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

[sArGe99]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

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

$$\omega = \frac {g (sin \alpha - k cos \alpha)}{cos \alpha + k sin \alpha}$$

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

 

[Doc Al]

At equilibrium, the frictional force equals the force which acts along the prism.
Equating, $$mg sin \alpha - m\omega cos\alpha$$ = k ($$m\omega sin \alpha + mg cos \alpha$$)

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.)

 

[hsarp71]

the maximum value of k is cot(alpha). So have you tried putting that in?

 

[sArGe99]

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.
I reckon I can use the sign system in any direction as long as I put the signs correctly?

 

[sArGe99]

Its given $$k < cot\alpha$$. 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..

 

[Doc Al]

I reckon I can use the sign system in any direction as long as I put the signs correctly?

Absolutely.

Its given $$k < cot\alpha$$. 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.

 

[sArGe99]

Sign of the expression for acceleration? I don't notice any particular change when $$k < cot\alpha$$! Where am I going wrong?

 

[Doc Al]

Sign of the expression for acceleration? I don't notice any particular change when $$k < cot\alpha$$! Where am I going wrong?

Take a look at your (corrected) expression. What happens to the sign of the denominator when k > cot α ?

 

[sArGe99]

The denominator turns out to be zero when $$k= cot\alpha$$.
Thanks for the help.

 

[sArGe99]

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?

 

[Doc Al]

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.