Max distance to place an object so it won't slide (torque and friction)

[Femme_physics]

Okay!

http://img651.imageshack.us/img651/7528/forces.jpg

 

[I like Serena]

Good morning again Fp!


There you go! Very good!

Uh, one thing, doesn't Fs have an y component?
Add that, and go grrl, go!

 

[Femme_physics]

Good morning ILS :)And damn it I knew I forgot something. I got used to the inclined coordinates. Add Fs sin(30) in sum of all forces on y.

 

[I like Serena]

Good morning ILS :)


And damn it I knew I forgot something. I got used to the inclined coordinates. Add Fs sin(30) in sum of all forces on y.

That's it!
Now do these 2 equations scare you?

 

[ehild]

Well, you know how the maximum of Fs is related to N. Use it in the equation to eliminate Fs. ehild

 

[Femme_physics]

http://img402.imageshack.us/img402/3127/82363220.jpg

http://img148.imageshack.us/img148/324/70065989.jpg

 

[I like Serena]

I like the flowers and the butterflies!
How does it feel?

Uhh, except that I'm afraid I was careless with my last hint. Sorry!

Can you check if the y component of Fs is up or down?
That is, can you check the sign of "Fs sin 30"?EDIT: Oh, and I seem to be missing a square around the omega...

 

[Femme_physics]

Can you check if the y component of Fs is up or down?
That is, can you check the sign of "Fs sin 30"?

But it can't be wrong, I already drew flowers and butterflies. It can't. Flowers and butterflies means I'm right. Sorry, it's a cosmological fact. Whatever you think about mechanics is wrong because of that. Okay, fine, I'll fix the sign and get the right figure, but that's it basically, that is the answer right? I'll have to do some reverse engineering after I'm done to better follow what you've been trying to tell me.

 

[I like Serena]

But it can't be wrong, I already drew flowers and butterflies. It can't. Flowers and butterflies means I'm right. Sorry, it's a cosmological fact. Whatever you think about mechanics is wrong because of that.


Okay, fine, I'll fix the sign and get the right figure, but that's it basically, that is the answer right? I'll have to do some reverse engineering after I'm done to better follow what you've been trying to tell me.

Yep! That's it!

Oh, and take note of the missing square around the omega.

 

[kjohnson]

Then how come I'm not considering moment of inertia, friction, etc, or whatever in my calculations of centripetal force? Shouldn't it matter if those are the things that REALLY cause centripetal force?

In kinematics/dynamics, we ignore friction by saying "smooth surface" and ignore moment of inertia by saying "point mass". I'm curious why aren't we told anything in questions about moment of inertia and friction if it so heavily affects the score. What if there was no friction, indeed? Would there be no centripetal force/much much lesser to an insignificant level?

Well you do consider all forces that act on the block (weight,normal,friction). But some of those forces do not have components in the radial direction such as weight. If you were to have no friction than the centripetal force is supplied purely by the component of the normal force that is in the radial direction.

The moment of inertial does not come into play unless the block is accelerating in a circle. Because just like F=ma, M=I(alpha) where 'M' is the net moment or torque, 'I' the moment of inertia, and 'alpha' the angular acceleration. Since you have no angular acceleration in this problem there is no torque applied to the block and no need to worry about the moment of inertia...

 

[Femme_physics]

Will this calculation FINALLY lead to my answer?

http://img718.imageshack.us/img718/9376/fxfxfxfxfxfxfxfx.jpg

Well you do consider all forces that act on the block (weight,normal,friction). But some of those forces do not have components in the radial direction such as weight. If you were to have no friction than the centripetal force is supplied purely by the component of the normal force that is in the radial direction.

The moment of inertial does not come into play unless the block is accelerating in a circle. Because just like F=ma, M=I(alpha) where 'M' is the net moment or torque, 'I' the moment of inertia, and 'alpha' the angular acceleration. Since you have no angular acceleration in this problem there is no torque applied to the block and no need to worry about the moment of inertia...

I think I somewhat understand you, I'd still love for a good source about centripetal force, the real forces that act, and how to get an understanding of it! I might have to make a different topic before it sinks in. I appreciate the feedback. :)

 

[I like Serena]

Will this calculation FINALLY lead to my answer?

http://img718.imageshack.us/img718/9376/fxfxfxfxfxfxfxfx.jpg

Looking good!

You did intend to take the square of 2.084 and not of R, I hope?

 

[Femme_physics]

Why but of course I did!

*sneaky shifty look*



Thank you ILS, and thank you everyone for helping me through this! :)

 

[I like Serena]

Why but of course I did!

*sneaky shifty look*



Thank you ILS, and thank you everyone for helping me through this! :)

Hey! Not so fast!

Do you have the answer for R?

 

[Femme_physics]

Sorry for the delay, but I always get the job done! ...eventually. :)

Is this the final equation I need to solve, then?

http://img231.imageshack.us/img231/8179/rrrrrrrrrrrrrrrro.jpg

 

[I like Serena]

Sorry for the delay, but I always get the job done! ...eventually. :)

Is this the final equation I need to solve, then?

Yep!

 

[Femme_physics]

Score! :) Thank you ILS, Everyone. Long overdue!

 

[I like Serena]

Score! :) Thank you ILS, Everyone. Long overdue!

Hey! I want to see more flowers and butterflies (or some other artistic expression)!

 

[ehild]

I would like to see solution for the minimal distance so as the object do not move downward.

ehild