Drag & Friction

Classical Mechanics Level 2

An object of mass M M is placed on an inclined plane with friction coefficient μ \mu as shown on the figure. The object is then given an initial speed of v 0 v_{0} parallel to the plane. On its subsequent motion, the object also experiences a drag force, F ( v ) F(v) , that satisfies :

F ( v ) = − b v \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad F(v)=-bv

After a very long time, what is the speed of the object? (in m / s ) m/s)

Details and Assumptions :

  • Assume that the inclined plane is very long

  • M = 1 k g M = 1 \ kg

  • θ = 3 0 o \theta = 30^o

  • b = 0.5 k g / s b = 0.5 \ kg/s

  • μ = 3 5 \mu = \frac{\sqrt{3}}{5}\

  • g = 10 m / s 2 g = 10 \ m/s^2


The answer is 4.

Jovan Alfian Djaja by Jovan Alfian Djaja , 18, Indonesia

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2 solutions

Steven Chase
Jan 8, 2021

Set the accelerating force equal to the retarding force to solve for the steady state velocity:

m g sin ⁡ ( θ ) = μ m g cos ⁡ θ + b v s s v s s = m g sin ⁡ ( θ ) − μ m g cos ⁡ θ b m g \sin(\theta) = \mu m g \cos \theta + b v_{ss} \\ v_{ss} = \frac{m g \sin(\theta) - \mu m g \cos \theta }{b}

4 Helpful 2 Interesting 6 Brilliant 0 Confused

Nice, i should've remembered this method earlier.

Jovan Alfian Djaja - 5 months ago

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It would be fun to solve a problem like this that focuses on the dynamics that occur before steady state

Steven Chase - 5 months ago

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I will post another version soon

Jovan Alfian Djaja - 5 months ago

Oh god it was so simple! I legit tried solving the differential equation 😂

Rohan Joshi - 5 months ago

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I know right! I forgot this method.

Jovan Alfian Djaja - 5 months ago

I will also post another version of this problem soon, stay tune.

Jovan Alfian Djaja - 5 months ago

This method is so simple, I like it! And it is analogous of the free fall with drag. I solved the ODE, but just because I need to practice it more haha And your solution is kind of a consequence of the solution for the ODE.

Victor Paes Plinio - 3 months, 3 weeks ago

The free body diagram of the system :

Where :

W x = m g sin ⁡ θ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad W_{x} = mg\sin\theta

N = W y = m g cos ⁡ θ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad N = W_{y} = mg\cos\theta

f = μ N = μ m g cos ⁡ θ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f = \mu N = \mu mg\cos\theta

Newton's 2 n d 2^{nd} law along the x axis :

m g sin ⁡ θ − μ m g cos ⁡ θ − b v = m x ¨ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad mg\sin\theta - \mu mg\cos\theta - bv = m \ddot{x}

m g ( sin ⁡ θ − μ cos ⁡ θ ) − b v = m x ¨ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad mg (\sin \theta - \mu \cos \theta) - bv = m \ddot{x}

Divide both sides with m m and let ( sin ⁡ θ − μ cos ⁡ θ ) g = A (\sin\theta -\mu\cos\theta)g = A to simplify calculations,

A − b m v = v ˙ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad A - \frac{b}{m} v = \dot{v}

Move − b m v -\frac{b}{m}v to the right side and multiply both sides with the integrating factor e ∫ b m d t = e b m t e ^{\int \frac{b}{m} dt} = e ^{\frac{b}{m}t} , resulting in :

A e b m t = d d t ( v e b m t ) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad Ae ^{\frac{b}{m}t} = \frac{d}{dt} (ve ^{\frac{b}{m}t})

Integrate both sides,

m A b e b m t + C = v e b m t \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{mA}{b}e ^{\frac{b}{m}t} + C = ve ^{\frac{b}{m}t}

Substitute t = 0 t = 0 and v = v 0 v = v_{0} to get C = v 0 − m A b C = v_{0} - \frac{mA}{b}

Multiply both sides by e − b m t e ^{-\frac{b}{m}t} ,

m A b + v 0 e − b m t − m A b e − b m t = v \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{mA}{b} + v_{0}e ^{-\frac{b}{m}t}-\frac{mA}{b}e ^{-\frac{b}{m}t} = v

v = m A b ( 1 − e − b m t ) + v 0 e − b m t \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad v = \frac{mA}{b}(1-e ^{-\frac{b}{m}t}) + v_{0}e ^{-\frac{b}{m}t}

Substitute A A to get :

v ( t ) = m g ( sin ⁡ θ − μ cos ⁡ θ ) b ( 1 − e − b m t ) + v 0 e − b m t \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \boxed {v(t) = \frac{mg (\sin \theta - \mu \cos \theta)}{b}(1-e ^{-\frac{b}{m}t}) + v_{0}e ^{-\frac{b}{m}t}}

Take the limit t → ∞ t \to \infty to get :

v = m g b ( sin ⁡ θ − μ cos ⁡ θ ) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \boxed {v = \frac{mg}{b} (\sin \theta - \mu \cos \theta)}

Substitute the numerical values to get :

v = 4 m / s \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \boxed {v = 4 \ m/s}

2 Helpful 1 Interesting 2 Brilliant 0 Confused

Nice! I did it the same way! Solving the ODE is not the simplest method, but I like to see the full equation of motion in the end. It give us the feeling of what is happening in the problem. The decaying exponential is something we might expect when the speed comes to a hold at some time in the future. And the shape of the curve approaches the no drag situation when the mass is very large or the drag is very small.

Victor Paes Plinio - 3 months, 3 weeks ago

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