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Vector notation work problem 
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#1
Feb2313, 02:21 PM

P: 51

1. The problem statement, all variables and given/known data
A force F=12NÎ10NĴ acts on an object. How much work does this force do as the object moves from the origin to the point R=13mÎ+11mĴ? 2. Relevant equations Work is the product of Force and Displacement (W=F*s) 3. The attempt at a solution Things I know, or think I know: The object moves from the origin to a point 13 units to the right and 11 units up. The square root of the those two numbers squared and added together gives the magnitude which if up and right are both positive directions is a positive quantity of 17 if only two significant figures are considered. The force acting on this object has one component that is in the same direction, producing positive work, and one component that is in the opposite direction, producing negative work. The vector notation puzzles me. Do I split this into two different work situations for each component? I'm not sure where to start besides the conceptual thinking part. I can come up with the correct answer by figuring out W_{Î}=F_{Î}*s_{Î} and then subtracting W_{Ĵ}=F_{Ĵ}*s_{Ĵ} which gives the correct answer of 46 Joules but I'm not sure if this process will get me into trouble later on down the road. How would you do this problem? I want to know other methods. Thank you so much in advance. 


#2
Feb2313, 02:35 PM

P: 937

Have learned about projection of vectors and the dot product of vectors yet?



#3
Feb2313, 02:40 PM

P: 51

Yes but the process doesn't immediately come to mind which is another reason I want to hear it from someone else other than my textbook. I do know that the dot product is the product of one vector(A) and then the amount of the other vector(B) that is parallel to the first vector (vector magnitude[A]*cos(theta)*vector magnitude[B]) but I don't remember the exact process of problem solving.



#4
Feb2313, 02:48 PM

P: 937

Vector notation work problem
The dot product of two vectors is [itex]\vec{u} \cdot \vec{v} = \vec{u}~\vec{v}\cos\theta[/itex] Or, [itex]u \cdot v = u_1v_1 + u_2v_2[/itex]
Work is defined as: [itex]W= \vec{F} \cdot \vec{r}[/itex] Where F is the force and r is the position vector. Do these stir any thoughts? 


#5
Feb2313, 02:55 PM

P: 51

Yes that all looks very familiar and it seems I might have done something of the sort in the second part (u1v1+u2v2) with the exception that I divided it into two different work calculations and then added them.



#6
Feb2313, 02:55 PM

P: 937

Although I am not certain about this next method (if someone could verify it for me): Also remember that work is defined as applying a force on a object, and pushing it through a displacement. However, we only care about the component of the force parallel to the displacement (in other words, in the same direction as the displacement. So, you could find the vector projection of the force on the displacement vector, and take the dot product of that projection and the displacement vector. This method here is actually what I was trying to point at in my previous post.



#7
Feb2313, 03:03 PM

P: 51

That method requires the process of figuring out the angle theta correct?



#8
Feb2313, 03:03 PM

Thanks
P: 5,687

Projections and angles only complicate things. The force and displacement vectors are given in terms of unit coordinate vectors. The dot product is very easy to do once these are known.



#9
Feb2313, 03:15 PM

P: 51

Ok so the ihat component of force is 12 and the ihat component of displacement is 13. The product of the two is work in the ihat direction. I do the same for the jhat components and then add them together, 156J + (110J) = 46J Is this the best way to do this?



#10
Feb2313, 03:38 PM

Thanks
P: 5,687

Let's say you have two vectors: aI + bJ and cI + dJ, where I and J as the unit coordinate vectors. Their dot product is (aI + bJ).(cI + dJ) = acI.I + adI.J + bcJ.I + bdJ.J = ac + bd, because I.I = J.J = 1 and I.J = J.I = 0.
So you just need to multiple the samedirection components and then sum the products. This might be less intuitive than the angle and projection stuff, but is usually far easier to deal with. 


#11
Feb2313, 03:46 PM

P: 51

I remember something about cross products being more difficult of matrices (which my math teachers always seemed to skip over, I've passed every math through Calc II with A's) I don't have the intuition for why I.I = 1 and I.J = 0 so I'm not sure I understand though I believe I need to. Thank you for your explanations.



#12
Feb2313, 03:55 PM

Thanks
P: 5,687

Using an XY plane, I is the unit vector in the X direction, and J in the Y direction. Now the dot product of any vector with itself is simply its length squared, which is 1 for unit vectors. That is why I.I = J.J = 1.
On the other hand, I and J are perpendicular, so a projection of one onto another is zero, thus I.J = J.I = 0. 


#13
Feb2313, 04:13 PM

P: 51

oh ok, that makes sense. I believe I'll be able to remember it that way, thank you. I'm pretty sure that is how it was taught in class but explained in a way I evidently didn't understand.



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