Calculate the work done by a force

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SUMMARY

The discussion focuses on calculating the work done by a force defined as F = b(1-x²/a²)j along a rectangular path with vertices at (0,0), (0,L), (a,L), and (a,0). The correct approach involves integrating the force dotted with the differential path vector dr for each segment of the path. The participants confirm that the differential vectors for each segment are dyj for the first segment, dxi for the second, -dyj for the third, and -dxi for the fourth. The conclusion indicates that the work done may not be zero, as the force is not necessarily conservative.

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  • Understanding of vector calculus, specifically line integrals.
  • Familiarity with conservative forces and their properties.
  • Knowledge of the dot product in vector mathematics.
  • Ability to perform definite integrals with variable limits.
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I am asked to calculate the work done by a force as it moves around a path. The force is F = b(1-x^2/a^2)j. The path is a rectangle with coordinates at (0,0); (0,L); (a,L); (a,0). The force moves clockwise around the path beginning at the origin. A diagram is attached.

I know work is the integral of F dot dr.
So for the first path I should have the the force F=b(1-x^2/a^2)j dotted with Lj (the path from the origin to point (0,L)). The integral is thus bL (1 - x^2/a^2) dy with limits from y=0 to y=L. Is this the right approach? If not, can someone please point me in the right direction??
 

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You have an extra factor of L that you don't need (look at the units). The work is the integral of \vec{F}\cdot d\vec{r} along the path. For example, the first segment has d\vec{r} = dy \hat{j}. You have to figure out what d\vec{r} is for each of the four segments.
 
So, for the first segment, dr = dyj. For the second segment, dr = dx i.
For the third segment, dr = -dyj. For the fourth segment, dr = -dx i. Is this correct? Are the limits on my integration correct as well?

Also, should the answer be 0 (closed path, conservative force...not sure if the force is conservative though)?
 
If the answer is zero then the force is conservative, but not all forces are conservative so you can't use that as a check here. (Your dr vectors are correct).

-Dale
 
I get an answer of 2bL (1- x^2/a^2). This does not seem correct to me, since it contains an x^2 term? Is this right? Is there a substitution I can make for x? x=a or x=L, for instance? This problem is driving me crazy...any help greatly appreciated!
 
You can eliminate two of the legs from your problem since the force is in the \hat{j} direction.

In segment 1, x=0. In segment 3, x=a.
 
Last edited:
Yes...I have the forces for the dx direction to be zero. I'm still doing something wrong though?
 
Did you substitute in the values for x that I just edited into my last post?
 

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