Block on an incline adjacent to a wall

[jaded18]

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?

 

[Doc Al]

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

 

[jaded18]

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

 

[Doc Al]

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.

 

[jaded18]

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...

 

[jaded18]

guys, i still don't have it ..

 

[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 don't seem to be applying any forces on each other.

I don't know.

 

[jaded18]

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

 

[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?

 

[jaded18]

for sure, it's NOT 0 ..lol

 

[Doc Al]

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...

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.

 

[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?