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How do I find Theta critical |
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| Oct8-07, 02:28 PM | #1 |
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How do I find Theta critical
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. |
| Oct8-07, 07:48 PM | #2 |
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Recognitions:
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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] |
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