# Question about force on a curve

1. Dec 6, 2008

### transgalactic

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 cant see the logic of these equations??

http://img177.imageshack.us/img177/7741/64565014ea9.gif

2. Dec 6, 2008

### Hootenanny

Staff Emeritus
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 spacial variable (vector).

Last edited: Dec 6, 2008
3. Dec 6, 2008

### Staff: Mentor

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!]

4. Dec 6, 2008

### transgalactic

so how do i apply this formula (dot product of force and distance)
to my case

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

5. Dec 6, 2008

### Staff: Mentor

Why don't you describe the problem you are trying to solve. (It's not obvious from your diagram.)

6. Dec 6, 2008

### Hootenanny

Staff Emeritus
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 $\bold{x} = x_1\hat{i} + x_2\hat{j}$

7. Dec 6, 2008

### transgalactic

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 cant follow their logic on the diagram ??

8. Dec 6, 2008

### Staff: Mentor

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.)

9. Dec 6, 2008

### transgalactic

i cant understand how they use angle theta
??

10. Dec 6, 2008

### Staff: Mentor

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. Dec 6, 2008

### transgalactic

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. Dec 6, 2008

### Staff: Mentor

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. Dec 6, 2008

### transgalactic

Rsinθ can meen a distance
but Rdθ is radius times angle
that is not a small distance
??

14. Dec 6, 2008

### Staff: Mentor

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. Dec 6, 2008

### transgalactic

so Rθ=Rsinθ
??

16. Dec 6, 2008

### Staff: Mentor

No. (That would imply that θ=sinθ, which is not true in general.)

17. Dec 14, 2008

### transgalactic

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 cant link it to work
even more if they use angles
??

18. Dec 15, 2008

### turin

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.