Difference between revisions of "Forest UCM NLM BlockOnIncline"

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::<math>\tan^{-1}(icx) = \tanh^{-1}(cx) = \tanh^{-1}\left (\frac{\left | b \right |}{a} x\right )</math>
 
::<math>\tan^{-1}(icx) = \tanh^{-1}(cx) = \tanh^{-1}\left (\frac{\left | b \right |}{a} x\right )</math>
 +
 +
:t = \frac{1}{\sqrt{gk\sin \theta}} \tan^{-1} \left ( \sqrt{\frac{k}{g \sin \theta}} \; \;v \right )
  
 
Solving for <math>v</math>
 
Solving for <math>v</math>

Revision as of 03:13, 19 August 2014

the problem

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

Ff=kmv2


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

N=|N|ˆj
Fg=|Fg|(sinθˆicosθˆj)=mg(sinθˆicosθˆj)
Ff=kmv2ˆi

Step 5: Used Newton's second law

in the ˆi direction

Fext=mgsinθmkv2=max=mdvxdt
dt=dvgsinθkv2

Integral table

dxa2+b2x2=1abtan1bxa


a2=gsinθ
b2=k
dvgsinθkv2=1gksinθtan1(kgsinθv)


i1
dvgsinθkv2=1gksinθitan1(kgsinθiv)
tan1(icx)=tanh1(cx)=tanh1(|b|ax)
t = \frac{1}{\sqrt{gk\sin \theta}} \tan^{-1} \left ( \sqrt{\frac{k}{g \sin \theta}} \; \;v \right )

Solving for v

v = \tan \left ( \sqrt{gk\sin \theta} i t \right )
=

Forest_UCM_NLM#Block_on_incline_with_friction