx 3.27 \text{ m/s}^2$

Alternative formula: $a = g(\sin\theta - \mu_k\cos\theta)$ $a = 10(0.5 - 0.2 \times 0.866)$ $a = 10(0.5 - 0.173)$ $a = 10(0.327) = 3.27 \text{ m/s}^2$

Part (b): Find speed after 20 m

Use kinematic equation: $v^2 = v_0^2 + 2ad$

$v^2 = 0^2 + 2(3.27)(20)$ $v^2 = 130.8$ $v = \sqrt{130.8} \approx 11.4 \text{ m/s}$

Answers:

• (a) Acceleration: 3.27 m/s² down the incline
• (b) Speed after 20 m: 11.4 m/s

Check: Without friction, $a = g\sin 30° = 5$ m/s². With friction, $a = 3.27$ m/s² < 5 m/s² ✓