lightburst Posted December 20, 2013 Share Posted December 20, 2013 (edited) I've been trying to compute for the Work done by the wind F using calculus. Haven't had progress at all. http://www.pennsbury.k12.pa.us/pennsbury/Staff%20Pages/Pennsbury%20High%20West/Paulson,%20Tim/Honors%20Physics%20Fall%202013/Resources%20and%20Links/Materials%20by%20Chapter/Chapter%206/Jane%20and%20Tarzan%20solution.pdf Edited December 20, 2013 by lightburst Link to comment Share on other sites More sharing options...

timo Posted December 20, 2013 Share Posted December 20, 2013 (edited) Generally, for a force F along a path P the work done by said force is [math] W = \int_{a}^b \! \vec F \cdot \frac{\partial \vec P}{\partial s}\, ds[/math], where s=a...b is a parameter describing the process along the path (e.g. time). Since the force is constant in this case, this simplifies to [math]W = \vec F \cdot \int_{a}^b \! \frac{\partial \vec P}{\partial s}\, ds = \vec F \cdot ( \vec P(b) - \vec P(a) )[/math]. Reason I am sketching this rather abstract calculation? Firstly, because it is what you asked in the headline. Secondly, because it demonstrates that the force of the wind in this case does not act on the vertical part of the path, only on the horizontal one. Also note that the force of wind in this example seems to be a conservative force. Edited December 20, 2013 by timo Link to comment Share on other sites More sharing options...

studiot Posted December 20, 2013 Share Posted December 20, 2013 (edited) You have mentioned potential energy in your equation (delta PE), but have not mentioned gravity in your list of forces? Is there not a change of elevation to take into account? Edited December 20, 2013 by studiot Link to comment Share on other sites More sharing options...

swansont Posted December 20, 2013 Share Posted December 20, 2013 You have mentioned potential energy in your equation (delta PE), but have not mentioned gravity in your list of forces? Is there not a change of elevation to take into account? That's included in the PE, as it is a conservative force. The non-conservative forces are mentioned because they can do work. Work is a dot product, as timo has pointed out. You have to integrate along the path, from start to finish. Write it out in terms of the angles. [math] W = \int_{\theta}^\phi \! \vec F \cdot \, ds[/math] Link to comment Share on other sites More sharing options...

lightburst Posted December 20, 2013 Author Share Posted December 20, 2013 Generally, for a force F along a path P the work done by said force is [math] W = \int_{a}^b \! \vec F \cdot \frac{\partial \vec P}{\partial s}\, ds[/math], where s=a...b is a parameter describing the process along the path (e.g. time). Since the force is constant in this case, this simplifies to [math]W = \vec F \cdot \int_{a}^b \! \frac{\partial \vec P}{\partial s}\, ds = \vec F \cdot ( \vec P(b) - \vec P(a) )[/math]. Reason I am sketching this rather abstract calculation? Firstly, because it is what you asked in the headline. Secondly, because it demonstrates that the force of the wind in this case does not act on the vertical part of the path, only on the horizontal one. Also note that the force of wind in this example seems to be a conservative force. Walk me through the math, please. I know the definition of Work along a curved path. I just don't know how I can apply it to this particular problem. My book isn't much help. (To be fair, this problem didn't even come from it) I figure I need to represent the path as a function (or a vector function). I thought P = <R cost, R sint> but that went down the drain. Link to comment Share on other sites More sharing options...

swansont Posted December 20, 2013 Share Posted December 20, 2013 The dot product is Fscos(theta) (theta being a generic angle between the force and displacement vectors, not the one given in the problem) So the path will be some distance along x that geometry/trig will give you, of length s, and the work is [math]Fs\int cos(\theta) [/math] Link to comment Share on other sites More sharing options...

studiot Posted December 20, 2013 Share Posted December 20, 2013 (edited) That's included in the PE, as it is a conservative force. The non-conservative forces are mentioned because they can do work. Work is a dot product, as timo has pointed out. You have to integrate along the path, from start to finish. Write it out in terms of the angles. We are clearly on a different wavelength here. Let us suppose that Jane is 0.75 metres above Tarzan and 1.00 metres above the nadir of the swing. (In the picture Jane is definitely shown as some distance above both.) Now if Jane starts with zero velocity and no wind she will execute pendulum motion with all the PE given by 50 . 1.g This will be just enough to raise her 0.75 metres above Tarzan on the other side of the river. However the wind is blowing and so Jane only reaches the level of Tarzan, ie 0.25 metres above the Nadir. So the difference in PE is lost in overcoming the wind resistance, ie 50. 0.75 . g This must equal the work done by the wind resistance If the wind is any stronger Jane will not reach Tarzan without an initial KE, to add extra energy. This extra KE will just compensate for the extra wind resistance. So the graph of launch speed v wind resistance does not pass through the origin. Edited December 20, 2013 by studiot Link to comment Share on other sites More sharing options...

swansont Posted December 20, 2013 Share Posted December 20, 2013 We are clearly on a different wavelength here. Let us suppose that Jane is 0.75 metres above Tarzan and 1.00 metres above the nadir of the swing. (In the picture Jane is definitely shown as some distance above both.) Now if Jane starts with zero velocity and no wind she will execute pendulum motion with all the PE given by 50 . 1.g This will be just enough to raise her 0.75 metres above Tarzan on the other side of the river. However the wind is blowing and so Jane only reaches the level of Tarzan, ie 0.25 metres above the Nadir. So the difference in PE is lost in overcoming the wind resistance, ie 50. 0.75 . g This must equal the work done by the wind resistance If the wind is any stronger Jane will not reach Tarzan without an initial KE, to add extra energy. This extra KE will just compensate for the extra wind resistance. So the graph of launch speed v wind resistance does not pass through the origin. I don't see how this is inconsistent with anything I said, or was mentioned in the OP. Conservative forces, such as gravity, go into the PE. Nonconservative forces are accounted for in the work. You would find the work by integrating Fds over the path. F is constant, so it's the integral of s cos(theta) over the path. Link to comment Share on other sites More sharing options...

studiot Posted December 21, 2013 Share Posted December 21, 2013 (edited) This is really difficult to discuss without doing the problem completely for lightburst. However I feel that discussing path integrals along curved paths is overkill for this question and someone should have said so. I have already indicated that it cannot be done without considering potential energy. The detail is just geometry, in terms of the variables given. To examine the work done by the wind in terms of the geometry and the variables given first realise that the weight (Tarzan + Jane) play no part in this, as stated by the problem. Presumably becuase their dimensions are small compared to the length of the rope L. So let us consider and elemental length of rope, dl, at a distance l from the pivot. The wind force per unit length = F/L and is horizontal. Each elemental length dl travels a horizontal distance lsin(theta) + lsin(phi) against (or with in part b) the force F/L dl so the work done is F/L * l(sin(theta) + sin(phi)) dl If we integrate this from l = 0 to l = L we have the total work done by the wind. OK lightburst, it is up to you to turn this into a polished solution. Note you cannot just claim a wind force F at some unknown point in the diagram as you have done, although you could replace my integral by a suitable averaging process and a single force, but you would then have to calculate that. Edited December 21, 2013 by studiot Link to comment Share on other sites More sharing options...

timo Posted December 21, 2013 Share Posted December 21, 2013 However I feel that discussing path integrals along curved paths is overkill for this question and someone should have said so. I fully agree that integrals along a curve are overkill. Saying so actually was the intended message of my previous post. But evidently so, that message didn't quite transfer. I blame the scary integral sign (despite the fact that I explicitly performed the integration). Link to comment Share on other sites More sharing options...

Endercreeper01 Posted December 21, 2013 Share Posted December 21, 2013 (edited) Angular work would be W = τθ, where θ is the angle traveled and τ is torque. Because τ = Iω, we can also write angular work as W = Iωθ We could use vector calculus, since torque is also equal to mv x r, where x represents the cross product, making W = θ(mv x r) Edited December 21, 2013 by Endercreeper01 Link to comment Share on other sites More sharing options...

lightburst Posted December 21, 2013 Author Share Posted December 21, 2013 I don't know what a Force per unit length is. I googled it and it seems to be related to electricity. I already solved the problem using conservation of energy. The Work done by the wind was simply FD (Force times horizontal distance). I just prefer knowing a detailed solution is all so I wanted to do it in calculus. I mean do I need to learn new math here? My experience is limited to single variable calculus. I mean I know how to do some of the more complicated stuff, but as far as being useful all I have is single var calc. [math]F\int \cos\theta\; ds[/math] I tried to use the arc length equation s = r(theta), but the damn r is there when I want a D. F/L * l(sin(theta) + sin(phi)) dl I know that l(sin(theta) + sin(phi)) = D and you end up with (F/L)D dl and when you integrate you lose the L and end up with FD which is what I expect. I just don't know what F/L is. Link to comment Share on other sites More sharing options...

studiot Posted December 21, 2013 Share Posted December 21, 2013 (edited) I know that l(sin(theta) + sin(phi)) = D and you end up with (F/L)D dl and when you integrate you lose the L and end up with FD which is what I expect. I just don't know what F/L is Does it and do you? How far does the fixed end of the vine move? I make it zero distance. So what is the work done at the fixed end by the wind? If F is the total force applied to a total length L of vine then we can say that each metre is subject to a force of F/L Since F is constant and L is constant F/L is constant or fixed. Please note I have used L for the fixed total length and l as a variable so that an integratiion can be performed. I am sorry if this has confused you. But remember that distance is also a function of l and work is the product of all the little forces acting over a variable distance. Work = a fixed force times a variable distance. = F times a function of l So the integration is [math]W = \int_0^L {\frac{F}{L}} (afunctionofl)dl = \frac{F}{L}\int_0^L {(afunctionofl)dl} [/math] Have another look at my post to see if you can sort out what that function is. Hint horizontal distance is due to the sine, vertical distance is due to the cos, but is not so simple to integrate. I'm away for a week now so I will have to let others sort out this question with you. good luck Edited December 21, 2013 by studiot Link to comment Share on other sites More sharing options...

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