Force that the wall exerts on the wedge

In summary: This is done by multiplying mg\cos(\theta) by sin(theta). This will give you the force the block is pushing against the wall.
  • #1
klm
165
0
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.

Find the magnitude, F_ww, of the force that the wall exerts on the wedge.
Express F_ww in terms of theta and m, along with any necessary constants.

2hmfvqw.jpg


ok i tried it and i got mgcos(theta)^2 . but it saying that is incorrect. i drew my free body diagram and can't figure out where i am going wrong. i found Fnet = mg sin theta as well
 
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  • #2
I think what you need to do is calculate the horizontal force exerted by the block on the wedge. And that would be the force that the wedge exerts on the wall and hence the force the wall exerts back on the wedge.

Did you do that?
 
  • #3
isn't the only force the box acting on the wedge is weight
 
  • #4
no i mean, like okay we know n=mg cos theta right. and that is the force the box is acting on the wedge and the same the wedge is acting on the box right?
 
  • #5
Hmm, it's been a while since I did mechanics, but since this is in an incline, there has to be a force in the horizontal vector as well as the vertical doesn't it?

So the horizontal vector would be found by F = mg sin theta
 
  • #6
I am really not sure though, best if someone else answers this.
 
  • #7
right i found mgsintheta for my fney x. but i don't understand how that is going to help...sorry!
 
  • #8
thanks for trying =)!
 
  • #9
Well what is the answer given?
 
  • #10
i don't know! i have to try to figure what the force of the wall exerts on the wedge.
 
  • #11
Well try putting in mg sin theta as the answer, and cya.
 
  • #12
i don't think it should be mg sin theta only
 
  • #13
nvm i think i figured it out. thanks guys
 
  • #14
What was it? I am interested to know :)
 
  • #15
What you are doing here is resolving components of forces twice. So you need to find the component of weight acting perpendicular to the incling which you have as [itex]mg\cos(\theta)[/itex]. Then you have to resolve this force into the component that pushes the block against the wall.
 

Related to Force that the wall exerts on the wedge

What is the definition of "force that the wall exerts on the wedge"?

The force that the wall exerts on the wedge is the amount of force that the wall applies on the wedge when the wedge is pushed against it. This force is perpendicular to the surface of the wall and depends on the material and angle of the wedge.

How is the force that the wall exerts on the wedge calculated?

The force can be calculated using the formula F = μFn, where F is the force, μ is the coefficient of friction between the wall and the wedge, and Fn is the normal force exerted by the wall on the wedge. The normal force is equal to the weight of the wedge in most cases.

What factors affect the force that the wall exerts on the wedge?

The force that the wall exerts on the wedge is affected by the angle of the wedge, the coefficient of friction between the wall and the wedge, and the weight of the wedge. Other factors such as the surface roughness and the material of the wall and the wedge may also have an impact.

Can the force that the wall exerts on the wedge be greater than the weight of the wedge?

Yes, the force that the wall exerts on the wedge can be greater than the weight of the wedge if the coefficient of friction between the wall and the wedge is high enough. This means that the wall is able to support the weight of the wedge and exert an additional force on it due to the friction between the two surfaces.

How does the force that the wall exerts on the wedge affect the motion of the wedge?

The force that the wall exerts on the wedge can either help or hinder the motion of the wedge. If the force is in the opposite direction of the motion, it will act as a resistance and make it harder for the wedge to move. However, if the force is in the same direction as the motion, it can assist in moving the wedge forward.

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