Net force an an infinitesimal string

In summary, the book I am using (Zwiebach on page 66) uses the expression dF_v = T_0 \frac{ \partial{y}}{\partial{x}} |_{x+dx} -T_0 \frac{ \partial{y}}{\partial{x}} |_{x} for the force on an infinitesimal length of string. We assume dy/dx is much less than 1. I am not sure how the author got these expressions for the tension at x and x+dx? Is he using a Taylor series approximation?
  • #1
ehrenfest
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Homework Statement


The book I am using (Zwiebach on page 66) uses the expression

[tex] dF_v = T_0 \frac{ \partial{y}}{\partial{x}} |_{x+dx} -T_0 \frac{ \partial{y}}{\partial{x}} |_{x} [/tex]

for the force on an infinitesimal length of string. We assume dy/dx is much less than 1. I am not sure how the author got these expressions for the tension at x and x+dx? Is he using a Taylor series approximation?

Homework Equations


The Attempt at a Solution

 
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  • #2
Hey there,

First, try splitting the tension acting along the string into the x and y direction (see attachment). We know that the tension along the x direction (I've called this [tex]T_{0}[/tex]) have got to cancel out, else the string will be accelerating along the x-direction. So, looking in the y-direction and using geometry, the tensions at [tex]x[/tex] and [tex]x+\Delta x[/tex] are:

At [tex]x[/tex] : [tex] T_{(x)y} = - T_{0} \tan{\theta_{x}}[/tex]

At [tex]x+\Delta x[/tex] : [tex]T_{(x + \Delta x)y} = T_{0} \tan{\theta_{x + \Delta x}}[/tex]

But, we know that, for small theta, [tex]\tan{\theta}\approx \frac{\partial {y}}{\partial {x}} [/tex]. So:

[tex] T_{(x)y} = - T_{0} \tan{\theta_{x}} = - T_{0} \frac{\partial{y}}{\partial{x}} |_{x} [/tex]

and

[tex]T_{(x + \Delta x)y} = T_{0} \tan{\theta_{x + \Delta x}} = T_{0}\frac{\partial{y}}{\partial{x}} |_{x + \Delta x} [/tex]

And the net force acting on the string is just the sum of these components, or;

[tex] dF_v = T_{x + \Delta x}_{y} - T_{x}_{y} = T_0 \frac{ \partial{y}}{\partial{x}} |_{x+dx} -T_0 \frac{ \partial{y}}{\partial{x}} |_{x} [/tex]
 

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  • #3
Archduke said:
Hey there,

First, try splitting the tension acting along the string into the x and y direction (see attachment). We know that the tension along the x direction (I've called this [tex]T_{0}[/tex] have got to cancel out, else the string will be accelerating along the x-direction.

But I think T_0 is defined in my book as the tension along the string, not in the x direction. Don't we want sin(theta) T_0 to get the force in the y-direction, not tan(theta) T_0?

Also, T_0 is defined in my book (and I should have written this in my first post) as the tension along the string. Thus, there is a small net force in the x-direction. However, it is negligable.

Also, when you say

[tex] T_{(x)y} = - F_{0} \tan{\theta_{x}}[/tex]

do you mean

[tex] T_{(x)y} = - T_{0} \tan{\theta_{x}}[/tex]

?

Because the first equation seems incorrect to me?
 
  • #4
ehrenfest said:
Also, when you say

[tex] T_{(x)y} = - F_{0} \tan{\theta_{x}}[/tex]

do you mean

[tex] T_{(x)y} = - T_{0} \tan{\theta_{x}}[/tex]
Yeah, sorry about that. I've changed it now.

As for the tension in the x-direction, there is some there, but it is canceled out by tension of the same magnitude, but opposite direction, at [tex]x + \Delta x[/tex].

As for the sin/tan problem, it appears we're both right, because we've chosen to call T_0 something different. However, I can't see how we've then got the same expression. :confused:

I'll be away for a few days, so hopefully someone else can resolve it for you. Sorry if I've just added to your confusion with the different notation!
 
  • #5
Archduke said:
ehrenfest said:
As for the tension in the x-direction, there is some there, but it is canceled out by tension of the same magnitude, but opposite direction, at [tex]x + \Delta x[/tex].

Actually the tension is not necessarily the same at x and x+dx because dy/dx is evaluated at each of these points and could be different.

And yes someone else please help clarify the sin/tan issue please!
 
  • #6
Inasmuch as you've assumed |dy/dx|<<1 it doesn't matter whether you use sin or tan. They are both the same to the order you are concerned about. And the tension is the same if the string is massless. It couldn't support a variable tension w/o producing infinite accelerations.
 
  • #7
Dick said:
Inasmuch as you've assumed |dy/dx|<<1 it doesn't matter whether you use sin or tan. They are both the same to the order you are concerned about.

I see. Thanks.

Dick said:
And the tension is the same if the string is massless. It couldn't support a variable tension w/o producing infinite accelerations.

If the string is massless, it could not support a net vertical tension either. So, it does have a density.
 
  • #8
ehrenfest said:
If the string is massless, it could not support a net vertical tension either. So, it does have a density.

Even if you are dealing with massive string (and if you are doing string theory, you are usually thinking of fundamental massless strings - really). One should be able to argue that the tension variation can be neglected to the order you are working at, just like the difference between sin and cos. But I'm having a hard time coming up with an argument simple enough to be convincing.
 
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  • #9
Dick said:
Even if you are dealing with massive string (and if you are doing string theory, you are usually thinking of fundamental massless strings - really). One should be able to argue that the tension variation can be neglected to the order you are working at, just like the difference between sin and cos. But I'm having a hard time coming up with an argument simple enough to be convincing.

Firstly, you probably mean sin and tan. Secondly, I am doing ST but I am in Zwiebach Chapter 4 which deals with nonrelativistic strings and he explicitly says that the string has a density. Thirdly, I am not sure what you mean about neglecting tension variation. The difference between the vertical tension at x and x + dx can be shown to be much larger than the difference between the horizontal and it is not negligable at least in the context of the problems I am doing.
 
  • #10
Sure, I mean sin and tan. And I am not talking about ignoring the difference between the components of tension, I'm talking about ignoring the absolute value of the tension variation. Which I'm having problems coming up with a good argument for. Can you help?
 
  • #11
I see. You're question is why can we use T_0 at every point on the string, right?
 
  • #12
Right. Ok. Here's my answer. In principle, you should also consider the x acceleration on a segment of string. If the tension is constant, then this of the order of (y')^2 (since it's a cos of a small angle) and can be safely ignored, since |y'|<<1. If you allow the tension to be variable then this is of the order of dT/dx and you can't ignore it. In fact, these variable tension modes correspond to longitudinal oscillations of the string. But that's not what this problem is about, it's about the transverse modes. If you want to do the full problem you would have to take x and y to both be functions of t and work from there. I think to really justify the approximation you would have to add a premise that the mass density of the string is sufficiently small. I think it best to keep a simple problem simple.
 

Related to Net force an an infinitesimal string

1. What is net force on an infinitesimal string?

The net force on an infinitesimal string is the sum of all the forces acting on the string.

2. How is net force calculated on an infinitesimal string?

To calculate the net force on an infinitesimal string, you need to add up all the individual forces acting on the string. This can be done using vector addition, taking into account the direction and magnitude of each force.

3. What factors affect the net force on an infinitesimal string?

The net force on an infinitesimal string can be affected by the tension of the string, the angle at which the forces act on the string, and the mass of the string.

4. How does the net force on an infinitesimal string affect its movement?

The net force on an infinitesimal string determines the acceleration and movement of the string. If the net force is zero, the string will remain stationary. If the net force is non-zero, the string will accelerate in the direction of the net force.

5. How is net force related to the motion of an infinitesimal string?

The net force on an infinitesimal string is directly related to its motion through Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that a larger net force will result in a greater acceleration and therefore a greater change in the string's motion.

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