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

Click For Summary
SUMMARY

The discussion focuses on calculating the work done on a cutting tool under microprocessor control, influenced by the force \(\vec{F} = -\alpha xy^2 \hat{j}\) with \(\alpha = 2.50 \, \text{N/m}^3\). The displacement from the origin to the point (3.00 m, 3.00 m) occurs along the line \(y = x\). The work done is derived using the equation \(W = \int \vec{F} \cdot d\vec{l}\), leading to the conclusion that \(\vec{F} \cdot d\vec{l} = -\alpha y^3 dy\) when substituting \(x = y\).

PREREQUISITES
  • Understanding of vector calculus and dot products
  • Familiarity with work-energy principles in physics
  • Knowledge of microprocessor control systems
  • Basic integration techniques for calculating work
NEXT STEPS
  • Study vector calculus applications in physics
  • Learn about work-energy theorem in mechanical systems
  • Explore microprocessor control in automation and robotics
  • Practice integration techniques with varying limits and functions
USEFUL FOR

Students in physics or engineering, particularly those studying mechanics and control systems, as well as professionals working with automated cutting tools and microprocessor applications.

burhan619
Messages
7
Reaction score
0

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.
 
Physics news on Phys.org
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
 

Similar threads

Replies
1
Views
1K
Replies
12
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 39 ·
2
Replies
39
Views
4K
Replies
5
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
1K