Forest UCM NLM BlockOnInclineWfriction

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The problem

Consider a block of mass m sliding down an infinitely long inclined plane shown below with a frictional force that is given by

[math]F_f = \mu mg[/math]

200 px

Find the blocks speed as a function of time.

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

200 px

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

[math]\vec{N} = \left | \vec{N} \right | \hat{j}[/math]
[math]\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]\vec{F_f} = - \mu mg \hat{i}[/math]

Step 5: Use Newton's second law

Motion in the [math]\hat j[/math] direction described by Newton's second law is:

[math]\sum F_{ext} = N - mg \cos \theta = ma_y = 0[/math]
[math]N = mg \cos \theta[/math]
[math]F_f \le \mu_s N = \mu_s mg \cos \theta[/math] where [math]\mu_s[/math] is the coefficent of STATIC friction

The [math]\le[/math] indicates that STATIC friction will be a force that is suficient to keep the block from moving. STATIC friction has a maximum value. If the sum of the other forces exceeds the static friction force, then the object will move, and the coeffiicent of kinetic friction [math](\mu_k)[/math] will be used to describe the motion.

What is the condition to satisfy for the object to move down the inclined plane?

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

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

if there is no acceleration then

[math] mg \sin \theta = F_f[/math]

If the object is not moving then

[math]F_f \le mg \sin \theta[/math]

The largest value for the frictional force of an object with no velocity is

[math]F_f = \mu_s N = \mu_s mg \cos \theta[/math]

The condition that must occur in order for an object with no initial velocity to begin moving with a non-zero accelleration is

[math]F_f = \mu_s N = \mu_s mg \cos \theta \lt mg \sin \theta[/math]

The critical angle of incline for the object to start moving becomes

[math]\theta_c = \tan^{-1}\left( {\mu_s mg}\right )[/math]

After the object exceeds the above condition is will have a non-zero velocity AND the coefficient of friction will decrease from [math]\mu_s[/math] to [math]\mu_k[/math]

[math]F_f = \mu_k N = \mu_k mg \cos \theta [/math]

Newton's second law under the condition that the object has a velocity becomes

[math]\sum F_{ext} = mg \sin \theta - F_f= ma_x = m \frac{dv_x}{dt}[/math]
[math]\sum F_{ext} = mg \sin \theta -\mu_k mg \cos \theta= ma_x = m \frac{dv_x}{dt}[/math]
[math]\int_0^t g \left ( \sin \theta - \mu_k\cos \theta \right ) dt = \int_{v_i}^{v} dv [/math]
[math]v= v_i - g \left ( \mu_k\cos \theta -\sin \theta \right ) t [/math]

The amount of time that lapses until the blocks final velocity is zero

[math]t= \frac{v_i}{\left ( \mu_k\cos \theta - \sin \theta \right ) }[/math]

After the above time the blocks speed is zero. The friction will change from being kinetic to static after the above time interval.

[math]v(t) =\left \{ {v_i - g \left (\mu_k\cos \theta -\sin \theta \right ) t \;\;\;\;\;\;\;\; t\lt \frac{v_i}{\left ( \mu_k\cos \theta - \sin \theta \right ) } \atop 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; t\gt = \frac{v_i}{\left ( \mu_k\cos \theta - \sin \theta \right ) }} \right .[/math]