What Is the Correct Direction of Normal Force in a Stick-in-Well Scenario?

[terryds]

Homework Statement

http://www.sumoware.com/images/temp/xzmbharheaqshjjf.png

A homogeneous stick with mass M is placed in a well with smooth surface (see picture). θ is the angle between the stick and the floor.
Determine the force the well gives to the stick.

Homework Equations

ΣF = 0
Στ = 0[/B]

The Attempt at a Solution

I draw the free body diagram but I'm not sure
http://www.sumoware.com/images/temp/xzgemxpxlpjnkafk.png
The first body-diagram has normal force 1 is perpendicular to the surface (like common ladder problem)

or

http://www.sumoware.com/images/temp/xzjmplgoxrepcxct.png
The second body diagram has normal force 1 is up-left in direction. I think it's because the stair push the wall down-right, it must give the reaction in opposite direction.
?

Which one is correct ? The first one or the second one ?

 

[ehild]

The end of the stick is curved, it can push the wall in any direction, but the vertical smooth wall can exert only horizontal force, perpendicular to the contact surface. As the wall exerts horizontal force on the stick, the stick must exert also horizontal force on the wall (Newton's third law) .

 

[terryds]

The end of the stick is curved, it can push the wall in any direction, but the vertical smooth wall can exert only horizontal force, perpendicular to the contact surface. As the wall exerts horizontal force on the stick, the stick must exert also horizontal force on the wall (Newton's third law) .

So, walls and floors can only give the reaction force perpendicular to the surface, right ?

Then,
∑F_y = 0
N2_y - W = 0
N2_y = W

∑F_x = 0
N2_x - N1 = 0
N2_x = N1

∑τ = 0
W cos Θ * 0.5 L - N1 sin Θ * L= 0
W cos Θ = N1 sin Θ * 2
W = 2 N1 tanΘ
N1 = W / 2tanΘ

Then, N2 = √((N2x)^2 + (N2y)^2) = √((W / 2tanΘ)^2 + (W)^2)
Right ?

 

[ehild]

N1 = W / 2tanΘ

Then, N2 = √((N2x)^2 + (N2y)^2) = √((W / 2tanΘ)^2 + (W)^2)
Right ?

It is right, but use parentheses in the denominators. N1 = W / (2tanΘ) and N2 = √((W / (2tanΘ))^2 + (W)^2) and pull out W from the square root.

 

[terryds]

It is right, but use parentheses in the denominators. N1 = W / (2tanΘ) and N2 = √((W / (2tanΘ))^2 + (W)^2) and pull out W from the square root.

Thanks a lot

 

[ehild]

You are welcome:)

 

[CWatters]

So, walls and floors can only give the reaction force perpendicular to the surface, right ?

Only if contact is frictionless.

Edit: Your diagram shows the bottom in the corner of the well so it wouldn't apply there.