Tension in the string and the normal force exerted by the wall

  • #1
A uniform solid disk of mass M and radius R hangs from a string of length l attached to a smooth wall. We want the tension in the string and the normal force exerted by the wall.
|
|\ <--- I call this angle A
|_\
|__\
|_(..) <--- that's a disk
|
|

So I have it set up so the tension points up and to the left, the normal exactly to the right and the weight exactly down. The answer appears to take the normal perpendicular to the tension, which quite confuses me. Anyway, I get
TsinA=N
TcosA=Mg
so N=MgtanA
and T=Mg/cosA

The book gets T=MgcosA
and N=MgsinA
 

Answers and Replies

  • #2
13
0
Change your coordinates to where T is straight up the vertical. This will give you the right answers.
 
  • #3
I see that it works for the tension. Doesn't the normal have to be perpendicular to the wall? If it is then mgsinA is not N.
 
Last edited:
  • #4
13
0
Yep

mg*Sin(A) is perpendicular to wall.
 
  • #5


Originally posted by MaxMoon
mg*Sin(A) is perpendicular to wall.
How can that be? There is no component of the weight that is perpendicular to the wall. MgsinA is perpendicular to the tension, which is not perpendicular to the wall.
 
  • #6
13
0
y

Here is something to think about: In order for the system to remain static, what must the normal force be?
 
  • #7
The normal force acts horizontally. In order for the system to have zero horizontal linear acceleration, the normal force must be equal in magnitude and opposite in sign to the sum of the other horizontal force. The weight acts down. Its horizontal component is 0. The tension has a horizontal component TsinA. We must have N-TsinA=0 or N=TsinA, which is not the correct result.
I'm very confused. Why do I get different answers for the tension doing the problem two different ways? My way seems perfectly right and the book's way seems perfectly right.
 
  • #8
Hurkyl
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Unless I'm missing something, the magnitude of tension cannot be less than the weight of the disk, because tension is the only force opposing gravity.

(incidentally, is the string attached to the center of the disk?)
 
  • #9
Originally posted by Hurkyl

(incidentally, is the string attached to the center of the disk?)
No. It's attached to the edge of the disk.
So IOW, the book must be wrong?
 
  • #10
Hurkyl
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Smooth walls can supply no frictional force to disks, right? If so, tension is the only force with an upwards component, and thus must be no less in magnitude than mg, so the book would have to be wrong...
 
  • #11
Doc Al
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Originally posted by StephenPrivitera
No. It's attached to the edge of the disk.
So IOW, the book must be wrong?
I don't know what you mean by "attached to the edge". The line of the string, if extended, passes through the center of the disk, right? If not, the string would exert a torque on the disk, rotating it.

In any case, your thinking seems correct. The book's answer does not.
 
  • #12
HallsofIvy
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Stephen, your picture seems to show the disk horizontal with, as yo say, the string at the edge. What keeps the disk from falling straight down? Is the disk attached to the wall also or is there a friction force there?
 
  • #13
Originally posted by HallsofIvy
Stephen, your picture seems to show the disk horizontal
Yes.
Originally posted by HallsofIvy
the string at the edge
I'm sorry that description wasn't very clear. The string, if extended past the edge, would go through the center. You can say it is attached to the center if you wish without affecting the result, except for the fact that when I say the string has length l, I mean the distance from the edge of the disk to the wall where the string is attached is l. The distance to the center is l+R.

Originally posted by HallsofIvy
Is the disk attached to the wall also or is there a friction force there?

The wall is smooth, no friction.

I have spoken with my professor, and he is also confused by the book's answer.
 

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