Applications of integration: work problems

In summary, for problem 1, there are some incorrect assumptions made in the solution attempt. The equation of the circular cross-section of the tank needs to be used, and the height each layer of gas is lifted is not a constant value. Also, the notation used is not standard and may have led to confusion.
  • #1
clairez93
114
0

Homework Statement



1. A cylindrical gasoline tank 3 feet in diameter and 4 feet long is carried on hte back of a truck and is used to fuel tractors. The axis of the tank is horizontal. Find the work done in pumping the entire contents of the fuel tank into a tract if the opening on the tractor tank is 5 feet above the top of the tank in the truck. Gasoline weighs 42 pounds per cubic foot.

2. Consider a 15-foot chain hanging from a winch 15 feet above ground level. Find the work done by the winch in winding up the specified amount of chain, if the chain weighs 3 pounds per foot.


Homework Equations





The Attempt at a Solution



1. I visualized it as being chopped into little chunks of rectangles:
dW = dF * x
dF = (weight)
weight = 42*4*1.5*dh
dF = 42*4*1.5*dy * (5-y)
[tex]\int^{3}_{0}42*4*1.5*dy * (5-y)[/tex] = 2646
Book Answer: 2457pi

2. dW = dF * x
dF = 3 dy
d = 10-y
[tex]\int^{10}_{0}3*(10-y) dy[/tex] = 150
Book answer: 300


Can anyone help me see my mistakes?
 
Physics news on Phys.org
  • #2
For number 1), It doesn't look like you're correctly taking the geometry of the tank into account. The axis of the tank is horizontal, so the volume of a tiny horizontal slice of dy thickness would change depending on y. I'd expect the expression inside the integral to be a much more complicated function than just (3-y). Although, I really don't think calculus is the way to go for this problem. There's a MUCH simpler way, if your prof. will let you take the easy way out.

I'll look at 2 tomorrow.
 
  • #3
clairez93 said:

Homework Statement



1. A cylindrical gasoline tank 3 feet in diameter and 4 feet long is carried on hte back of a truck and is used to fuel tractors. The axis of the tank is horizontal. Find the work done in pumping the entire contents of the fuel tank into a tract if the opening on the tractor tank is 5 feet above the top of the tank in the truck. Gasoline weighs 42 pounds per cubic foot.

2. Consider a 15-foot chain hanging from a winch 15 feet above ground level. Find the work done by the winch in winding up the specified amount of chain, if the chain weighs 3 pounds per foot.


Homework Equations





The Attempt at a Solution



1. I visualized it as being chopped into little chunks of rectangles:
dW = dF * x
dF = (weight)
weight = 42*4*1.5*dh
dF = 42*4*1.5*dy * (5-y)
[tex]\int^{3}_{0}42*4*1.5*dy * (5-y)[/tex] = 2646
Book Answer: 2457pi

2. dW = dF * x
dF = 3 dy
d = 10-y
[tex]\int^{10}_{0}3*(10-y) dy[/tex] = 150
Book answer: 300


Can anyone help me see my mistakes?
For #1, you are assuming that all of the "layers" of gasoline are 1.5 ft wide, which isn't true. You need to use the equation of the circular cross-section of the tank. Since you are integrating from y = 0 to y = 3, you are assuming that the origin is at the bottom of the tank, which has an impact on the equation of the circular cross-section of the tank. This assumption also impacts the expression that represents the height each layer of gas is lifted. For example, the slab at a level y has to be lifted to a height of 8 ft (the top of the tank is at (0, 3) according to your coordinate system, and has to be lifted to (0, 8)). This means that this typical layer has to be lifted (8 - y) ft.
 
  • #4
I'd also like to add that your notation seems really odd to me. First off, usually work is expressed as F(x)*dx, not dF*x. It makes a difference--you're dividing the work up into increments of x (thin layers of gasoline, tiny links of chain), not increments of force. You also write this 'd', and I can't tell if you're thinking of it as distance, or misusing the d in dF or dx.

I'll check this forum tomorrow, so please don't hesitate to ask more questions if you're confused!
 

Related to Applications of integration: work problems

1. What are some real-world applications of integration in work problems?

Integration is commonly used to solve problems related to work, such as finding the amount of work done by a force or the displacement of an object. Some examples include calculating the work done by a moving object against friction, finding the total force exerted on an object, and determining the work done by a varying force over a distance.

2. How is integration used to determine work done by a force?

Integration is used to find the work done by a force by taking the integral of the force with respect to the displacement. This allows us to calculate the total amount of work done, taking into account the varying force over a specific distance.

3. Can integration be used to find the displacement of an object in a work problem?

Yes, integration can be used to find the displacement of an object in a work problem. By taking the integral of the velocity of the object with respect to time, we can find the distance traveled by the object.

4. How does integration relate to the concept of energy in work problems?

Integration is closely related to the concept of energy in work problems. The work done by a force can be thought of as the change in energy of an object. By using integration, we can find the total amount of work done and therefore the change in energy of the object.

5. Are there any limitations to using integration in work problems?

While integration is a powerful tool for solving work problems, it does have some limitations. It may not be applicable in situations where the force is constantly changing and cannot be expressed as a function. It also assumes that the work is being done in a straight line and that there are no external forces acting on the object.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
608
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
Back
Top