Introduction to Inclined Planes

Main ideas and lessons (inclined planes)

1) Resolve forces on an incline (key geometry + components)

A block on an incline experiences gravity downward. That force can be decomposed into:

• Normal force (perpendicular to the surface)
• Parallel component of weight (along the incline), which tends to slide the block

To define directions, the speaker uses a right-triangle projection:

• The hypotenuse corresponds to the magnitude of gravity-related force (labeled as (mg))
• The incline angle (\theta) is the same angle used in the triangle for decomposition

Using SOHCAHTOA:

• Normal-force component (adjacent side): [ \cos\theta=\frac{x}{mg}\quad\Rightarrow\quad x=mg\cos\theta ] Since there is no acceleration perpendicular to the surface: [ N = mg\cos\theta ]

• Down-slope component (opposite side): [ \sin\theta=\frac{y}{mg}\quad\Rightarrow\quad y=mg\sin\theta ] The force that accelerates the block down the incline is: [ f_g \;(\text{component of gravity down the ramp}) = mg\sin\theta ]

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2) Acceleration down a frictionless incline

Choose axes with:

• (x): along the incline
• (y): perpendicular to the incline

In the frictionless case, the only force along (x) is the downslope gravitational component (mg\sin\theta).

Using Newton’s 2nd law: [ \sum F_x = ma ] [ mg\sin\theta = ma ] Mass cancels: [ a = g\sin\theta ]

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3) Incline with friction (kinetic and static)

If the block slides down the incline:

• (f_g = mg\sin\theta) acts down the incline (positive (x))
• Kinetic friction opposes motion, so it acts up the incline (negative (x))

Kinetic friction magnitude: [ f_k = \mu_k N ]

Static friction condition (mentioned): [ f_s \le \mu_s N ]

Substitute (N=mg\cos\theta): [ f_k = \mu_k mg\cos\theta ]

Net force along the incline: [ ma = mg\sin\theta - \mu_k mg\cos\theta ]

Final acceleration (sliding down with kinetic friction): [ a = g\sin\theta - \mu_k g\cos\theta ]

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4) Sliding up the incline (sign changes)

If the block is sliding up the incline:

• The downslope component (mg\sin\theta) points opposite the motion → negative (x)
• Kinetic friction also opposes motion → negative (x)

So: [ a = -g\sin\theta - \mu_k g\cos\theta ]

If the incline is reversed direction (so the force directions relative to the chosen (x) axis flip), the speaker states the acceleration becomes positive: [ a = g\sin\theta + \mu_k g\cos\theta ]

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Worked physics problems (as presented)

Example: Block slides down a (30^\circ) frictionless incline from rest

Given

• Angle: (\theta = 30^\circ)
• Initial speed: (v_i = 0)
• Start from rest
• Frictionless (only (mg\sin\theta) acts along the incline)

Part a) Acceleration

Use: [ a = g\sin\theta ] With (g=9.8) and (\sin 30^\circ = 1/2): [ a = 9.8(1/2)=4.9\ \text{m/s}^2 ]

Part b) Final speed after traveling 200 m

Kinematics: [ v_f^2 = v_i^2 + 2ad ] Substitute (v_i=0), (a=4.9), (d=200): [ v_f^2 = 0 + 2(4.9)(200)=1960 ] [ v_f=\sqrt{1960}\approx 44.27\ \text{m/s} ]

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Example: Block travels up a (25^\circ) incline with initial speed 14 m/s (gravity component slowing it)

Part a) Acceleration

Use (gravity component only, as shown): [ a_x = -g\sin\theta ] With (\theta=25^\circ): [ a_x = -9.8\sin(25^\circ)\approx -4.14166\ \text{m/s}^2 ]

Part b) Distance up the incline until it stops

Stopping means (v_f=0). Use: [ v_f^2 = v_i^2 + 2ad ] Substitute (v_f=0), (v_i=14), (a=-4.14166): [ 0 = 14^2 + 2(-4.14166)d ] [ 0 = 196 - 8.28332\, d \Rightarrow d \approx 23.662\ \text{m} ]

Part c) Time to stop

Use: [ v_f = v_i + at ] Substitute (v_f=0), (v_i=14), (a=-4.14166): [ 0 = 14 + (-4.14166)t \Rightarrow t = \frac{-14}{-4.14166}\approx 3.38\ \text{s} ]

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Speakers / sources featured

• Speaker/Instructor: Not named in the subtitles (the video includes personal references like “I like to call it…” and mentions other resources)
• Source mentioned: “Kinematics organic chemistry tutor” (YouTube channel/video name)
• Platform mentioned: Patreon (creator’s page mentioned for the “full version” of a related kinematics video)