What Happens to Normal Force on a Slanted Plank?

AI Thread Summary
The discussion focuses on the normal force acting on a plank leaning against a wall, emphasizing that when the plank is stationary, the normal force from the ground equals the weight of the plank (MG) in the vertical direction, while the wall exerts a horizontal normal force. It is noted that if the plank begins to slide, the vertical normal force remains constant, but the horizontal normal force changes, potentially leading to loss of contact with the wall. The conversation also touches on the role of torque and moments in maintaining the plank's position. Participants express confusion about the dynamics involved when the plank starts to slide and the implications for normal forces. The thread concludes with a suggestion to seek further clarification from educational resources or a teacher.
thejinx0r
Messages
27
Reaction score
0
This has always puzzled me, the normal force, and I think I might understand this now, or at least part's of it.

So here's the case: there's a plank of wood leaning against a wall.

If it's not slipping/sliding or any of the sort, then there is no movement in the x or y direction.
Then, there must be a normal force from the floor holding it up. There must also be a normal force on the side that keeps it from falling.

That get's tricky here for me.

So, if it was perfectly balanced and not leaning on anything, then the normal force would just be equal to MG in the y-direction.

Now, if it is leaning on a wall, then the normal force coming from the wall would be only in the x direction. Thus, the horizontal normal force should not affect the vertical one. So, the normal force from the ground must also be equal to MG in this case.

I'm also learning about angular momentum right now. So, would the horizontal normal force be just a torque such that it prevents it from rotating down?

Now, uncertain territory begins:
Suppose the plank began to slide without friction.

The vertical normal force would then be a constant right?
And the horizontal normal force would change? But there is one little problem.
I've always noticed that my hockey stick (i.e a plank of wood) always seem to make a big bang sound as though the whole stick hit the floor and not just one edge.

So, does that mean that at some point, it loses contact with the wall?
 
Physics news on Phys.org
thejinx0r said:
This has always puzzled me, the normal force, and I think I might understand this now, or at least part's of it.

So here's the case: there's a plank of wood leaning against a wall.

If it's not slipping/sliding or any of the sort, then there is no movement in the x or y direction.
Then, there must be a normal force from the floor holding it up.
Not just. There might be a frictional component along the wall working in the y-direction.
There must also be a normal force on the side that keeps it from falling.
Not quite. The normal force from the wall is strictly horizontal, and prevents the plank from going into the wall.


So, if it was perfectly balanced and not leaning on anything, then the normal force would just be equal to MG in the y-direction.
correct.
Now, if it is leaning on a wall, then the normal force coming from the wall would be only in the x direction. Thus, the horizontal normal force should not affect the vertical one.
Indeed correct.
So, the normal force from the ground must also be equal to MG in this case.
Nope, answered above!
 
arildno said:
Not quite. The normal force from the wall is strictly horizontal, and prevents the plank from going into the wall.

Well, I see your point, but wouldn't it be considered falling if the wall was not there?
 
thejinx0r said:
Well, I see your point, but wouldn't it be considered falling if the wall was not there?

Nope. If the GROUND was witout friction, then the normal force would exert a horizontal force upon the plank so that the plank would start sliding down along the wall (the C.m of the plank would accelerate away from the wall.)
Thus, it is NOT the normal force from the wall by itself that normally prevents falling, but a combination of it and the friction along the ground (and, as asubsidiary point, some friction along the wall, as noted).
 
So, would the horizontal normal force be just a torque such that it prevents it from rotating down?

yes...for a stationary rigid plank of the type you describe the sum of the moments about any point is zero...a moment is the same thing as a torque...you can find complete explanations in many introductory physics texts...
 
Naty1 said:
yes...for a stationary rigid plank of the type you describe the sum of the moments about any point is zero...a moment is the same thing as a torque...you can find complete explanations in many introductory physics texts...

Ya, sort of true. I have a Kleppner and Kolenkow and there's nothing about it.
If there's one that you can recommend, then it would help me tremendously.

But anyways, I do understand the stationary part.
The part that I don't understand is when it starts to slide.

It's at the end of the first post, but I guess no one cared to comment.

I tried to go from first principles and what I knew about normal forces, but still not any better than yesterday,

Worst comes to worst, I'll just ask a teacher on Tuesday. (Tomorrow is Canadian thanks giving)
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top