Work done by a distributed force on a string

[Eidos]

Hi All

I'd like to know how I could calculate the work done by a distributed force on a string.

Let's say the force at a point $$x$$ at a time $$t$$ is given by

$$F(x,t)$$.

Now the instantaneous amplitude of the string is given by $$y(x,t)$$, say

I think that the work done by the force in changing the configuration of the string from some $$y(x_0,t_0)$$ to $$y(x,t)$$ should be something like
$$\int_{y(x_0,t_0)}^{y(x,t)} F(x,t) dy$$

I'll use the total derivative on $$y(x,t)$$ which gives

$$dy=y_t dt+y_x dx$$

where

$$\frac{\partial y}{\partial x}=y_x$$

Now the integral becomes something like

$$\int F(x,t) y_t dt + \int F(x,t) y_x dx$$

My concerns here are the limits I need to put in each integral.

Any help would be greatly appreciated :D

 

[omoplata]

I would like to ask some questions to understand this better.

How does the force $$\vec{F}(x,t)$$ act on the string? I'm guessing it's perpendicular to the $$x$$ direction and along the $$y$$ direction?

If it is perpendicular, why are you integrating from $$y(x_{0},t_{0})$$ to $$y(x,t)$$ ? Shouldn't it be integrated from $$y(x,t_{0})$$ to $$y(x,t)$$ ? (i.e. instead of $$\int_{y(x_{0},t_{0})}^{y(x,t)} F(x,t) dy$$ , I think it should be $$\int_{y(x,t_{0})}^{y(x,t)} F(x,t) dy$$ ). I mean, why does the $$x$$ coordinate change?

 

[Eidos]

I've attached the picture of how the force enters the system.

I agree with your assertion that the limits should include changes in $$t$$ only.
How will this effect my change of variable then?

That is when I take the total derivative of $$dy$$,
how do I exclude $$y_x dx$$ from the total derivative?

 

[omoplata]

If the tension in the string is completely neglected, my approach would be

• Find the infinitesimal work $$dW(x)$$ done between $$x$$ and $$x + dx$$ by $$dW(x) = \int_{y(x,t_{0})}^{y(x,t)} F(x,t) dy$$. If $$F$$ doesn't cahnge with time, just take it as $$F(x)$$. If it does, then find $$y(t)$$ using Newton's second law for the part of string between $$x$$ and $$x+dx$$ with mass $$m$$ by $$F(x,t) = m \frac{ \partial^{2} y }{ \partial t^{2} }$$

• Find the total work by $$W = \int_{x=x_{1}}^{x_{2}} dW(x) dx$$

But a string in real life would have elasticity and tension and that analysis would be different. Then there would also be the extra work done to stretch the string.

 

[PJC01]

I would first clarify what is meant by "work" in this scenario. In physics, work is defined as the force applied to an object multiplied by the distance it moves in the direction of the force. In this case, the "object" is the string and the force is distributed along its length. So the work done by the force on the string would be the integral of the force multiplied by the displacement of each infinitesimal element of the string along its length.

To calculate this, we can break the string into small segments of length dx. The force acting on each segment is F(x,t) and the displacement of that segment is y(x,t) - y(x+dx,t). So the work done by that segment would be F(x,t)(y(x,t) - y(x+dx,t))dx.

To find the total work done by the distributed force, we would integrate this over the entire length of the string, from x = a to x = b. This would give us the following expression:

W = \int_{a}^{b} F(x,t)(y(x,t) - y(x+dx,t))dx

This integral can be simplified using the total derivative of y(x,t) as shown in the original post. However, the limits of integration would remain the same, as they represent the starting and ending points of the string.

I would also note that in this scenario, the force is not acting on a point mass but on a string with varying tension and displacement. Therefore, the work done by the force would not result in a change in kinetic energy as it would for a point mass. Instead, it would result in a change in the potential energy of the string. This is an important distinction to make in understanding the physical implications of this calculation.

I hope this helps clarify the concept of work done by a distributed force on a string. Please let me know if you have any further questions or concerns.