Finding the work done on a cutting tool

  • #1

Homework Statement


A cutting tool under microprocessor control has several forces acting on it. One force is F⃗ =−αxy2j^, a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is α = 2.70N/m3 . Consider the displacement of the tool from the origin to the point x= 2.95m , y= 2.95m .

Part A
Calculate the work done on the tool by F⃗ if this displacement is along the straight line y=x that connects these two points.

Part B
Calculate the work done on the tool by F⃗ if the tool is first moved out along the x-axis to the point x= 2.95m , y=0 and then moved parallel to the y-axis to x= 2.95m , y= 2.95m .


The Attempt at a Solution


So I was simply trying to use the W=force*distance, and I tried to use the vector equation for force by plugging in the coordinate that it would end up at and then for distance I used the distance from the origin to the point (2.95, 2.95) for the first part, and then I broke Part B into two parts, calculating work from origin to (2.95, 0) and then work from there to (2.95, 2.95).

It didn't work, and it is due by 11:59 PM. I am really confused. (Please, if you use calculus in your replies, please explain in detail. My calculus is very weak right now...)
 

Answers and Replies

  • #2
haruspex
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Please post your working. It is hard to say where you went wrong from merely an outline of what you did.
Also, this is very much an international forum. I don't know what time zone you are in.
 
  • #3
Hello

I am in the Eastern Time Zone by the way, so about 51 minutes left (!)

For the first one, I took W=F*D and took F⃗ =−αxy^2 j-hat, plugged in α=2.70, x=2.95, y=2.95, and then multiplied that by the sqrt(2*2.95^2). I got -289.

For the second one, I did the same thing but in two parts. The first part I plugged in α=2.70, x=2.95, y=0, and then multiplied that by 2.95. Then I added the product of the next part, which I plugged in α=2.70, x=2.95, y=2.95, and multiplied that by 2.95.

The homework system tells me I am wrong. Please correct me.
 
  • #4
haruspex
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Hello

For the first one, I took W=F*D and took F⃗ =−αxy^2 j-hat, plugged in α=2.70, x=2.95, y=2.95, and then multiplied that by the sqrt(2*2.95^2). I got -289.
Substituting those coordinates in the expression for F gives you the force at those coordinates. You need to integrate the force along the path to those coordinates.
 
  • #5
I do not know how to integrate the force along the path...Could you explain in detail?
 
  • #6
haruspex
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I do not know how to integrate the force along the path...Could you explain in detail?
Suppose you move a distance dx in the x direction. To stay on that path you must also move dx in the y direction. Express that as a vector. The work done by the force is the dot product of F with that vector. Compute that dot product and integrate over the range of x.
 
  • #7
Please look at my math

Could I write it as the integral from 0 to 2.95 of (-2.7*x*y^2) dx?
 
  • #8
Or would it be the integral of -2.7*y^2*dx from 0 to 2.95?
 
  • #9
haruspex
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Please look at my math

Could I write it as the integral from 0 to 2.95 of (-2.7*x*y^2) dx?
In the OP you wrote ##\alpha xy2\hat j##. I gues that meant y2. If so, yes, but remember that y is a known function of x along this path.
 
  • #10
I do not know what you mean by y being a known function of x. Was my first integral correct, and then if I plug in the value of that integral into the W=F*D equation, then it works?
 
  • #11
haruspex
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I do not know what you mean by y being a known function of x.
What does it tell you about the path in part A?
 
  • #12
OH

so y=x. Then for my final integral, it is the integral of x^3 dx from 0 to 2.95? (I may have that completely wrong.)
 
  • #13
haruspex
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OH

so y=x. Then for my final integral, it is the integral of x^3 dx from 0 to 2.95? (I may have that completely wrong.)
Yes, not forgetting alpha.
 
  • #14
So it would be the integral of -2.7x^3 dx from 0 to 2.95

Then for part B,

is the first integral -2.7*0*dx from 0-2.95? and then the second one...since it is dy, then I could do -2.7*x*2y*dy from 0 to 2.95?
 
  • #15
haruspex
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Then for part B,

is the first integral -2.7*0*dx from 0-2.95? and then the second one...since it is dy, then I could do -2.7*x*2y*dy from 0 to 2.95?
I don't understand the 2y. What happened to y2?
 
  • #16
I thought since it was dy, I would not so anything to the x, but the change would happen to the y part, unless I was supposed to change it to (1/3) y^3? My calculus is really rusty
 
  • #17
I calculated the integral for the first part and got -51.12, and then multiplied that by the sqrt(2*(2.95^2)) and got -213.3. It is still wrong. I do not know what to do.
 
  • #18
haruspex
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I thought since it was dy, I would not so anything to the x, but the change would happen to the y part, unless I was supposed to change it to (1/3) y^3? My calculus is really rusty
For this part of the question, it's the same algebraic form of F, but as you move up from y=0 to the finishing point x is constant. So, yes, you should get y3/3 in the result.
 
  • #19
Okay then. But for part A, was the answer that I got in joules? Am I supposed to do another calculation to get joules?
 
  • #20
haruspex
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I calculated the integral for the first part and got -51.12, and then multiplied that by the sqrt(2*(2.95^2))
Why did you multiply it by anything? What quantity do you think the result of the integration is giving you?
 
  • #21
I thought that it was giving me F that I would plug into the W = F*D. I am supposed to give W in terms of joules
 
  • #22
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I thought that it was giving me F that I would plug into the W = F*D. I am supposed to give W in terms of joules
You need to put aside W=F*D. That is for constant force in the same direction as the displacement. With variable forces we have to use the integral version, ##W=\int F.dx##. And since they're not necessarily in the same direction, F and dx there are vectors, and the dot means dot product. That is the integral you performed. Job done. And yes, it's Joules.
 
  • #23
I got the first part! three minutes left

I took the integral of -2.7*2.95*(1/3)y^3 from 0 to 2.95 and I got that wrong. I figured the first part gives me 0, so I didn't do anything with it
 
  • #24
haruspex
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I got the first part! three minutes left

I took the integral of -2.7*2.95*(1/3)y^3 from 0 to 2.95 and I got that wrong. I figured the first part gives me 0, so I didn't do anything with it
You took the integral of that? Don't you mean that was the result of integrating?
 
  • #25
Time is up, and I have to do move on to other things. Thank you for helping me. I will go over it again soon!
 

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