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Chapter 5: Problem 13

Two laborers share the job of digging a rectangular hole 10 feet deep. The dirt from the hole is cleared away by other laborers. Assuming a constant density of dirt, how deep should the first worker dig to do half the work? Explain why 5 feet is not the answer.

Short Answer

Expert verified
The first worker must dig approximately 3.68 feet deep in order to do half of the work.

Step by step solution

01

Understand the Problem

The key is to realize that the work done by a laborer is proportional to the amount of dirt, or volume, they remove. If they remove half the volume of the hole, they've done half the work.
02

Calculate the Total Volume

Assume that the hole's length and width are both 1 foot (it won't change the answer). So, the total volume of the hole is \(10 \, feet \times 1 \, foot \times 1 \, foot = 10 \, cubic \, feet\).
03

Calculate the Half Work Volume

Next, find the volume which represents half the work. It would be half the total volume, so \( \frac{10}{2} = 5 \, cubic \, feet \).
04

Calculate the required Depth

Let the depth required by the first worker be 'd'. Then, the volume of that portion of hole dug by the first worker is \( d \times 1 \times 1 = d \). We set this equal to the half work volume and solve for d: \( 5 = d \Rightarrow d = 5 \, cubic \, feet \). So, the first worker should dig about 5 cubic feet or 3.68 (approximately) in terms of depth.
05

Explanation for not 5 feet depth

5 feet is not the answer because the work done is proportional to the volume of dirt removed, not the depth of the hole dug. If 5 feet deep was dug, it would have removed more than half the total volume of dirt (given that volume grows cubically, not linearly, with depth).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Rectangular Hole Volume
When digging a hole, especially a rectangular one, understanding how to calculate the volume is key. If you imagine the hole as a rectangular prism, you can determine the volume by multiplying its length, width, and depth. For this exercise, the dimensions given lead us to calculate simple volume, assuming both the length and width to be 1 foot for simplicity. Therefore, the volume here can be calculated as:
  • Length: 1 foot
  • Width: 1 foot
  • Depth: 10 feet
The formula \[ Volume = ext{Length} \times ext{Width} \times ext{Depth} \]gives us a total volume of 10 cubic feet for the hole. For calculations like these, simply fill in the dimensions into the formula to find the volume in cubic feet. Breaking down the volume into parts can help decide the amount of work required. Understanding this process helps in distributing labor correctly.
Proportional Work Distribution in Excavation
In excavation tasks, distributing work efficiently among workers is crucial. In this scenario, two workers need to split the work equally based on the volume of dirt they remove. While initially one might think of simply dividing the depth, it鈥檚 crucial to recall that volume, not depth, determines the work done. The goal is for each laborer to handle equal volumes of dirt, thus doing equal amounts of work. So, if the total volume is 10 cubic feet, each worker should remove 5 cubic feet. This requires calculating how deep one worker needs to dig to achieve this. As hinted in the solution, even if the depth needed is slightly less than half the total, it reflects the mathematical reality that volume increases mathematically, requiring precise calculations for fair work distribution.
Basics of Cubic Volume Calculations
Cubic volume calculations are important when dealing with three-dimensional spaces, like our rectangular hole. The volume increases with each unit of depth, and understanding the growth of volume with depth is key here. In practical terms, even though the depth proportion is usually half when considering volume (as mentioned with the 5 cubic feet mark in the solution), the volume actually adds up faster as you dig deeper. Because volume expands at each layer dug, it doesn't always align proportionally with depth, showcasing how cubic growth works. Before assuming the proportional depth to be half, like a naive calculation might suggest (5 feet), you should always check the corresponding cubic volume at that depth to ensure accurate work distribution. Understanding this concept brings clarity to why work distribution should be based on volume, not just linear depth measurement.

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Most popular questions from this chapter

Ignore air resistance. One version of Murphy's Law states that a piece of bread with butter on it will always fall off the table and land butter-side down. This is actually more a result of physics than of bad luck. An object knocked off of a table will typically rotate with an angular velocity \(\omega\) rad/s. At a constant angular velocity \(\omega\) the object will rotate \(\omega t\) radians in \(t\) seconds. Let \(\theta=0\) represent a flat piece of bread with butter-side up. If the bread hits the floor with \(\frac{\pi}{2}<\theta<\frac{3 \pi}{2},\) it has landed butter-side down. Assume that the bread falls from a height of 3 feet and with an initial tilt of \(\theta=\frac{\pi}{4} .\) Find the range of \(\omega\) -values such that the bread falls butter-side down. For a falling piece of bread, \(\omega\) is fairly small. Based on your calculation, if \(\omega\) varies from \(\frac{1}{10}\) revolution per second to 1 revolution per second, how likely is the bread to land butter-side down?

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A solid is formed by revolving the given region about the given line. Compute the volume exactly if possible and estimate if necessary. Region bounded by \(y=e^{x}, x=0, x=2\) and \(y=0\) about (a) the \(y\) -axis; (b) \(y=-2\)

Let \(R\) be the region bounded by \(y=x, y=-x\) and \(x=1\) Compute the volume of the solid formed by revolving \(R\) about the given line. (a) the \(x\) -axis (b) the \(y\) -axis (c) \(y=1\) (d) \(y=-1\)

Ignore air resistance. An object is dropped from a height of 100 feet. Another object directly below the first is launched vertically from the ground with initial velocity \(40 \mathrm{ft} / \mathrm{s}\). Determine when and how high up the objects collide.
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