Calculating Work Done by a 2-D Force

[jl9999]

"Work Done by 2-D Force"
What work is done by a force = ((1 N/m)x) + (4 N), with x in meters, that moves a particle from a position 1 = (2 m) + (3 m) to a position 2 = - (4 m) - (3 m)?

I figured all I would have to do is find the displacement of the x and y coordinates after the force had acted on them and then multiply the displacement vectors by their corresponding x and y force vectors, then use some trig to get the total work done. The answer I came up with was 35J, which of course was wrong. What am I missing? any help would be appreciated.

 

[K^2]

The x-component has to be integrated, since force isn't a constant.

The exact formula for work done in higher dimensions along some curve is this.

[tex]W = \int_{r_0}^{r_1}F(r) \cdot dl[/tex]

Note the scalar product between the force and displacement along curve.

Is this notation something you can make sense of, or do you need help breaking it down into geometric terms?

 

[jl9999]

Yeah, I can't make complete sense of the notation. I'm guessing it's something about an integral and a dot product but I might need a little extra push. thanks

 

[K^2]

Well, in the problem you posted, the first simplification is noting that force is conservative. (You might want to look that up if you don't know how I know this.) That means that the work is the same regardless of the path taken. So let us move in just the X direction first, and then in just the Y direction.

The dot product means that we are only interested in force component along displacement. Since we are going to move along x first, only F_x is of interest. The integral becomes.

[tex]W_x = \int_{x_1}^{x_2} F_x(x, y) dx = \int_{2m}^{-4m} 1\frac{N}{m} x dx = \frac{1}{2}\frac{N}{m}\left((-4m)^2 - (2m)^2\right) = 6 Nm = 6J[/tex]

Note that if F_x actually depended on y, you'd have to use y₁, since no displacement in y took place yet. In contrast, on the second step, when moving in y direction, you'd be keeping x=x₂. Of course, you could have started with y first, but again, in this case, it does not matter. And second step, similarly.

[tex]W_y = \int_{y_1}^{y_2} F_y(x, y) dy = \int_{3m}^{-3m} 4N dy = 4N\left((-3m) - 3m\right) = -24Nm = -24J[/tex]

The total work is just the sum of the two, W = -18J.

I don't know how much you know about the integration, but it's basically area under the curve. If the curve goes negative, these regions are counted as negative area. Similarly, integrating from right to left gives the same answer as from left to right, but with a minus sign. Sketching F_x as function of x and F_y as function of y, and looking at the areas from x₁ to x₂ and from y₁ to y₂ respectively, should tell you where the integration results come from.

 

[jl9999]

Now that I look at it again with your explanation in my mind it seems very easy. Anyway, thanks for your help and the quick tutorial.