Work done by a vector force

In summary: Could you please help me understand?In summary, the homework statement is trying to find out what work is done by a force when an object moves from point A to point B. The results showed that the force is conservative and does not require any work to be done.
  • #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.
 
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  • #2
mattyc33 said:

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
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
mattyc33 said:
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
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
mattyc33 said:
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
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
mattyc33 said:
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
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
mattyc33 said:
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
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
mattyc33 said:
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
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 I am looking for are theta = 0 and pi.

I don't understand what the limits for this integral would be though. As well as what I am doing with these values of theta.
 
  • #14
mattyc33 said:
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 I am looking for are theta = 0 and pi.

I don't understand what the limits for this integral would be though. As well as what I am 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
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 doesn't seem right to me, did I misplace a number that I can substitute in for b?
 
  • #16
sorry 2b*
 
  • #17
mattyc33 said:
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 doesn't 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
I actually got help from a TA but thank you so much for your help until now.
 

1. What is work done by a vector force?

The work done by a vector force is a physical quantity that measures the amount of energy transferred to an object by a force as it moves through a certain distance in the direction of the force. It is calculated by multiplying the magnitude of the force by the distance moved in the direction of the force.

2. How is work done by a vector force different from work done by a scalar force?

Work done by a vector force takes into account both the magnitude and direction of the force, while work done by a scalar force only considers the magnitude. This means that the direction of the force must be taken into account when calculating work done by a vector force.

3. What is the unit of measurement for work done by a vector force?

The SI unit for work is joules (J). This is the same unit used for both work done by a vector force and work done by a scalar force.

4. Can the work done by a vector force be negative?

Yes, the work done by a vector force can be negative. This occurs when the force and the displacement are in opposite directions, meaning that the force is actually acting against the direction of motion. In this case, the work done is considered to be negative.

5. How does the angle between the force and displacement affect the work done by a vector force?

The angle between the force and displacement affects the work done by a vector force through the use of the dot product. The work done is equal to the magnitude of the force multiplied by the magnitude of the displacement, and then multiplied by the cosine of the angle between them. This means that the work done is greatest when the force and displacement are in the same direction (angle of 0 degrees) and zero when they are perpendicular (angle of 90 degrees).

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