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Chapter 11: Problem 44

A steel cable with cross-sectional area \(3.00 \mathrm{~cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8} \mathrm{~Pa}\). Find the maximum upward acceleration that can be given a \(1200 \mathrm{~kg}\) elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

Short Answer

Expert verified
The maximum upward acceleration that can be given to a 1200 kg elevator supported by the steel cable without exceeding one-third of the cable's elastic limit is 10.2 m/s虏.

Step by step solution

01

Calculate the Maximum Stress

First, calculate the maximum stress the steel cable can endure which is one-third of the elastic limit. Stress is defined as \(\text{stress} = \frac{F}{A}\) where F is the force applied and A is the area. So maximum stress can be computed as: \(\text{Max Stress} = \frac{1}{3} \times 2.40 \times 10^{8} \mathrm{Pa}\) = \(8.00 \times 10^{7} \mathrm{Pa}\).
02

Calculate the Tension Force

The tension force in the cable when the elevator is at rest is equal to the weight of the elevator. The force can be calculated by the equation \( F = m \times g \), where m is the mass of the elevator and g is the acceleration due to gravity. Substituting the given values, we find \( F = 1200 kg \times 9.8 m/s^{2} = 11760 N \).
03

Calculate the maximum additional force

The maximum additional force can be calculated once the maximum stress and tension force are known. Minor rearrangement of the stress formula gives us the maximum force that the cable can withstand: \( F = Stress \times Area \). Given that the cross-sectional area is 3.00 cm虏 and converting it to m虏 (since we are using SI units), we have \( A = 3.00 \times 10^{-4} m虏 \). The maximum tension force the cable can withstand is then \( F_{max} = Stress \times A = 8.00 \times 10^{7} \mathrm{Pa} \times 3.00 \times 10^{-4} \mathrm{m虏} = 24000 N \). Therefore, the maximum additional force that can be applied is \( F_{additional} = F_{max} - F = 24000 N - 11760 N = 12240 N \).
04

Derive the maximum acceleration

The last step is to calculate the maximum acceleration of the elevator based on the maximum additional force. The formula for acceleration is \( a = \frac{F}{m} \), where F is the force applied and m is the mass. So, the maximum acceleration is: \( a = \frac{F_{additional}}{m} = \frac{12240 N}{1200 kg} = 10.2 m/s^2 \).

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Key Concepts

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

Elastic Limit
The elastic limit of a material is the maximum stress that it can withstand while still being able to return to its original shape once the force is removed. It's a critical property for materials that are used in structures that undergo stress but are expected to maintain their integrity, such as bridges, buildings, and, in our exercise, elevator cables.

When a material exceeds its elastic limit, it enters the plastic deformation phase, where it will begin to permanently deform. By staying within one-third of the steel cable's elastic limit, as in the exercise, we are ensuring the safety and longevity of the cable.
Maximum Stress Calculation
Maximum stress calculation is vital for determining how much load a material can support before reaching its elastic limit. In engineering, calculations like these are essential for ensuring structures are safe and effective. The maximum stress the elevator cable can endure is one-third of the elastic limit, which prevents it from reaching a state of permanent deformation or failure. The formula given in the exercise,
\[\text{Max Stress} = \frac{1}{3} \times 2.40 \times 10^{8} \mathrm{Pa}\],
guides us to quantify the maximum stress that the cable can safely carry without risking damage.
Tension Force
Tension force refers to the force that is transmitted through a string, cable, or chain when it is pulled tight by forces acting from opposite ends. It's a force that tries to elongate the material. In the context of our exercise, the tension force in the elevator cable is equivalent to the weight of the elevator when it's at rest. This must be factored into calculations when determining how much additional force the cable can handle, as in the step:
\[ F = m \times g \ F = 1200 \, \text{kg} \times 9.8 \, \text{m/s}^{2} = 11760 \, \text{N} \].
The tension force is essentially the baseline, as any additional tension (from acceleration, for example) is added to this original force.
Maximum Acceleration
Maximum acceleration is the greatest acceleration that can be applied to an object without surpassing the maximum force it can withstand. In the elevator scenario, calculating the maximum acceleration that the cable can sustain ensures that it does not experience stress beyond its safe limit. The formula used to find maximum acceleration from the exercise,
\[ a = \frac{F_{additional}}{m} \],
clearly details how the mass of the elevator and the maximum additional force (which accounts for the stress limit of the cable) play into determining how quickly the elevator can safely accelerate. This ensures not just the safety of the cable, but also of the occupants, as the elevator operates within the material's stress boundaries.

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Most popular questions from this chapter

A uniform \(300 \mathrm{~N}\) trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges.

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

You are a construction engineer working on the interior design of a retail store in a mall. A 2.00 -m-long uniform bar of mass \(8.50 \mathrm{~kg}\) is to be attached at one end to a wall, by means of a hinge that allows the bar to rotate freely with very little friction. The bar will be held in a horizontal position by a light cable from a point on the bar (a distance \(x\) from the hinge) to a point on the wall above the hinge. The cable makes an angle \(\theta\) with the bar. The architect has proposed four possible ways to connect the cable and asked you to assess them: $$ \begin{array}{lllll} \text { Alternative } & \text { A } & \text { B } & \text { C } & \text { D } \\\ \hline x(\mathrm{~m}) & 2.00 & 1.50 & 0.75 & 0.50 \\ \theta(\text { degrees }) & 30 & 60 & 37 & 75 \end{array} $$ (a) There is concern about the strength of the cable that will be required. Which set of \(x\) and \(\theta\) values in the table produces the smallest tension in the cable? The greatest? (b) There is concern about the breaking strength of the sheetrock wall where the hinge will be attached. Which set of \(x\) and \(\theta\) values produces the smallest horizontal component of the force the bar exerts on the hinge? The largest? (c) There is also concern about the required strength of the hinge and the strength of its attachment to the wall. Which set of \(x\) and \(\theta\) values produces the smallest magnitude of the vertical component of the force the bar exerts on the hinge? The largest? (Hint: Does the direction of the vertical component of the force the hinge exerts on the bar depend on where along the bar the cable is attached?) (d) Is one of the alternatives given in the table preferable? Should any of the alternatives be avoided? Discuss.

A 3.00-m-long, \(190 \mathrm{~N}\), uniform rod at the \(\mathrm{zoo}\) is held in a horizontal position by two ropes at its ends (Fig. E11.21). The left rope makes an angle of \(150^{\circ}\) with the rod, and the right rope makes an angle \(\theta\) with the horizontal. A \(90 \mathrm{~N}\) howler monkey (Alouatta seniculus) hangs motionless \(0.50 \mathrm{~m}\) from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle \(\theta\). First make a free-body diagram of the rod.

Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff (Fig. \(\mathbf{P 1 1 . 4 9}\) ). Suppose that an \(82.0 \mathrm{~kg}\) climber, who is \(1.90 \mathrm{~m}\) tall and has a center of gravity \(1.1 \mathrm{~m}\) from his feet, rappels down a vertical cliff with his body raised \(35.0^{\circ}\) above the horizontal. He holds the rope \(1.40 \mathrm{~m}\) from his feet, and it makes a \(25.0^{\circ}\) angle with the cliff face. (a) What tension does his rope need to support? (b) Find the horizontal and vertical components of the force that the cliff face exerts on the climber's feet. (c) What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff?
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