A heavy rope, 80 ft long, weighs 0.3 lb/ft and hangs over the edge of a building 180 ft high.

How much work *W* is done in pulling 30 ft of the rope to the top of the building?

To solve for the work \( W \) done in pulling 30 feet of the rope to the top of the building, we need to consider the weight of each small portion of the rope and how far it is pulled. The work done in pulling a portion of the rope is calculated by multiplying the force (weight) by the distance over which that portion is pulled.

### Step-by-Step Solution:

1. **Weight per foot of the rope**: The rope weighs \( 0.3 \, \text{lb/ft} \). So, for any small portion of the rope of length \( dx \), the weight is \( 0.3 \, dx \) lb.

2. **Distance each portion is pulled**: We are pulling up 30 feet of rope. Let \( x \) represent the length of the rope hanging down after pulling up part of it. When \( x = 80 \), the rope is fully hanging down, and when \( x = 50 \), we have pulled up 30 feet of the rope, leaving 50 feet hanging.

   If we pull a small portion of the rope at a distance \( x \), this portion has to be pulled a distance of \( 180 - x \) feet to the top.

3. **Work for a small portion**: The work \( dW \) to pull up a small portion of the rope of length \( dx \) hanging at \( x \) is:
   \[
   dW = \text{force} \times \text{distance} = 0.3 \, dx \times (180 - x)
   \]

4. **Total Work**: To find the total work, we integrate from \( x = 50 \) (after pulling up 30 feet) to \( x = 80 \) (when the rope is fully hanging):
   \[
   W = \int_{50}^{80} 0.3 (180 - x) \, dx
   \]

5. **Solving the integral**:
   \[
   W = 0.3 \int_{50}^{80} (180 - x) \, dx
   \]
   Now, integrate \( (180 - x) \):
   \[
   W = 0.3 \left[ 180x - \frac{x^2}{2} \right]_{50}^{80}
   \]

   Evaluating at the bounds:
   \[
   W = 0.3 \left[ \left( 180(80) - \frac{80^2}{2} \right) - \left( 180(50) - \frac{50^2}{2} \right) \right]
   \]
   \[
   W = 0.3 \left[ (14400 - 3200) - (9000 - 1250) \right]
   \]
   \[
   W = 0.3 \left[ 11200 - 7750 \right]
   \]
   \[
   W = 0.3 \times 3450 = 1035 \, \text{foot-pounds}
   \]

### Final Answer:
The total work done in pulling 30 feet of the rope to the top of the building is \( 1035 \, \text{foot-pounds} \).

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Do you need further details or have any questions?

Here are 5 related questions to expand on this topic:
1. How would the work change if the rope were heavier (e.g., 0.5 lb/ft)?
2. What is the work done to pull the entire 80-foot rope to the top?
3. How would you calculate the work done if the building were only 100 feet tall?
4. If the rope were not uniform (i.e., varying weight per foot), how would the problem change?
5. What if instead of pulling vertically, the rope were being dragged horizontally?

**Tip:** When solving work problems involving ropes, always consider how far each small portion of the rope is moved and its weight to set up your integral correctly.

A heavy rope, 80 ft long, weighs 0.3 lb/ft and hangs over the edge of a building 180 ft high. How much work W is done in pulling 30 ft of the rope to the top of the building?

Calculate the work done to pull 30 feet of a heavy rope, weighing 0.3 lb/ft, over the edge of a 180 ft building using definite integrals. This problem covers concepts from calculus, including work-energy principles and integral setup.

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