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

In summary, the book's answer is that N=MgtanA, where M is the mass of the disk, T is the tension in the string, A is the angle at which the tension points, and g is the acceleration of gravity. However, when I do the problem differently, using T=MgcosA, where M is the mass of the disk, N=MgsinA, and g is the acceleration of gravity, I get different results. My thinking seems correct, but my professor is still confused.
  • #1
StephenPrivitera
363
0
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
 
Physics news on Phys.org
  • #2
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
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
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
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
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
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
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.
 

What is tension in a string?

Tension in a string is the force that is exerted on the string when it is pulled or stretched. It is a measure of the resistance of the string to being deformed.

How is tension in a string calculated?

Tension in a string can be calculated using the formula T=mg, where T is tension, m is the mass of the object attached to the string, and g is the acceleration due to gravity.

What factors affect tension in a string?

The tension in a string is affected by several factors, including the force applied to the string, the length and thickness of the string, and the material properties of the string.

What is the normal force exerted by a wall?

The normal force exerted by a wall is the force that the wall applies to an object in contact with it, perpendicular to the surface of the wall. It is equal in magnitude and opposite in direction to the force applied by the object onto the wall.

How is the normal force exerted by a wall related to tension in a string?

The normal force exerted by a wall is directly related to the tension in a string. As the tension in the string increases, the normal force exerted by the wall also increases, as the string pulls harder against the wall.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
381
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
856
  • Introductory Physics Homework Help
2
Replies
42
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
603
  • Introductory Physics Homework Help
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
866
Back
Top