The work integral (general question)

[ZanyCat]

I'm sure you're all familar with this forumla for work, $W=\int_{s_i}^{s_f} \mathrm{\vec{F}}\cdot\,\mathrm{d}\vec{s}$

I don't understand how to evaluate this integral. How do you antidifferentiate in terms of a vector? How do you evaluate the dot product when ds isn't an actual value?

Thanks! :)

 

[NaOH]

It's done with something called vector calculus, more specifically, line integrals.
http://en.wikipedia.org/wiki/Line_integral#Vector_calculus

Depending on whether it's a scalar or vector field, it is calculated slightly differently. Nevertheless, we will do a dot product inside the integral while manipulating the integral into something more familiar, so only parallel components will count.

 

[ZanyCat]

So would we do the dot product of F and delta-s? If so, that would give us a scalar value (or perhaps a function), and what would we integrate that in terms of?

 

[SteamKing]

Remember that ds can be broken down into components dx and dy just like F can be broken down into components Fx and Fy (Fx and Fy do not represent partial derivatives here).

 

[Nickie]

I can understand your confusion about evaluating the work integral. Let me break it down for you.

Firstly, the work integral is a mathematical representation of the physical concept of work. It represents the amount of energy transferred to or from an object by a force acting on it over a certain distance. The integral symbol (∫) is used to represent the addition of an infinite number of small amounts of work, which is why it is also known as the "area under the curve" in calculus.

Now, let's look at the components of the integral. The integral sign is followed by the limits of integration, s_i and s_f, which represent the initial and final positions of the object. The function inside the integral is the dot product of the force vector (F) and the displacement vector (ds). The dot product is a mathematical operation that results in a scalar value, which can be thought of as a magnitude or a numerical value. This value is then multiplied by the infinitesimal displacement vector (ds) and integrated over the distance traveled by the object.

To evaluate the integral, you first need to know the function for the force (F) acting on the object. This could be a constant force, a variable force, or a function of position. Once you have the function for the force, you can calculate the dot product with the displacement vector (ds), which is a vector with both magnitude and direction. The dot product of two vectors is calculated by multiplying their magnitudes and the cosine of the angle between them. This will result in a scalar value that can then be integrated over the distance traveled by the object.

It may seem complex, but with practice and understanding of the underlying concepts, you will be able to evaluate work integrals easily. I hope this helps to clarify your doubts. Keep exploring and learning!