Find work done using force in two dimensions

In summary: The problem statement asks you to find the amount of work done along each of the three legs of the path, and then to add them up. The first leg of the path is particularly tricky because both x and y are changing simultaneously along this leg. The total length of this leg is 2 sqrt(5). What are x and y expressed as functions of s along this leg of the path?
  • #1
shanepitts
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Homework Statement


Screenshot_2015-09-24-10-55-40-1.png


Homework Equations


F⋅dr=W

The Attempt at a Solution


Screenshot_2015-09-24-10-55-15-1.png
 
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  • #2
The problem statement asks you to find the amount of work done along each of the three legs of the path, and then to add them up. The first leg of the path is particularly tricky because both x and y are changing simultaneously along this leg. Let s be the distance along this leg of the path measured from the initial point at the origin. The total length of this leg is ##\sqrt{4^2+2^2}=2\sqrt{5}##. What are x and y expressed as functions of s along this leg of the path?

Chet
 
  • #3
shanepitts said:

Homework Statement


View attachment 89256

Homework Equations


F⋅dr=W

The Attempt at a Solution


View attachment 89257
If r2 = x2 + y2 + z2, then what is dr? In general, dr ≠ dx + dy + dz

I believe a better definition for the work performed is W = ∫C F ⋅ ds, where the path C is the triangle specified in the OP.
 
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  • #4
Thank you,

But how can I calculate the work down on each leg of the triangle?

Shall I integrate ∫C F⋅dr along each line, using the limits as the length of each leg, and then sum them up?
 
  • #5
shanepitts said:
Thank you,

But how can I calculate the work down on each leg of the triangle?

Shall I integrate ∫C F⋅dr along each line, using the limits as the length of each leg, and then sum them up?

Well, generally finding the work performed involves evaluating a path integral, which is a little different from evaluating a "regular" definite integral.

For example, two of the legs of the triangle C are aligned with the x and y coordinate axes, making for some simplifications in evaluating ∫ F ⋅ ds on these paths.

The link below discusses and illustrates methods for evaluating path integrals:

http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx
 
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  • #6
If f is the fraction of the distance between (0,0,0) and (4,-2,0) along the path, then x = 4f, y = -2 f, and z = 0f. In terms of the fractional distance f, what is the force vector F? In terms of the fractional distance f, what is the differential vector dr along the path? What is F dotted with dr?

Chet
 
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Related to Find work done using force in two dimensions

1. How is work defined in two dimensions?

Work in two dimensions is defined as the product of the magnitude of the force and the displacement in the direction of the force. It is a scalar quantity measured in joules (J).

2. What is the formula for calculating work in two dimensions?

The formula for calculating work in two dimensions is W = F * d * cosθ, where W is work, F is the force applied, d is the displacement, and θ is the angle between the force and displacement vectors.

3. What does the angle between the force and displacement vectors represent in work calculations?

The angle between the force and displacement vectors represents the amount of work that is being done in the direction of the force. A greater angle means that less work is being done, while a smaller angle means that more work is being done.

4. Can you have negative work in two dimensions?

Yes, negative work is possible in two dimensions. Negative work occurs when the force and displacement vectors are in opposite directions, meaning that the force is working against the displacement. This results in a negative value for work.

5. How can you determine the direction of work in two dimensions?

The direction of work in two dimensions can be determined by looking at the angle between the force and displacement vectors. If the angle is less than 90 degrees, the work is positive and in the direction of the force. If the angle is greater than 90 degrees, the work is negative and in the opposite direction of the force.

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