# Moving Wedge

1. Aug 30, 2008

### Dschumanji

1. The problem statement, all variables and given/known data
A wedge with mass M rests on a frictionless table. A block with mass m is place on the wedge. There is no friction between the block and wedge. When the system is released, what is the acceleration of the wedge and the acceleration of the block. Give the accelerations in terms of components.

2. Relevant equations
F=ma

3. The attempt at a solution
I used the table as the frame of reference. The positive x direction is to the right, and the positive y direction is up. The wedge makes an angle $$\beta$$ with the table from the negative x direction towards the positive y direction.

Here are the equations I came up with for the accelerations:
Wedge
$$\vec{a}$$ = -$$\frac{mSin(\beta)Cos(\beta)}{M}$$g$$\hat{i}$$

Block
$$\vec{a}$$ = gSin($$\beta$$)Cos($$\beta$$)$$\hat{i}$$ - gSin2($$\beta$$)$$\hat{j}$$

They seem to make sense when M is extremely massive compared to m, and when $$\beta$$ varies from 0 to 90 degrees; however, they seem to fail when m $$\geq$$ M.

Do my equations look correct?

2. Aug 30, 2008

### Dschumanji

/Bump

3. Aug 31, 2008

### Dschumanji

Argh someone help me out here, this problem is torturing me.

For both the block and wedge I used a coordinate system where positive x is to the right and positive y is up. Here are the net forces on the block and wedge:

Block
$$\Sigma$$Fx=nSin($$\beta$$)=max
$$\Sigma$$Fy=nCos($$\beta$$)-mg=may

Wedge
$$\Sigma$$Fx=-nSin($$\beta$$)=MAx
$$\Sigma$$Fy=N-nCos($$\beta$$)-Mg=0

A is the acceleration of the wedge, a is the acceleration of the block, n is the normal force from the contact of the wedge and block, and N is the normal force exerted by the table. I have no idea where to go from here. I get confused since the block is sliding down the wedge, which is a non inertial frame of reference. I don't know whether a is the acceleration of the block with respect to the table or the wedge.

4. Aug 31, 2008

### kaeltemittel

Hi!

Honestly, i´m a bit tired now; nevertheless I´ll try to write something conclusive…

For both the block and wedge you used a coordinate system where positive x is to the right and positive y is up. Not so clever. Attach your x-y-system to the wedge and let the horizontal coordinate X of the wedge lie in the “table” system (laboratory system). This reduces the number of equations for the wedge´s movement to one:

M A = N sin beta.

Then you can describe the block´s motion by two other equations:

x: m(a + A cos beta) = - mg sin beta

y: -m A sin beta = N - mg cos beta

You see, you don´t have two normal forces anymore, your n and N become the same…with these three equations the problem is settled (at least I hope so)

Greetings from Germany!

Last edited: Aug 31, 2008
5. Aug 31, 2008

### Dschumanji

My equations show that the movement of the wedge is in one direction. The net force in the y direction was included to show that I assumed, with good reason, that the wedge doesn't accelerate in the verticle direction.

6. Sep 2, 2008

### Dschumanji

Moving Wedge Solved!

I posted a problem a couple days ago concerning a moving wedge and block system. I'm so excited to have finally solved it! I didn't get much help from the last thread, my guess is because it is such a tough problem (atleast for us physics noobs). Because I am so excited I thought it would be fun to share! Here is the problem statement and some work:

A wedge with mass M is place on a frictionless table. A block with mass m is placed on top of the wedge. There is no friction between the block and wedge. What are the accelerations of the wedge and block when the system is released? Give your answer in terms of components.

We assume that the table is our main frame of reference since it is inertial. The positive x direction is the right and positive y direction is up. The wedge makes an angle $$\alpha$$measured from the negative x axis towards the positive y axis. With this information we can figure out the forces acting on the wedge and block. Here are the net forces:

Wedge
$$\Sigma$$Fx=-nSin($$\alpha$$)=MAx

Block
$$\Sigma$$Fx=nSin($$\alpha$$)=max
$$\Sigma$$Fy=nCos($$\alpha$$)-mg=may

We assume that the y acceleration of the wedge is zero and don't include it in the equations above. The three equations above indicate we have four unknown variables, n, Ax, ax, and ay. We need to find a fourth equation that relates some of the variables to be able to solve for the important variables: Ax, ax, and ay.

The fourth equation comes from the kinematic description of the block's motion in the non-inertial frame of the wedge. The x and y position of the block are given by the following:

x = x0+voxt+0.5(ax - Ax)t2
y = y0+voyt+0.5ayt2

We can assume that the block starts at the origin and know that the initial velocity of the block is zero. The equations then reduce to:

x = 0.5(ax - Ax)t2
y = 0.5ayt2

Solving for t2 and setting the equations equal to each other we arrive at:

$$\frac{y}{a_{y}}$$ = $$\frac{x}{a_{x} - A_{x}}$$

If we divide each side by x we can substitute y/x by -Tan($$\alpha$$). We use -Tan($$\alpha$$) because we originally measured $$\alpha$$ from the negative x axis to the positive y axis; in the case of the wedge as a frame of reference, $$\alpha$$ is measured from the positive x axis to the negative y axis since it is moving down the wedge. We then arrive at our fourth equation:

-Tan($$\alpha$$) = $$\frac{a_{y}}{a_{x} - A_{x}}$$

After doing tons of algebra you find that the accelerations are:

Ax = $$\frac{-mgTan($$\alpha$$)}{(m+M)Tan^{2}($$(\alpha)$$)+M}$$
ax = $$\frac{MgTan($$\alpha$$)}{(m+M)Tan^{2}($$\alpha$$)+M}$$
ay = $$\frac{-(m+M)gTan^{2}($$\alpha$$)}{(m+M)Tan^{2}($$\alpha$$)+M}$$

7. Sep 2, 2008

### Dschumanji

Re: Moving Wedge Solved!

LaTeX seems to be spazzing out. The three images not showing up are:

-mgTan($$\alpha$$)
MgTan($$\alpha$$)
-(m+M)gTan2($$\alpha$$)