Difference between revisions of "Forest UCM NLM BlockOnInclineWfriction"

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=Step 5: Use Newton's second law=
 
=Step 5: Use Newton's second law=
  
Motion in the  <math>\hat i</math> direction governed by Netwon's second law is:
+
==Motion in the  <math>\hat j</math> direction described by Newton's second law is:==
  
:<math>\sum F_{ext} = mg \sin \theta - \mu mg= ma_x = m \frac{dv_x}{dt}</math>
+
:<math>\sum F_{ext} = N - mg \cos \theta = ma_y = 0</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>
+
::<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 < 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
 
The amount of time that lapses until the blocks final velocity is zero
  
<math>t= \frac{v_i}{\left ( \mu - \sin \theta \right ) }</math>
+
<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.
 
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< \frac{v_i}{\left ( \mu - \sin \theta \right ) } \atop  0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; t>= \frac{v_i}{\left ( \mu - \sin \theta \right ) }} \right .</math>
+
:<math>v(t) =\left \{  {v_i - g \left (\mu_k\cos \theta -\sin \theta \right ) t \;\;\;\;\;\;\;\; t< \frac{v_i}{\left ( \mu_k\cos \theta - \sin \theta \right ) } \atop  0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; t>= \frac{v_i}{\left ( \mu_k\cos \theta - \sin \theta \right ) }} \right .</math>
  
  
 
[[Forest_UCM_NLM#Block_on_incline_with_friction]]
 
[[Forest_UCM_NLM#Block_on_incline_with_friction]]

Latest revision as of 13:31, 21 August 2014

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]


Forest_UCM_NLM#Block_on_incline_with_friction