Gareth56 Posted February 7, 2009 Share Posted February 7, 2009 I understand that Work is not a vector quantity but a scalar. The definition of Work = Force x displacement. So if I apply a force of magnitude X to an object causing a displacement in a certain direction doesn't that constitute a vector? Link to comment Share on other sites More sharing options...
ydoaPs Posted February 7, 2009 Share Posted February 7, 2009 Are you using a cross product or a dot product? Link to comment Share on other sites More sharing options...
D H Posted February 7, 2009 Share Posted February 7, 2009 The proper definition is [math]W=\oint {\mathbf F}\cdot d{\mathbf l}[/math] The "[math]\cdot[/math]" symbol indicates inner product. For a constant force and a straight line path this simplifies to [math]W={\mathbf F}\cdot {\mathbf d}[/math] Only in the very simple case where the force and displacement are aligned do you get W=F*d. These simplifying cases are used in high school physics because high school physics students (and unfortunately, their teachers) do not have the necessary mathematical background to understand the general formula. Link to comment Share on other sites More sharing options...
Klaynos Posted February 7, 2009 Share Posted February 7, 2009 Displacement and force are both vectors, you've actually written the wrong equation there force x displacement Is different to: [math]Work = force \cdot displacement[/math] They are two different ways of multiplying vectors the second (the dot product) results in a scalar value. Whereas the first (the cross product) results in a vector orthogonal to both of the starting vectors. Link to comment Share on other sites More sharing options...
ydoaPs Posted February 7, 2009 Share Posted February 7, 2009 (edited) DH, use [math][\math] tags for [math]\LaTeX[\math]. Edited February 7, 2009 by swansont Fix LaTex tags, or maybe not Link to comment Share on other sites More sharing options...
Gareth56 Posted February 7, 2009 Author Share Posted February 7, 2009 Displacement and force are both vectors, you've actually written the wrong equation there force x displacement Is different to: [math]Work = force \cdot displacement[/math] They are two different ways of multiplying vectors the second (the dot product) results in a scalar value. Whereas the first (the cross product) results in a vector orthogonal to both of the starting vectors. What I meant when I wrote WD = Force x Dispacement ....was Force multiplied by Dispalcement. It seems I need to read upon Dot products and Cross products. Thanks as always. Link to comment Share on other sites More sharing options...
fatoumata Posted February 16, 2009 Share Posted February 16, 2009 well, W= force* distance moved in the direction of the force since distance is a scalar and U consider the absolute value of F, Work is always a scalar. Link to comment Share on other sites More sharing options...
D H Posted February 16, 2009 Share Posted February 16, 2009 well, W= force* distance moved in the direction of the force since distance is a scalar and U consider the absolute value of F, Work is always a scalar. As mentioned above, [math]W=F*d[/math] is a simplification of the concept of work. It assumes that the force is constant and that the displacement is a straight line path parallel to the force. How much work is done in making a 1 kg mass move 1 meter horizontally? The gravitational force acts on the mass throughout. Since [math]{\boldsymbol F}_g = m {\boldsymbol g} \approx m*9.81\,\text{m}/\text{s}^2[/math] downward, that simple formula suggests that the work done is about 9.81 joules. It is in fact zero. Why? The correct interpretation in this case is [math]W= {\boldsymbol F} \cdot {\boldsymbol d}[/math], where both the force and displacement are vectors. The work is identically zero in the above example since the displacement vector is normal to the force vector. If the path is not a straight line or if the force is not constant one must resort to the integral definition, [math]W=\oint {\boldsymbol F} \cdot d{\boldsymbol l}[/math]. Link to comment Share on other sites More sharing options...
Gareth56 Posted February 17, 2009 Author Share Posted February 17, 2009 Sorry to appear dim here but I still cannot see why something such as Work which has both a magnitude and a direction can be a scalar! Link to comment Share on other sites More sharing options...
D H Posted February 17, 2009 Share Posted February 17, 2009 A scalar is any real number, not a non-negative real number. Link to comment Share on other sites More sharing options...
swansont Posted February 17, 2009 Share Posted February 17, 2009 Sorry to appear dim here but I still cannot see why something such as Work which has both a magnitude and a direction can be a scalar! Work doesn't have a direction. Force and displacement do, but the dot product gets rid of direction information. Link to comment Share on other sites More sharing options...
D H Posted February 17, 2009 Share Posted February 17, 2009 Work doesn't have a direction. Force and displacement do, but the dot product gets rid of direction information. I gather that Gareth is assuming that since work can be positive or negative it cannot be a scalar. The answer is that the word "scalar" has multiple meanings, only one of which is a quantity with magnitude only. Another meaning: A vector can be multiplied by a scalar (google "algebra over a field"). For example, the vector (1,2,3) can be scaled by 2, yielding the vector (2,4,6) or by -2, yielding the vector (-2,-4,-6). Here a scalar is any real number, and that is the context in which work is a scalar. Note that for a complex vector space, complex numbers act as scalars. Physicists have yet another definition of scalar: Something that isn't changed by rotation of coordinates. A vector doesn't satisfy this, but work does. Link to comment Share on other sites More sharing options...
Gareth56 Posted February 17, 2009 Author Share Posted February 17, 2009 Work doesn't have a direction. Force and displacement do, but the dot product gets rid of direction information. If I push a car with a certain Force in a certain Direction for a certain Distance then wouldn't I have performed Work on said car? And as the Force has been applied in a certain direction (say easterly) isn't that a vector? Link to comment Share on other sites More sharing options...
iNow Posted February 17, 2009 Share Posted February 17, 2009 Yes, you performed work, but it was not the work which had the direction. Work is not a concept with direction. Force is. That is where your vector information lives. Work is scaler because you can perform work lifting a tractor or pushing a car or hammering a nail... the concept of work is not dependent on direction. You're mixing the idea of force into it, and force is where the vector is accounted for. Force and work are not the same thing. Force is vector, work is scalar. Link to comment Share on other sites More sharing options...
swansont Posted February 17, 2009 Share Posted February 17, 2009 If I push a car with a certain Force in a certain Direction for a certain Distance then wouldn't I have performed Work on said car? And as the Force has been applied in a certain direction (say easterly) isn't that a vector? The direction of the force compared to the displacement will affect the amount of work done, but at the end of it all the only thing you can say is whether you added energy to the system or removed it. The dot product tells you the extent the force and displacement are in the same direction, but doesn't actually tell you their direction. If you take your example and rotate it by an arbitrary angle, you'll get the same answer for the work. Link to comment Share on other sites More sharing options...
Bignose Posted February 17, 2009 Share Posted February 17, 2009 If you take your example and rotate it by an arbitrary angle, you'll get the same answer for the work. I.e. it doesn't matter if you pushed you car 100 m to the North, or 100 m to the East, or 100 m to the West, (of course, assuming a perfectly flat, uniform piece of land), you have done the same amount of work. Similarly, if you calculated how much work you did by pushing your car 100 m to the North. Then, the next day, everyone in the world agreed to start calling "North", "East", it doesn't change the amount of work performed. I know that that example is pretty farcical, but the idea is to show that if you choose one direction as the x coordinate, and then rotated the coordinates 90 degrees so that what was x is now y, it doesn't change the amount of work that was performed. I.e., the specific direction doesn't matter. Link to comment Share on other sites More sharing options...
Gareth56 Posted February 18, 2009 Author Share Posted February 18, 2009 Couldn't that argument/reasoning be applied to a quantity of Force? Link to comment Share on other sites More sharing options...
swansont Posted February 18, 2009 Share Posted February 18, 2009 If you move 10 meters east vs 10 meters north you are in different locations, even if you relabel the axes. Link to comment Share on other sites More sharing options...
D H Posted February 18, 2009 Share Posted February 18, 2009 Bignose physically rotated the vectors, rather than rotating the reference frames but leaving the vectors the unchanged. Those are very different concepts. Physically rotating the vectors: It takes a whole lot more work to lift a car vertically 100 meters than to roll it horizontally 100 meters. Rotating the reference frames: Whether you decide your x, y, and z unit vectors are directed north, east, and down versus east, north, and up versus whatever does not change the amount of work needed to push a car 100 meters north, but it sure does change the representation of the force vector needed to do accomplish that goal. Link to comment Share on other sites More sharing options...
Gareth56 Posted February 18, 2009 Author Share Posted February 18, 2009 Rotating vectors and frames of reference, oh dear now it's getting really confusing for a layman whose interested in physics. I'll just have to accept that Work is not a vector. Thanks to all. Link to comment Share on other sites More sharing options...
iNow Posted February 18, 2009 Share Posted February 18, 2009 Rotating vectors and frames of reference, oh dear now it's getting really confusing for a layman whose interested in physics. I'll just have to accept that Work is not a vector. Thanks to all. Gareth, The way I look at those comments is that sometimes work will increase if you're moving upward (since you have to fight gravity). That's where vector becomes a factor, but it still has nothing to do with "type" of work. Only "amount" of work. And that's really the biggest point. The crucial issue is "amount" of work, not direction of work or "type" of work. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now