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Work done by a vector force

  • Thread starter mattyc33
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  • #1
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Homework Statement


Determine the work done by the Force F = yi(hat) - xj(hat) on an object that travels from point A (a,0) to point B (-a,0)

a) along an elliptical path described by x=acosθ, y=bsinθ
b) along a straight line from A to B
c) From these results, can we determine whether or not the force is conservative?

Homework Equations



W=∫F ° dl
dl=dxi(hat) + dyj(hat)

The Attempt at a Solution



a) W=∫(a to -a) bsinθi(hat)dl - ∫(0 to 0) acosθj(hat)dl
so that just gives me ∫(a to -a) bsinθi(hat)dl which equals 0. This doesn't seem right to me. What's mainly confusing me is that the y is corresponding to the i(hat) vectors and vice versa for the x. I don't really know what to do with the b.

b) dl = dxj(hat)
W = ∫(a to -a) xdy = 0

c) I know that you can tell if a force is conservative if the work depends on the path. Therefore this is probably not conservative.
 

Answers and Replies

  • #2
gabbagabbahey
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Homework Statement


Determine the work done by the Force F = yi(hat) - xj(hat) on an object that travels from point A (a,0) to point B (-a,0)

a) along an elliptical path described by x=acosθ, y=bsinθ
b) along a straight line from A to B
c) From these results, can we determine whether or not the force is conservative?

Homework Equations



W=∫F ° dl
dl=dxi(hat) + dyj(hat)

The Attempt at a Solution



a) W=∫(a to -a) bsinθi(hat)dl - ∫(0 to 0) acosθj(hat)dl
so that just gives me ∫(a to -a) bsinθi(hat)dl which equals 0. This doesn't seem right to me. What's mainly confusing me is that the y is corresponding to the i(hat) vectors and vice versa for the x. I don't really know what to do with the b.
Use your two relevant equations, what is [itex]\mathbf{F}\cdot d\mathbf{l}[/itex] if [itex]\mathbf{F} = y\mathbf{i} - x\mathbf{j}[/itex] and [itex]d\mathbf{l} = dx\mathbf{i}+ dy\mathbf{j}[/itex]?
 
  • #3
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So then I would get ∫ydx - xdy.

How would I integrate y with dx and x with dy?
I'm sort of behind on this stuff and this may seem simple and I'm sorry for that.
 
  • #4
gabbagabbahey
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So then I would get ∫ydx - xdy.

How would I integrate y with dx and x with dy?
I'm sort of behind on this stuff and this may seem simple and I'm sorry for that.
You have equations for both [itex]x[/itex] and [itex]y[/itex] in terms of a single parameter, [itex]\theta[/itex], use them and remember that [itex]dx =\frac{dx}{d\theta}d\theta[/itex]
 
  • #5
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So I would get an equation that looks like ∫bsinθ(dx/dθ)dx - ∫acosθ(dx/dθ)dθ

which yields:

-bcosθ - bsinθ

Would that be my answer? Or are there further steps pertaining to points A and B?
 
  • #6
gabbagabbahey
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So I would get an equation that looks like ∫bsinθ(dx/dθ)dx - ∫acosθ(dx/dθ)dθ
Don't you mean [itex]\int_{\theta(x=a, y=0)}^{\theta(x=-a,y=0)}\left(b\sin\theta \frac{dx}{d\theta} - a\cos\theta \frac{dy}{d\theta} \right)d\theta[/itex]?

which yields:

-bcosθ - bsinθ
I get something different, can you post your steps?

Would that be my answer? Or are there further steps pertaining to points A and B?
Work done is the path integral from A to B, so you should have a definite integral where the limits are the value of [itex]\theta[/itex] at each endpoint.
 
  • #7
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Sorry I seemed to have misplaced a dx with what should have been dθ.

Continuing on:

I split the integrals:
∫bsinθ(dx/dθ)dθ - ∫acosθ(dx/dθ)dθ with the limits a to -a.

If I integrate bsinθ I get -bcosθ and if I integrate acosθ I get a sinθ (another mistake in my previous answer)

Therefore the expression would be:

-bcosθ - asinθ from the limits a to -a

resulting in:

(-bcos(-a) - asin(-a)) - (-bcos(a) - asin(a))

I think I may be integrating the original equation incorectly with the due to confusion by the dθ's and dx's.
 
  • #8
gabbagabbahey
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Sorry I seemed to have misplaced a dx with what should have been dθ.

Continuing on:

I split the integrals:
∫bsinθ(dx/dθ)dθ - ∫acosθ(dx/dθ)dθ with the limits a to -a.

If I integrate bsinθ I get -bcosθ and if I integrate acosθ I get a sinθ (another mistake in my previous answer)
What happened to [itex]\frac{dx}{d\theta}[/itex] and [itex]\frac{dy}{d\theta}[/itex]? Why would the limits be a to -a? What is the value of [itex]\theta[/itex] at the point A? What about at point P?
 
  • #9
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I made another mistake in writing the final equation down so the final equation from before would be:

∫bsinθ(dx/dθ)dθ - ∫acosθ(dy/dθ)dθ with limits a to -a

This doesn't really change much though.

Well technically it's θ to θ but since there is no y component wouldn't the limits just be the x components?

I'm extremely confused at this point on how I would find the θ in point A or B because I have no idea what "a" means.
 
  • #10
gabbagabbahey
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I made another mistake in writing the final equation down so the final equation from before would be:

∫bsinθ(dx/dθ)dθ - ∫acosθ(dy/dθ)dθ with limits a to -a

This doesn't really change much though.

Well technically it's θ to θ but since there is no y component wouldn't the limits just be the x components?

I'm extremely confused at this point on how I would find the θ in point A or B because I have no idea what "a" means.
"a" and "b" are just numbers. Plot out your parametric equations as you range theta from 0 to 2pi... what value of theta gives you x=a and y=0? What value of theta gives you x=-a and y=0?

As for dx/dθ and dy/dθ, you will need to calculate them...
 
  • #11
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Well for sinθ y=0 at 0, ∏ and 2∏ Therefore would "a" be pi/2 and theta be 0, 180 or 360?

for cosθ y=0 at ∏/2 and 3∏/2 so b would be pi/2 or 3pi/2 and theta be 90 or 270...

I am just really confused at this point as I feel like I need to review basic geometry however this question is one of the only ones I don't understand in my textbook chapter.
 
  • #12
gabbagabbahey
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Well for sinθ y=0 at 0, ∏ and 2∏ Therefore would "a" be pi/2 and theta be 0, 180 or 360?
Why ask me? You have [itex]y(\theta)=b\sin\theta[/itex], so what is [itex]y\left( \frac{\pi}{2} \right)[/itex]?

for cosθ y=0 at ∏/2 and 3∏/2 so b would be pi/2 or 3pi/2 and theta be 90 or 270...
Huh?:confused:

You have [itex]x(\theta)=a\cos\theta[/itex], what are [itex]x\left( 0 \right)[/itex], [itex]x\left( \frac{\pi}{2} \right)[/itex], [itex]x\left( \pi \right)[/itex], etc?

I am just really confused at this point as I feel like I need to review basic geometry however this question is one of the only ones I don't understand in my textbook chapter.
I think you should review linear algebra, especially curves and parameterization.
 
  • #13
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Then y=b at theta = pi/2.
y=0 at theta = 0 as well




x=a at theta = 0
x=-a at theta = pi

Therefore the values that im looking for are theta = 0 and pi.

I don't understand what the limits for this integral would be though. As well as what im doing with these values of theta.
 
  • #14
gabbagabbahey
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Then y=b at theta = pi/2.
y=0 at theta = 0 as well




x=a at theta = 0
x=-a at theta = pi

Therefore the values that im looking for are theta = 0 and pi.

I don't understand what the limits for this integral would be though. As well as what im doing with these values of theta
You are integrating along an eliptical path from point A to point B. When you parametrize that path in terms of [itex]\theta[/itex], you find that [itex]\theta=0[/itex] corresponds to point A and [itex]\theta=\pi[/itex] corresponds to point B, so those values of [itex]\theta[/itex] are your lower and upper limits.
.
 
  • #15
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So if I complete the integral:

∫(bsinθdx/dθ - acosθdy/dθ)dθ From 0 to pi

= -bcosθ - asinθ from 0 to pi

= b-1

Would b-1 be my final answer for work? That doesnt seem right to me, did I misplace a number that I can substitute in for b?
 
  • #16
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sorry 2b*
 
  • #17
gabbagabbahey
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So if I complete the integral:

∫(bsinθdx/dθ - acosθdy/dθ)dθ From 0 to pi

= -bcosθ - asinθ from 0 to pi

= b-1

Would b-1 be my final answer for work? That doesnt seem right to me, did I misplace a number that I can substitute in for b?
No, [itex]\int_0^{\pi} b\sin\theta \frac{dx}{d\theta} d\theta \neq -b\cos\theta[/itex], what happened to [itex]\frac{dx}{d\theta}[/itex]?
 
  • #18
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I actually got help from a TA but thank you so much for your help until now.
 

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