Max \omega for Stationary Bar on Prism 1 with Coeff. k < cot\alpha

  • Thread starter Thread starter sArGe99
  • Start date Start date
  • Tags Tags
    Max Prism
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
10 replies · 3K views
sArGe99
Messages
133
Reaction score
0

Homework Statement



Prism 1 with bar 2 of mass m placed on it gets a horizontal acceleration [tex]\omega[/tex] directed to the left. At what maximum value of this acceleration [tex]\omega[/tex] will the bar be still stationary relative to the prism, if the coefficient of friction between them [tex]k < cot\alpha[/tex]

th_untitled3.jpg


Homework Equations




[tex]mg sin \alpha - m \omega cos \alpha[/tex] = Force trying to move the body along the prism.
k ([tex]m\omega sin \alpha + mg cos \alpha[/tex]) = Frictional force.


The Attempt at a Solution


Since the prism is accelerating to the left, the bar experiences a pseudo force [tex]m\omega[/tex] directed to the right. Resolving it has components
[tex]m \omega cos \alpha[/tex] opposed to [tex]mg sin \alpha[/tex] and the component [tex]m\omega sin \alpha[/tex] acting along the direction of [tex]mg cos \alpha[/tex]

At equilibrium, the frictional force equals the force which acts along the prism.
Equating, [tex]mg sin \alpha - m\omega cos\alpha[/tex] = k ([tex]m\omega sin \alpha + mg cos \alpha[/tex])
Further simplifying,

[tex]\omega = \frac {g (sin \alpha - k cos \alpha)}{cos \alpha + k sin \alpha}[/tex]

This is not the correct answer. Can you please point out where I've made a mistake?
 
Physics news on Phys.org

Discussion

sArGe99 said:
At equilibrium, the frictional force equals the force which acts along the prism.
Equating, [tex]mg sin \alpha - m\omega cos\alpha[/tex] = k ([tex]m\omega sin \alpha + mg cos \alpha[/tex])
Looks like you made a sign error. Which way does the friction force act? Which direction are you calling positive?

(I'd use ΣF = 0 parallel to the wedge, coupled with a consistent sign convention.)
 
the maximum value of k is cot(alpha). So have you tried putting that in?
 
Doc Al said:
Looks like you made a sign error. Which way does the friction force act? Which direction are you calling positive?

(I'd use ΣF = 0 parallel to the wedge, coupled with a consistent sign convention.)

Oh. Thanks for that correction. I've got the right answer now. :biggrin:
I reckon I can use the sign system in any direction as long as I put the signs correctly?
 
Its given [tex]k < cot\alpha[/tex]. What is the importance of this data in this question? I got the correct answer although I didn't have this point in mind. So I'm thinking why its given..
 
Last edited by a moderator:
sArGe99 said:
I reckon I can use the sign system in any direction as long as I put the signs correctly?
Absolutely.
sArGe99 said:
Its given [tex]k < cot\alpha[/tex]. What is the importance of this data in this question? I got the correct answer although I didn't have this point in mind. So I'm thinking why its given..
Consider the sign of your expression for the acceleration if that condition were not met. :wink:
 
Sign of the expression for acceleration? I don't notice any particular change when [tex]k < cot\alpha[/tex]! :frown: Where am I going wrong?
 
sArGe99 said:
Sign of the expression for acceleration? I don't notice any particular change when [tex]k < cot\alpha[/tex]! :frown: Where am I going wrong?
Take a look at your (corrected) expression. What happens to the sign of the denominator when k > cot α ?
 
Last edited:
The denominator turns out to be zero when [tex]k= cot\alpha[/tex]. :-p
Thanks for the help.
 
Last edited:
I see that the tendency of the body 2 is to move up. Why is it like that, usually it always tends to move down, right?
 
sArGe99 said:
I see that the tendency of the body 2 is to move up. Why is it like that, usually it always tends to move down, right?
The block would tend to move down the prism, if the prism weren't accelerating towards it. That changes things by introducing additional forces.