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Chapter 6: Problem 9

Lifting an elevator cable An electric elevator with a motor at the top has a multistrand cable weighing 4.5 \(\mathrm{lb} / \mathrm{ft}\) . When the car is at the first floor, 180 \(\mathrm{ft}\) of cable are paid out, and effectively 0 \(\mathrm{ft}\) are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top?

Short Answer

Expert verified
The motor does 72,900 ft-lb of work in lifting the elevator cable.

Step by step solution

01

Understand the Problem

You are asked to find the work done by the elevator motor to lift a cable from the first floor (where 180 ft of cable is out) to the top floor (where 0 ft of cable is out). The cable weighs 4.5 lb/ft.
02

Set Up the Variables

Define the length of the cable, which varies from 0 ft to 180 ft. The weight of the cable per foot is 4.5 lb/ft. Let \( x \) represent the length of the cable out at any point from the ground.
03

Write the Expression for Force

The force exerted by the cable at any point \( x \) feet is its weight in pounds, which is the weight per foot multiplied by the length of the cable remaining. The force is \( F(x) = 4.5 \times x \).
04

Set Up the Expression for Work

The work \( W \) done in lifting the cable a tiny distance \( dx \) is \( dW = F(x) \times dx \). Here, \( F(x) = 4.5x \). Integrate this expression from 0 to 180 to calculate the total work.
05

Integrate to Find Total Work

Integrate the force function from 0 to 180 feet: \[W = \int_{0}^{180} 4.5x \; dx.\]Calculating this integral gives:\[W = 4.5 \times \left[ \frac{x^2}{2} \right]_{0}^{180}.\]
06

Evaluate the Definite Integral

Calculate the definite integral:\[W = 4.5 \times \left[ \frac{180^2}{2} - \frac{0^2}{2} \right].\]This simplifies to:\[W = 4.5 \times \frac{32400}{2} = 4.5 \times 16200.\]
07

Compute the Result

Multiply out the expression to find the total work:\[W = 72900 \text{ ft-lb }.\] This is the work done by the motor in lifting the cable from the first floor to the top.

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Key Concepts

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

work done by a force
The concept of "work done by a force" in physics refers to the energy transferred when a force causes an object to move. It鈥檚 essential to note that work is only done when movement occurs in the direction of the force.
  • If you push an object and it moves, you鈥檝e done work on it.
  • The formula to calculate work is: \( W = F \times d \),
  • where \( W \) is the work done, \( F \) is the force applied, and \( d \) is the distance over which the force is applied.
In our elevator problem, the motor exerts force to lift the cable vertically. Since the force needed is upwards and opposite to gravity, work is being done to lift the cable.
integration
Integration is a fundamental concept in calculus used to find the whole from the parts. If you think of a graph, integration helps you find the area under the curve.
  • It's the reverse process of differentiation.
  • Integration can be thought of as "adding up" pieces or slices to find a total.
When finding work, integration is used to add up the infinitesimally small amounts of work done as each section of cable is lifted.
definite integral
A definite integral is used to compute the area under a curve over a specific interval. In other words, it helps find the accumulated quantity between two boundaries.
  • Expressed as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the bounds.
  • The process involves evaluating the integral at these bounds and finding the difference.
For the elevator cable problem, we use the definite integral to find the total work done lifting the cable from 0 to 180 feet.
physics applications in calculus
Calculus is not just about numbers; it is a powerful tool for solving real-life problems in physics and engineering. Let's see how calculus is applied in physics:
  • It helps in understanding motion, forces, and energy.
  • In our elevator problem, we model the continuous distribution of the cable using calculus.
By using integration, we calculate the total work effectively, which would be difficult using basic algebra. Understanding these applications makes calculus a valuable skill in various scientific and engineering fields.

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Most popular questions from this chapter

In Exercises \(31-36,\) use a CAS to perform the following steps for the given curve over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(6.24 . )\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your appoximations for \(n=2,4,8\) with the actual compare= given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$ x=\frac{1}{3} t^{3}, \quad y=\frac{1}{2} t^{2}, \quad 0 \leq t \leq 1 $$

Set up, but do not evaluate, an integral that represents the area of the surface obtained by rotating the curve \(x=a(t-\sin t)\) \(y=a(1-\cos t), 0 \leq t \leq 2 \pi,\) about the \(x\) -axis.

Use the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines in Exercises \(y=\sqrt{x}, \quad y=0, \quad y=2-x\)

Use a theorem of Pappus to find the volume generated by revolving about the line \(x=5\) the triangular region bounded by the coordinate axes and the line \(2 x+y=6 .\) As you saw in Exercise 29 of Section \(6.4,\) the centroid of a triangle lies at the intersection of the medians, one-third of the way from the midpoint of each side toward the opposite vertex.)

In Exercises \(1-8 :\) a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve to see what it looks like. If you can, graph the surface, too. c. Use your grapher's or computer's integral evaluator to find the surface's area numerically. \(y=\tan x, \quad 0 \leq x \leq \pi / 4 ; \quad x\) -axis
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