# Pseudo Force ~

1. Mar 19, 2009

### sArGe99

1. The problem statement, all variables and given/known data

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 $$k < cot\alpha$$

2. Relevant 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.

3. 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?

2. Mar 19, 2009

### Staff: Mentor

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.)

3. Mar 19, 2009

### hsarp71

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

4. Mar 19, 2009

### sArGe99

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?

5. Mar 19, 2009

### 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..

Last edited by a moderator: Mar 19, 2009
6. Mar 19, 2009

### Staff: Mentor

Absolutely.
Consider the sign of your expression for the acceleration if that condition were not met.

7. Mar 19, 2009

### sArGe99

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

8. Mar 19, 2009

### Staff: Mentor

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

Last edited: Mar 19, 2009
9. Mar 19, 2009

### sArGe99

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

Last edited: Mar 19, 2009
10. Mar 19, 2009

### 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?

11. Mar 20, 2009

### Staff: Mentor

The block would tend to move down the prism, if the prism weren't accelerating towards it. That changes things by introducing additional forces.