What is the acceleration of a wedge with an inclined mass resting on it?

[putongren]

A block of mass m rests on a frictionless wedge that has an inclination of θ and mass M. Find the acceleration of the wedge M to the right such that the block m remains stationary relative to the wedge

1. Homework Statement
a, $\theta$, M, m

Homework Equations

The Attempt at a Solution

Draw a FBD centered on m. There is the force due to gravity, the normal force due to contact between M and m. There is also a force from M that imparts to m. Am I missing anything else?

 

[haruspex]

There is also a force from M that imparts to m.

You already listed a normal force from M on m. What is this other force?

 

[IAN 25]

I would find the component of acceleration (due to gravity) acting on m down the slope first. Then consider the component in direction of a.

 

[putongren]

I realize that there is a force that from M that imparts to m due to M's acceleration. I'm not sure what the direction and magnitude of this force. Is the direction of this force just parallel to the movement of M towards the right? So how many force vectors are there altogether for the FBD? 3?

 

[J Hann]

The only force the plane exerts on m is the normal force.
Have you tried resolving the normal force into its components and relating these
components to the motion of the block?

 

[haruspex]

I realize that there is a force that from M that imparts to m due to M's acceleration

M accelerates to the right. If it were to exert no force on m then m would move inside M. A normal force between surfaces is that force which opposes interpenetration. The force that M 'imparts' to m due to M's acceleration is the normal force.

 

[putongren]

I just realized that there should be a force parallel to the plane that points up along the slope? Is that true?

The y component of the normal force is mg cos theta and the x component is mg sin theta

 

[haruspex]

I just realized that there should be a force parallel to the plane that points up along the slope? Is that true?

What would supply such a force? The slope is frictionless.

The y component of the normal force is mg cos theta and the x component is mg sin theta

Don't confuse normal force with gravitational force. Are you defining the Y direction as vertical or normal to the wedge slope?
Whichever you choose, try to answer this question: if the normal force is N, what is its component in the Y direction?
(To answer this you do not need to consider gravity or mass or acceleration.)

 

[putongren]

The force that points up along the slope is probably s fictitious force. But that's beyond the scope of the course I'm taking.

I'm defining the y direction as vertical to the wedge slope. It's component in the y direction is simply cos theta.

 

[haruspex]

The force that points up along the slope is probably s fictitious force. But that's beyond the scope of the course I'm taking.

Yes.

I'm defining the y direction as vertical to the wedge slope. It's component in the y direction is simply cos theta.

No, it's either 'vertical' or 'normal to the slope'; which?
What direction does the normal force act in? (The clue is in the name.)

 

[putongren]

Alright, the direction is vertical.

 

[haruspex]

Alright, the direction is vertical.

OK, we can agree that the component of the normal force N in the y direction is N cos(theta), yes?
What other forces act in the y direction?
What is the acceleration of the block in the y direction?
What equation does that give you?

 

[putongren]

There's gravity (mg) that acts in the y direction. Since the block is static, then the equation should be N cos(theta) - mg = 0

 

[haruspex]

There's gravity (mg) that acts in the y direction. Since the block is static, then the equation should be N cos(theta) - mg = 0

Well, it's not static, but you are right that it has no vertical acceleration, so yes, that is the right equation.

 

[putongren]

So is the equation for the x direction N sin(theta) = a?

 

[haruspex]

So is the equation for the x direction N sin(theta) = a?

Yes.

 

[putongren]

so a = mg tan(theta)?

 

[haruspex]

so a = mg tan(theta)?

Yes.

 

[putongren]

Thanks, I think that was the answer.

 

[putongren]

Actually, the answer is a = tan (theta), since I forgot to cancel out the m. Close enough.

 

[haruspex]

Actually, the answer is a = tan (theta), since I forgot to cancel out the m. Close enough.

Let's compromise on g tan(theta).