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

• putongren
In summary, the block of mass M rests on a frictionless wedge that has an inclination of θ and mass M. The acceleration of the wedge M to the right such that the block m remains stationary relative to the wedge is mg cos(theta) - mg = 0.
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

## 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?

putongren said:
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?

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.

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?

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?

putongren said:
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.

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

putongren said:
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.
putongren said:
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.)

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.

putongren said:
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.
putongren said:
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.)

Alright, the direction is vertical.

putongren said:
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?

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

putongren said:
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.

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

putongren said:
So is the equation for the x direction N sin(theta) = a?
Yes.

so a = mg tan(theta)?

putongren said:
so a = mg tan(theta)?
Yes.

Thanks, I think that was the answer.

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

putongren said:
Actually, the answer is a = tan (theta), since I forgot to cancel out the m. Close enough.
Let's compromise on g tan(theta).

## 1. What is the Accelerating Wedge Problem?

The Accelerating Wedge Problem is a physics problem that involves a wedge-shaped object sliding down a ramp at an angle. The goal is to determine the acceleration of the wedge as it moves down the ramp.

## 2. What factors affect the acceleration of the wedge in this problem?

The acceleration of the wedge is affected by several factors, including the angle of the ramp, the mass of the wedge, and the coefficient of friction between the wedge and the ramp.

## 3. How is the Accelerating Wedge Problem solved?

To solve the Accelerating Wedge Problem, we use Newton's Second Law of Motion, which states that the net force on an object is equal to its mass times its acceleration. By analyzing the forces acting on the wedge, we can determine its acceleration.

## 4. Can the Accelerating Wedge Problem be applied to real-life situations?

Yes, the Accelerating Wedge Problem can be applied to various real-life situations, such as determining the acceleration of a car moving down a sloped road or the acceleration of an object sliding down a frictionless ramp.

## 5. What are some limitations of the Accelerating Wedge Problem?

The Accelerating Wedge Problem assumes that the surface of the ramp is smooth and that there is no air resistance. In real-life situations, these factors may affect the acceleration of the wedge and make the problem more complex.

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