What is the Dot Product of a Cutting Tool Under Microprocessor Control?

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burhan619
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Homework Statement



A cutting tool under microprocessor control has several forces acting on it. One force is [itex]\vec{F}[/itex]=-αxy2[itex]\hat{j}[/itex], a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is α=2.50 N/m3. Consider the displacement of the tool from the origin to the point x=3.00m, y=3.00m.

(a) Calculate the work done on the tool by [itex]\vec{F}[/itex] if this displacement is along the straight line y=x that connects these two points.


Homework Equations



W=∫[itex]\vec{F}[/itex][itex]\cdot[/itex]d[itex]\vec{l}[/itex]

The Attempt at a Solution



I'm trying to use the equation above, so here's what I know:

d[itex]\vec{l}[/itex]=dx[itex]\hat{i}[/itex]+dy[itex]\hat{j}[/itex]
[itex]\vec{F}[/itex]=-αxy2[itex]\hat{j}[/itex]

Since it's the dot product,
[itex]\vec{F}[/itex][itex]\cdot[/itex]d[itex]\vec{l}[/itex]=dx+-αxy2dy.

I'm confused as to why the right side of that equation is equal to -αy3dy, as the textbook solution suggests. Any help is appreciated.
 
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burhan619 said:
I'm confused as to why the right side of that equation is equal to -αy3dy, as the textbook solution suggests. Any help is appreciated.
If x = y then [itex]-axy^2 = -ay^3[/itex]

The dot product of the two vectors is:

[itex]\vec{a} \cdot \vec{b} = ab\cos\theta[/itex]

where [itex]\theta[/itex] is the angle between the two vectors.

AM