Friction and Inclined Planes

Static and kinetic friction, motion on inclines with calculus analysis

🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

Friction and Inclined Planes

Friction Forces

Static friction: fsμsNf_s \leq \mu_s N

Kinetic friction: fk=μkNf_k = \mu_k N

where NN is the normal force and typically μs>μk\mu_s > \mu_k.

Motion with Friction

For an object sliding with kinetic friction:

ma=Fappliedfk=Fappliedμkmgma = F_{applied} - f_k = F_{applied} - \mu_k mg

dvdt=Fappliedmμkg\frac{dv}{dt} = \frac{F_{applied}}{m} - \mu_k g

Stopping Distance

Object sliding on horizontal surface with initial velocity v0v_0:

a=μkga = -\mu_k g

vdvdx=μkgv\frac{dv}{dx} = -\mu_k g

v00vdv=μkg0ddx\int_{v_0}^0 v \, dv = -\mu_k g \int_0^d dx

12v02=μkgd-\frac{1}{2}v_0^2 = -\mu_k gd

d=v022μkgd = \frac{v_0^2}{2\mu_k g}

Inclined Planes

For angle θ\theta from horizontal:

Coordinate system: x-axis along incline (positive down), y-axis perpendicular

Weight components:

  • Parallel to incline: F=mgsinθF_{\parallel} = mg\sin\theta
  • Perpendicular: F=mgcosθF_{\perp} = mg\cos\theta

Normal force: N=mgcosθN = mg\cos\theta (when no vertical acceleration)

Sliding Down Incline

With kinetic friction:

ma=mgsinθμkmgcosθma = mg\sin\theta - \mu_k mg\cos\theta

a=g(sinθμkcosθ)a = g(\sin\theta - \mu_k\cos\theta)

Condition for sliding: sinθ>μkcosθ\sin\theta > \mu_k\cos\theta, or tanθ>μk\tan\theta > \mu_k

Velocity after distance dd: v2=v02+2ad=v02+2gd(sinθμkcosθ)v^2 = v_0^2 + 2ad = v_0^2 + 2gd(\sin\theta - \mu_k\cos\theta)

Critical Angle for Static Friction

Object on the verge of sliding:

fs=μsNf_s = \mu_s N

mgsinθc=μsmgcosθcmg\sin\theta_c = \mu_s mg\cos\theta_c

tanθc=μs\tan\theta_c = \mu_s

Motion with Time-Dependent Force

Force F(t)F(t) applied up an incline:

mdvdt=F(t)mgsinθμkmgcosθm\frac{dv}{dt} = F(t) - mg\sin\theta - \mu_k mg\cos\theta

v(t)=v0+0tF(t)mdtgt(sinθ+μkcosθ)v(t) = v_0 + \int_0^t \frac{F(t')}{m} \, dt' - gt(\sin\theta + \mu_k\cos\theta)

Example: Exponential Force

If F(t)=F0et/τF(t) = F_0e^{-t/\tau}:

v(t)=v0+F0τm(1et/τ)gt(sinθ+μkcosθ)v(t) = v_0 + \frac{F_0\tau}{m}(1 - e^{-t/\tau}) - gt(\sin\theta + \mu_k\cos\theta)

Blocks Connected on Incline

Two blocks (masses m1m_1, m2m_2) connected by rope over pulley:

Block 1 on incline at angle θ\theta, block 2 hanging vertically.

Constraint: a1=a2=aa_1 = a_2 = a (magnitude)

Block 1 (up incline positive): m1a=Tm1gsinθfkm_1a = T - m_1g\sin\theta - f_k

Block 2 (down positive): m2a=m2gTm_2a = m_2g - T

Adding equations (T cancels): a=m2gm1gsinθμkm1gcosθm1+m2a = \frac{m_2g - m_1g\sin\theta - \mu_k m_1g\cos\theta}{m_1 + m_2}

Energy Considerations with Friction

Work done by friction: Wf=fkd=μkmgdcosθW_f = -f_k \cdot d = -\mu_k mgd\cos\theta

(negative because friction opposes motion)

This work is converted to thermal energy (non-conservative force).

📚 Practice Problems

No example problems available yet.