Forest UCM NLM BlockOnInclineWfriction

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]

[math]\int_0^t g \left ( \sin \theta - \mu \right ) dt = \int_{v_i}^{v} dv [/math]

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

Forest_UCM_NLM#Block_on_incline_with_friction