Forest UCM NLM BlockOnIncline

the problem

Consider a block of mass [math]\displaystyle{ m }[/math] is sliding down the inclined plane shown below with a frictional force that is given by

[math]\displaystyle{ F_f = kmv^2 }[/math]

How long does it take to fall a distance [math]\displaystyle{ x }[/math]?

File:TF UCM InclinedPlaneWfriction.png

Step 1: Identify the system

The block is the system with the following external forces, A normal force, a gravitational force, and the force of friction.

Step 2: Choose a suitable coordinate system

A coordinate system with one axis along the direction of motion may make solving the problem easier

Step 3: Draw the Free Body Diagram

File:TF UCM FBD InclinedPlaneWfriction.png

Step 4: Define the Force vectors using the above coordinate system

[math]\displaystyle{ \vec{N} = \left | \vec{N} \right | \hat{j} }[/math]

[math]\displaystyle{ \vec{F_g} = \left | \vec{F_g} \right | \left ( \sin \theta \hat{i} - \cos \theta \hat{j} \right )= mg \left ( \sin \theta \hat{i} - \cos \theta \hat{j} \right ) }[/math]

[math]\displaystyle{ \vec{F_f} = - kmv^2 \hat{i} }[/math]

Step 5: Used Newton's second law

Motion in the [math]\displaystyle{ \hat i }[/math] direction described by Newton's second law is:

[math]\displaystyle{ \sum F_{ext} = mg \sin \theta -mkv^2 = ma_x = m \frac{dv_x}{dt} }[/math]

Notice a terminal velocity [math]\displaystyle{ v_t }[/math] exists when [math]\displaystyle{ a_x =0 }[/math]

[math]\displaystyle{ mg \sin \theta -mkv^2 = ma_x = 0 }[/math]

[math]\displaystyle{ \Rightarrow v_t^2 = \frac{g \sin \theta}{k} }[/math]

This means that the block does not stop sliding but instead it reaches a minimal (terminal) velocity.

Insert the terminal velocity constant into Newton's second law

[math]\displaystyle{ \sum F_{ext} = mk \left ( v_t^2 - v^2 \right) = ma_x = m \frac{dv_x}{dt} }[/math]

[math]\displaystyle{ \int dt = \int \frac{dv}{k \left ( v_t^2 - v^2 \right)} }[/math]

before we had [math]\displaystyle{ (v dv) }[/math] in the numerator and could do a [math]\displaystyle{ u du }[/math] substitution, but not this time, use an integral table.

Integral table [math]\displaystyle{ \Rightarrow }[/math]

[math]\displaystyle{ \int \frac{dx}{a^2 + b^2x^2} = \frac{1}{ab} \tan^{-1} \frac{bx}{a} }[/math]

[math]\displaystyle{ a^2 = v_t^2 = \frac{g \sin \theta}{k} }[/math]

[math]\displaystyle{ b^2= -1 = i^2 }[/math]

[math]\displaystyle{ \int \frac{dv}{g\sin \theta - kv^2} = \frac{1}{ikv_t} \tan^{-1} \left ( \frac{iv}{v_t} \right ) }[/math]

[math]\displaystyle{ = \frac{-i}{kv_t} \tan^{-1} \left ( \frac{iv}{v_t} \right ) }[/math]

Identities

[math]\displaystyle{ i\tan^{-1}(icx) = -\tanh^{-1}(cx) = -\tanh^{-1}\left (\frac{\left | b \right |}{a} x\right ) }[/math]

[math]\displaystyle{ \tan^{-1}(z) = \frac{i}{2} \log \left ( \frac{i + z}{i-z}\right ) }[/math]

[math]\displaystyle{ \tanh^{-1}(z) = \frac{1}{2} \log \left ( \frac{1 + z}{1-z}\right ) }[/math]

[math]\displaystyle{ \tan^{-1}(ix) = \frac{i}{2} \log \left ( \frac{i + ix}{i-ix}\right )=\frac{i}{2} \log \left ( \frac{1 + 1x}{1-x}\right ) = i\tanh^{-1}(x) }[/math]

substituting

[math]\displaystyle{ \int dt = \int \frac{dv}{k \left ( v_t^2 - v^2 \right)} }[/math]

[math]\displaystyle{ t = \frac{-i}{kv_t} \tan^{-1} \left ( \frac{iv}{v_t}\right ) = \frac{1}{kv_t} \tanh^{-1} \left ( \frac{v}{v_t} \right ) }[/math]

Solving for [math]\displaystyle{ v }[/math]

[math]\displaystyle{ v = v_t \tanh \left ( k v_t t \right ) = \frac{dx}{dt} }[/math]

[math]\displaystyle{ \int dx = v_t \int \tanh \left ( k v_t t \right ) dt }[/math]

Integral table [math]\displaystyle{ \Rightarrow }[/math]

[math]\displaystyle{ \int \tanh (x) dx = \ln \left ( \cosh (x) \right ) }[/math]

[math]\displaystyle{ x = \frac{1}{k} \ln \left ( \cosh (k v_t t) \right ) }[/math]

solving for the fall time

[math]\displaystyle{ t = \frac{\cosh^{-1} \left ( e^{kx} \right)}{kv_t}= \frac{\cosh^{-1} \left ( e^{kx} \right)}{\sqrt{kg \sin \theta}} }[/math]

Forest_UCM_NLM#Block_on_incline_with_friction