What Is Theta Critical for an Object on an Inclined Plane?

  • Thread starter Thread starter pyrojelli
  • Start date Start date
  • Tags Tags
    Theta
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 7K views
pyrojelli
Messages
16
Reaction score
0
We have Uk and/or Us and angle of inclination (theta)


Are these eq-ns relevant? sin(theta critical)=Us(tan(theta critical))=Us
and sin(theta static)/cos(theta static)=Uk(tan(theta static)=Uk



How do I even interpret these eq-ns? Exam is tomorrow and I need to know how to find the angle of inclination that allows an object to start sliding (theta critical) and the angle of inclination so that the object will slide w/o accelaration. This is for an object on an inclined plane w/ friction.
 
Physics news on Phys.org
I'm confused by the equations you wrote... can you write them exactly as they are?

The moment when sliding occurs is when the static frictional force becomes [tex]\mu_s*F_n[/tex].

Take the equation perpendicular to the plane... [tex]F_n - mgcos(\theta) = 0[/tex], so [tex]F_n = mgcos(\theta)[/tex]

The equation parallel to the plane is: [tex]mgsin(\theta) - f = 0[/tex]

so this is while the block is not sliding...

ie: [tex]f = mgsin(\theta)[/tex] (1)

so this equation is always true while the block is not sliding... you will notice that as theta becomes larger (the incline becomes steeper)... f becomes larger... this is all while the block is still not sliding... but there is a limit to how long this can go on... the limit occurs when f becomes [tex]\mu_s*F_n = \mu_s*mgcos(\theta)[/tex].

so to find the angle at which this limit occurs substitute [tex]f = \mu_s*mgcos(\theta)[/tex] into (1)

so you get:

[tex]\mu_s*mgcos(\theta) = mgsin(\theta)[/tex]

[tex]\mu_s = tan(\theta)[/tex]