Accelerating wedge with friction

[Patta1667]

Homework Statement

A block rests on a wedge inclined at angle $$\theta$$. The coefficient of friction is $$\mu$$. The wedge is then given a horizontal acceleration a. Assuming that $$\tan (\theta) < \mu$$, find the minimum acceleration for the block to remain on the wedge without sliding.

Homework Equations

The Attempt at a Solution

I've been working angles like crazy and can't even get the theory down, much less come up with the equation. The block would slide off the wedge if not given an acceleration, and once the min. acceleration to keep the block on the wedge is reached the force of friction changes direction and will now be pointed down the wedge if acceleration increases.

I don't want a worked solution, only some help explaining the forces at work in this problem.

 

[rock.freak667]

Draw all the forces acting up the plane and down the plane. Then draw in the acceleration a.

For the wedge to not movie, what should the resultant acceleration be?

 

[dx]

If the block is not sliding, then it must be that the friction force is giving the block the same acceleration as the wedge. What is the maximum horizontal component of the friction force, and the corresponding maximum acceleration that friction can give the block?

 

[Patta1667]

If the block is not sliding, then it must be that the friction force is giving the block the same acceleration as the wedge. What is the maximum horizontal component of the friction force, and the corresponding maximum acceleration that friction can give the block?

Okay, the maximum friction force along the surface is $$\mu N$$, so the horizontal component is $$f \cos \theta = \mu N \cos \theta = \mu m g \cos^2 \theta$$. This means the block has an acceleration $$a = \mu g \cos^2 \theta$$ horizontally if friction alone is responsible.

Unfortunately, using $$\theta = \pi/2$$ yields $$a = \mu g/2$$. The correct answer is $$a_{min} = g \frac{1 - \mu}{1 + \mu}$$. Any ideas?

 

[dx]

Sorry I misread the question. Give me a minute to post a correct response.

 

[dx]

The correct answer is $$a_{min} = g \frac{1 - \mu}{1 + \mu}$$. Any ideas?

I keep getting the answer

$$a_{min} = \frac{g(\sin \theta - \mu \cos \theta)}{\cos \theta + \mu \sin \theta}$$

...

 

[Patta1667]

I keep getting the answer

$$a_{min} = \frac{g(\sin \theta - \mu \cos \theta)}{\cos \theta + \mu \sin \theta}$$

...

Ooops, let $$\theta = \pi/4$$ to get that answer... sorry! Well, could you lead me in the right direction please on how you obtained that?

 

[dx]

Ok, it's a little messy working in the original frame, so switch to a frame moving with the wedge. Since the wedge is accelerating, this frame is not inertial, so you will have to add a horizontal force ma to the block. So you have the weight, friction force, normal force and inertial force. The normal force is found in the usual way using the condition that the block is not accelerating perpendicular the incline. Then you find the friction force from the normal force. You find a_min by equating the component of the total force parallel to the incline to zero, i.e. imposing the condition that the block is not sliding.

 

[Patta1667]

Thank you! I had taken N = mgcos(theta) in all my equations, I completely forgot that the inertial force should be considered. The answer comes right out now, thanks!

 

[dx]

No problem.