Why Is Work on a Curve Described with Integrals and Differentials?

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Homework Help Overview

The discussion revolves around understanding the concept of work in physics, particularly in the context of a constant force applied to a mass moving along a curved path. Participants express confusion regarding the transition from the basic formula for work to the more complex integral and differential forms used for curved motion.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of work and its application to curved paths, questioning how to apply the integral form of work. There are discussions about the role of angles and the relationship between force and displacement in the context of curved motion.

Discussion Status

The conversation is ongoing, with various participants attempting to clarify the mathematical expressions involved in calculating work. Some have provided insights into the use of integrals and the significance of angles, while others are still grappling with the logic behind these concepts. There is no clear consensus yet, as multiple interpretations and approaches are being explored.

Contextual Notes

Participants have noted the need for additional context regarding the specific problem being addressed, as well as the assumptions about the force and its direction. There is a mention of potential confusion stemming from the use of angles in the calculations.

transgalactic
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i know that the formula of work is W=F*X

but when i saw how they describe the work of a constant force applied on a mass moving at a curve it looks very different (integrals,differentials)

i can't see the logic of these equations??

http://img177.imageshack.us/img177/7741/64565014ea9.gif
 
Last edited by a moderator:
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transgalactic said:
i know that the formula of work is W=F*X

but when i saw how they describe the work of a constant force applied on a mass moving at a curve it looks very different (integrals,differentials)

i can't see the logic of these equations??

http://img177.imageshack.us/img177/7741/64565014ea9.gif
The expressions can be derived from the general definition of work done by a force:

W = \int_\gamma \bold{F}\left(\bold{x}\right)\bold{\cdot}d\bold{x}

Where the integral is taken over the path \gamma which the force acts and x is the spatial variable (vector).
 
Last edited by a moderator:
transgalactic said:
i know that the formula of work is W=F*X
That's OK as a basic definition, but it only works if the force is constant and in the same direction as the displacement. A more general definition is:
dW = \vec{F}\cdot d\vec{s}

[ah... Hoot beat me to it while I was getting coffee!]
 
so how do i apply this formula (dot product of force and distance)
to my case

<br /> W = \int_\gamma \bold{F}\left(\bold{x}\right)\bold{\cdot}d\bold{x}<br />?
 
Why don't you describe the problem you are trying to solve. (It's not obvious from your diagram.)
 
transgalactic said:
so how do i apply this formula (dot product of force and distance)
to my case

<br /> W = \int_\gamma \bold{F}\left(\bold{x}\right)\bold{\cdot}d\bold{x}<br />?
For the straight section of the path, between A & B:

F(x) = F0 \hat{i}

And

dx = dx1 \hat{i}

So

\bold{F}\left(\bold{x}\right)\bold{\cdot}d\bold{x} = F_0 dx_1 \underbrace{\left(\hat{i} \cdot \hat{i} \right)}_{1}

Hence,

W_{AB} = \int^B_A F_0 dx_1 = F_0\int_B^A dx_1 = F_0\left[x_1\right]^B_A

W_{AB} = F_0\left(B-A\right) = F_0\alpha

Do you follow?

Note that I am using the notation \bold{x} = x_1\hat{i} + x_2\hat{j}
 
i want to build the equations of work
from point B to D of my diagram

the equation that i was given as solution are odd
i can't follow their logic on the diagram ??
 
transgalactic said:
i want to build the equations of work
from point B to D of my diagram

the equation that i was given as solution are odd
i can't follow their logic on the diagram ??
But you haven't told us why you can't follow the logic or what you think the answer should be.

If I assume, like Hoot did, that the force F is constant and in the +x direction, then I disagree with the answer. Note that under this assumption no integration is needed to get the answer. (But it makes a good exercise.)
 
i can't understand how they use angle theta
??
 
  • #10
transgalactic said:
i can't understand how they use angle theta
??
First things first: Is it true that the force is constant and directed toward +x?

In finding the work done, you must take the component of the force in the direction of the displacement (or vice versa). That involves the angle theta.
 
  • #11
the force is constant

i need to build an integral of force with small displacement
the displacement cannot be an angle

its suppose to be a small distance
??
 
  • #12
Assuming that curved section is part of a circle, a small displacement (ds) can be written in terms of a small angle (dθ) via ds = Rdθ.
 
  • #13
Rsinθ can meen a distance
but Rdθ is radius times angle
that is not a small distance
??
 
  • #14
When the angle is measured in radians, Rθ is an arc length. When θ is small, as in dθ, the length (Rdθ) is small. (Review the concept of radian and arc length.)
 
  • #15
theta is in radians

so Rθ=Rsinθ
??
 
  • #16
transgalactic said:
theta is in radians

so Rθ=Rsinθ
??
No. (That would imply that θ=sinθ, which is not true in general.)
 
  • #17
can you recommend a manual for that kind of questions
explaining the integral thing with the use of angles

for me an integral is a way to calculate an area
i can't link it to work
even more if they use angles
??
 
  • #18
Have you taken multivariable calculus? You should find this stuff in a calculus text. BTW, the work integral can be thought of as an area, but you have to imagine that it is out of the page. It is sort of like the area of a ribbon that you are looking edge on, so it can curve along the path and all you see is a line. (But this analogy only works in 2 dimensions.) Check the Wikipedia article on "line integral". Wikipedia even has the work integral as the prototypical example in the introduction.
 

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