Work done by a non constant force over a non constant angle

[CHACHA 14]

TL;DR

Hi everyone, I was struggling with a homework where we need to calculate the work done over a particle moving along a path where the force acting on this path and the angle between these two are changing everytime. I tried to set up a double integral since the force is not constant and dependent on the distance from the original source of the force and also for the angle which is changing.

I was just wondering if I needed to set up this double integral or not. I've never seen anywhere people setting up double integrals for calculating the work.

 

[A.T.]

I was just wondering if I needed to set up this double integral or not.

It's a single integral:
https://en.wikipedia.org/wiki/Work_(physics)#Work_done_by_a_variable_force

 

[CHACHA 14]

It's a single integral:
https://en.wikipedia.org/wiki/Work_(physics)#Work_done_by_a_variable_force

But in the case showed in Wikipedia the angle is either 0 degrees or 180 that is why they don't consider the angle. But in my case, the angle between the force and the displacement is not the same and the force is not constant. Do I still just need to set up one integral for just the force ? But inside the integral it is a dot product where we have cos teta so I would need also to figure it out the angle which is changing.

 

[A.T.]

But in the case showed in Wikipedia the angle is either 0 degrees or 180

In the second formula of the section "Work done by a variable force", F(t) and v(t) are completely arbitrary vector functions of time, with an arbitrary angle between them.

... that is why they don't consider the angle.

The angle is handled by the dot-product between F(t) and v(t).

Do I still just need to set up one integral for just the force ?

One integral, but not of "just the force" but of the dot-product. The integral includes everything up to "dt".

 

[PeroK]

But in the case showed in Wikipedia the angle is either 0 degrees or 180 that is why they don't consider the angle. But in my case, the angle between the force and the displacement is not the same and the force is not constant. Do I still just need to set up one integral for just the force ? But inside the integral it is a dot product where we have cos teta so I would need also to figure it out the angle which is changing.

The line integral $\int \vec F \cdot d\vec r$ covers all cases. Given that it's a line integral it must admit parameterization using a single variable, such as time or arc length.

 

[PeroK]

PS a double integral generally represents integration over a surface, and not along a line.

 

[Dale]

TL;DR Summary: Hi everyone, I was struggling with a homework where we need to calculate the work done over a particle moving along a path where the force acting on this path and the angle between these two are changing everytime. I tried to set up a double integral since the force is not constant and dependent on the distance from the original source of the force and also for the angle which is changing.

I was just wondering if I needed to set up this double integral or not. I've never seen anywhere people setting up double integrals for calculating the work.

Chances are you are not supposed to evaluate this integral. You probably are supposed to use the conservation of energy.

 

[CHACHA 14]

In the second formula of the section "Work done by a variable force", F(t) and v(t) are completely arbitrary vector functions of time, with an arbitrary angle between them.

The angle is handled by the dot-product between F(t) and v(t).

One integral, but not of "just the force" but of the dot-product. The integral includes everything up to "dt".

So how can I get rid off the cos teta since it is a dot product inside my integral ?
Should I only consider the variable as my distance r initial and r final even though the angle between the displacement and the force is changing ?

 

[Dale]

You should post the actual homework problem in the homework section. I suspect you are doing this the hard way.

You don’t get rid of the cosine. It is important. The dot product includes the cosine inside it. It may hide the cosine, but it doesn’t remove it

 

[CHACHA 14]

I need to prove that the electric force is a conservative force. For that I need to calculate the work done on one particle for two different paths. The first one is simple because the angle between the force and the displacement is 0 for the entire path and thus cos teta is just equal to 1. I just need to integrate over the distance the electron is traveling because the electric force is not a constant force. The other one is more complicated because the force is doing an angle with the path. On one hand the force is not constant and on the other hand the force is doing an angle with the path which is changing everytime. How can I figure it out or hide this cos teta if teta is changing over the distance the particle is traveling ?

 

[Dale]

How can I figure it out or hide this cos teta if teta is changing over the distance the particle is traveling ?

Break all the vectors into components. Then the dot product is $\vec a \cdot \vec b = (a_x,a_y)\cdot (b_x,b_y) = a_x b_x + a_y b_y$

You may also find it easier to evaluate $$\int_{t_i}^{t_f} \vec F(t) \cdot \vec v(t) \ dt$$ instead of $$\int_{s_i}^{s_f} \vec F(s) \cdot d\vec s$$