# Thread: Work Done By A Force

1. ## Work Done By A Force

I've been racking my brain on this question for hours now, and I can't seem to get it. It's as follows:

"Find the work done by the force F(x,y) = (x + 2y)i + (6y - 2x)j acting counterclockwise once along the triangle with vertices (1,1), (3,2) and (1,4)."

A classmate thinks this should equal out to zero since it's a closed triangle and there isn't any ultimate displacement, but we cannot get it to work when we try solving it.

Can somebody offer any help?

2. Originally Posted by Katzenjammer
I've been racking my brain on this question for hours now, and I can't seem to get it. It's as follows:

"Find the work done by the force F(x,y) = (x + 2y)i + (6y - 2x)j acting counterclockwise once along the triangle with vertices (1,1), (3,2) and (1,4)."

A classmate thinks this should equal out to zero since it's a closed triangle and there isn't any ultimate displacement, but we cannot get it to work when we try solving it.

Can somebody offer any help?

Hi , Have you consider the Green's Theorem ?

$\displaystyle \int_{closed C} F \cdot dr = {\int\int}_D ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} ) ~dydx$ ?

but you should not expect that it must be a conservative field ...

Here $\displaystyle P = x + 2y ~ , ~ Q= 6y-2x$ then

$\displaystyle \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$

$\displaystyle = -2 - 2 = -4$

so the work done is $\displaystyle {\int\int}_D (-4)~dydx$

$\displaystyle = (-4) \cdot ~ ( Area ~ of ~the~ triangle)$

my answer is -12

3. That does make sense.

I think I was making it harder than it had to be. I was trying to find piecewise parametrizations of each line, C1 C2 and C3 and I wasn't getting anywhere.

Thank you so much!

4. Originally Posted by Katzenjammer
That does make sense.

I think I was making it harder than it had to be. I was trying to find piecewise parametrizations of each line, C1 C2 and C3 and I wasn't getting anywhere.

Thank you so much!

If the path is closed , i will think of the Green's Therorem first