# Homework Help: Block on an incline adjacent to a wall

1. Oct 9, 2007

A wedge with an inclination of angle theta rests next to a wall. A block of mass m is sliding down the plane, as shown. There is no friction between the wedge and the block or between the wedge and the horizontal surface.
http://session.masteringphysics.com/problemAsset/1010942/31/MFS_cf_11.jpg
Find the magnitude, F_ww, of the force that the wall exerts on the wedge.
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i know that the net force on the block is just = mgsin(theta) .. and i also know that the net horizontal force on the wedge is zero, but the problem asks for the magnitude of the force that the wall exerts on the wedge .. so i'm getting nowhere. help?

2. Oct 9, 2007

### Staff: Mentor

Hint: What force does the block exert on the incline?

3. Oct 9, 2007

i'm not sure i understand where you are getting at, but the block exerts a force of mg on the incline .. no?

4. Oct 9, 2007

### Staff: Mentor

No. mg is the weight of the object. If it were resting on the incline, then you'd be correct. But it's not. Hint: Think normal force.

5. Oct 9, 2007

normal force is mgcos(theta) so the force the the block exerts on the incline is just the negative of that ?? -mgcos(theta) ?? i'm sorry i'm just guessing here...

6. Oct 10, 2007

guys, i still don't have it ..

7. Oct 10, 2007

### pooface

no mgcos(theta) is correct. that is the FN on the block by the incline and vice versa, they must equalize.

I suspect the force the wall on the wedge is zero because they are just next to each other not leaning and dont seem to be applying any forces on each other.

I don't know.

8. Oct 10, 2007

yeah NET force is 0 but obviously that's not the answer it's looking for ..

9. Oct 10, 2007

### pooface

My best guess would be the horizontal component of the frictional force. but since there is no friction, there is no frictional force. So my guess would be 0N. The wall does not exert any force on the wedge. That is what the question is asking for right?

Last edited: Oct 10, 2007
10. Oct 10, 2007

for sure, it's NOT 0 ..lol

11. Oct 10, 2007

### Staff: Mentor

You shouldn't have to guess about the normal force--or Newton's 3rd law. Yes, the normal force of incline on block is mgcos(theta) pointing out of the incline surface, so the force of block on incline is mgcos(theta) pointing into the incline. Find its x and y components.

12. Oct 6, 2010

### OssumPawesome

The normal force is mgcos(theta), but that is at an angle. you need to take the horizontal component of that force, mgcos(theta)*sin(theta). (or mgsin(2theta)/2, if you want to cut down on trig terms)

Think about it this way... the force by the wall on the wedge has to be 0 at 2 points, 0 degrees of wedge (block is flat) and 90 degrees (block doesn't touch wedge)... does this statement work for both of those?