Problem 16 - College Physics

Question

A high-speed lifting mechanism supports an \(800-\mathrm{kg}\) object with a steel cable that is \(25.0 \mathrm{~m}\) long and \(4.00 \mathrm{~cm}^{2}\) in cross-sectional area. (a) Determine the elongation of the cable. (b) By what additional amount does the cable increase in length if the object is accelerated upward at a rate of \(3.0 \mathrm{~m} / \mathrm{s}^{2} ?\) (c) What is the greatest mass that can be accelerated upward at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) if the stress in the cable is not to exceed the elastic limit of the cable, which is \(2.2 \times 10^{8} \mathrm{~Pa}\) ?

Step-by-Step Solution

Answer

The initial elongation of the cable is 0.098m (a). When the object is accelerated upwards at \(3.0 m/s^2\), the cable increases in length by another 0.012m (b). Finally, the greatest mass that can be accelerated upward at \(3.0 m/s^2\) without exceeding the elastic limit of the cable is about 8571kg (c).

Step 1: Determine the initial elongation of the cable

We know that stress \(= \frac{Force}{Area}\) and Hooke’s law states that \(Stress = E \cdot Strain\), where E is the elastic modulus of the material. Therefore, we can find the strain on the cable using \(Strain = \frac{Force}{Area \cdot E}\). In this case, the force is caused by the weight of the 800-kg object, so we have \(Force = mass \cdot g = 800kg \cdot 9.8m/s^2\). The area of the cable is given as \(4.00 cm^2 = 4.00 \times 10^{-4} m^2\). Elastic modulus (E) for steel is \(200 \times 10^9 N/m^2\). Substitute these values into the formula to calculate the strain, which represents the elongation per unit length.

Step 2: Calculate the total elongation of the cable

Once we have the strain, we can obtain the elongation of the cable by multiplying the strain by the initial length of the cable, \(Length_{increase} = strain \cdot initial\ length\), with the initial length being 25.0m.

Step 3: Find the additional elongation due to upward acceleration

When the object is accelerated upward, the effective force is increased, so we need to re-calculate the force including the acceleration (\(Force = mass \cdot (g + acceleration)\)). The strain can be re-calculated using the new force, the area and E. Then, the additional elongation \(Length_{increase} = strain \cdot initial\ length\) can be obtained using the new strain.

Step 4: Determine the maximum mass that can be lifted without exceeding the cable's elastic limit

If the stress is not to exceed the elastic limit of the cable, we can determine the maximum force the cable can withstand using the equation \(Max\ force = Elastic\ limit \times Area\). The maximum mass can then be found by dividing this force by the sum of \(g + acceleration\) to account for the upward acceleration, \(Max\ mass = \frac{Max\ force}{g + acceleration}\). The elastic limit is given as \(2.2 \times 10^8 Pa\), the area is \(4.00 \times 10^{-4} m^2\), \(g = 9.8m/s^2\), and the acceleration is \(3.0m/s^2\).

Key Concepts

Hooke's Law

Hooke's Law is fundamental in understanding how materials deform under force. It states that, within the elastic limit, the amount by which a material body is deformed (the strain) is directly proportional to the force causing the deformation (the stress). Mathematically, it can be expressed as \( Stress = k \times Strain \) where \( k \) is the spring constant that depends on the material's properties and dimensions.

For the steel cable lifting an object, as presented in the problem, Hooke's Law allows us to establish a relationship between the force applied (due to the object's weight and additional upward acceleration) and the consequent elongation of the cable. By knowing the spring constant (or modulus of elasticity) for steel, we can determine the exact amount of elongation for a given force.

Stress-strain relationship

The stress-strain relationship is key to understanding material behavior when forces are applied. Stress is defined as the force applied per unit area of a material, while strain is the measure of deformation representing the displacement between molecules relative to the original length of the material.

Stress (usually measured in Pascals, Pa) can be computed using the formula \( Stress = \frac{Force}{Area} \) and strain is a dimensionless quantity calculated as \( Strain = \frac{Change\underline{\phantom{xxx}}in\underline{\phantom{xxx}}length}{Original\underline{\phantom{xxx}}length} \). In the textbook problem, calculating the stress induced by the weight of the object and its acceleration, and relating it to strain through the modulus of elasticity, helps predict how the cable will elongate.

Modulus of Elasticity

The Modulus of Elasticity, often symbolized as \( E \), is intrinsic to a material and quantifies its stiffness. This value describes how much strain results from the applied stress. The higher the Modulus of Elasticity, the less a material will deform under a given stress.

In practical terms, for the given steel cable in our problem, its Modulus of Elasticity, \( E = 200 \times 10^9 N/m^2 \), tells us that steel is very stiff, meaning it resists deformation strongly. By using this constant, we can calculate the resultant strain from the known stress, thus predicting how much the cable will stretch when lifting the object.

Mechanical properties of materials

Materials have mechanical properties, such as ductility, toughness, strength, and hardness, defining their behavior under various conditions. These properties are derived from the stress-strain relationship and the material's response up to its elastic limit and beyond it into plastic deformation.

Understanding these properties is vital for engineers when designing objects and structures. In the context of the cable, knowledge of its mechanical properties, particularly strength and ductility, allows for calculating safety limits such as the maximum load before it starts to undergo permanent deformation—a crucial consideration for any lifting mechanism.

Elastic limit

The elastic limit of a material is the maximum stress that can be applied without causing permanent deformation. Once the stress on a material exceeds this limit, the material will not return to its original shape after the force is removed, thus deforming plastically.

For the 800 kg object being lifted by the steel cable, determining the elastic limit is essential for ensuring the safety and integrity of the lifting system. We calculate the maximum permissible force as \(Elastic\underline{\phantom{xxx}}limit \times Area\) to ensure that the stress in the cable never exceeds this threshold, hence preventing any permanent and unsafe changes to the cable's length and strength.