# Wedge On A Frictionless Surface

1. Nov 14, 2007

### mjkr1988

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

A wedge with mass rests on a frictionless horizontal table top. A block with mass is placed on the wedge. There is no friction between the block and the wedge. The system is released from rest.

2. Relevant equations

(-gm)/((M+m)tan(alpha)+(M/tan(alpha)))

3. The attempt at a solution

The question is to calculate the vertical and horizontal components of the BLOCKS acceleration.

I know that the above equation is the acceleration of the wedge, so i know that the acceleration of the block should be equal an opposite. Is this actually true? I'm not too sure. Also, would I swap the tan functions for cosine and sine for horizontal and vertical acceleration respectively?

Could some one please point me in the right direction! :) that would be much appreciated!

2. Nov 15, 2007

### Staff: Mentor

What does Newton's 3rd law tell you? Considering the block and wedge together, what external horizontal forces act?
Why would you think that? (Did you derive the above equation or just find it in a book?)

As always, identify the forces acting on each and apply Newton's 2nd law.

3. Nov 15, 2007

### mjkr1988

I derived the frist equation myself with great difficulty. Now I have to find the horizontal and vertical acceleration of the block. So I thought the acceleration of the block is equal to the acceleration of the wedge (as you pointed out Newton's Third Law - opposite and equal reactions).

The reasoning behind the swapping of cos and sin for tan is from simple trig identities, sine of an angle is the vertical and cosine is the horizontal. (Thats if its a right angle triangle, which it is, and they have given the angle)

So would the acceleration of the block be equal and opposite, and then would I multiply these values with cos(alpha) and sine(alpha) to derive the horizontal and vertical, or would I have to derive a new equation from scratch?

4. Nov 15, 2007

### Staff: Mentor

Newton's 3rd law talks of force, not acceleration. The force that the block exerts on the wedge is equal and opposite to the force that the wedge exerts on the block. Take advantage of the fact that there's no friction between block and wedge to figure out the net force on the block and its resultant acceleration.

5. Nov 16, 2007

### mjkr1988

Right ok, Ive figured out the horizonal acceleration of the block, swapping the negative for a postive, (as the accelerations of each are away from each other). Then times the equation by M, divide through by small m to get a, acceleration of block.

So how would I go about getting the vertical component?

6. Nov 16, 2007

### Staff: Mentor

Figure out the normal force between block and wedge. Then you'd be able to find the net force on the block and its acceleration.

7. Nov 16, 2007

### mjkr1988

Ive got mg as one component, and I know the horizonal so what other components should I include?

8. Nov 16, 2007

### mjkr1988

so is it mg - the vertical value for the normal force?

9. Nov 16, 2007

### mjkr1988

and is the normal force the vector sum of Mg and the orginal equation i quoted for the acceleration of the wedge?

10. Nov 16, 2007

### Staff: Mentor

What's the only horizontal force acting on the wedge? Use that to figure out the normal force. (What's the direction of the normal force?)