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Onc cnd of a nylon rope of length \(4.5 \mathrm{~m}\) and diamcter \(6 \mathrm{~mm}\) is fixed to a trec-limb. A monkey weighing \(100 \mathrm{~N}\) jumps to catch the free end and stays there. Find the elongation of the rope and the corresponding change in the diameter. Young's modulus of nylon \(=4.8 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\) and Poisson's ratio of nylon \(=0.2\).

Short Answer

Expert verified
The elongation of the rope is approximately 0.033 mm, and the diameter decreases by about 0.00887 mm.

Step by step solution

01

Calculate the Cross-sectional Area

Diameter: \(d = 6 \text{ mm} = 0.006 \text{ m}\). The radius \(r\) is half of the diameter, \(r = \frac{d}{2} = 0.003 \text{ m}\). Calculate the cross-sectional area \(A\) using the formula:\[A = \pi r^2 = \pi (0.003)^2 = 2.827 \times 10^{-5} \text{ m}^2\]
02

Use Hooke’s Law for Elongation

Young’s modulus \(E = 4.8 \times 10^{11} \text{ N/m}^2\). Force \(F = 100 \text{ N}\). Use Hooke's Law, \(F = EA\varepsilon\), where \(\varepsilon\) is the strain. Rearrange for \(\varepsilon\): \[ \varepsilon = \frac{F}{A \cdot E} = \frac{100}{2.827 \times 10^{-5} \times 4.8 \times 10^{11}} \approx 7.39 \times 10^{-6}\] The elongation \(\Delta L\) is given by \(\Delta L = L \times \varepsilon\), where \(L = 4.5 \text{ m}\). \[\Delta L = 4.5 \times 7.39 \times 10^{-6} = 3.33 \times 10^{-5} \text{ m}\]
03

Calculate Change in Diameter Using Poisson’s Ratio

Poisson’s ratio \(u = 0.2\). The change in diameter \(\Delta d\) is related to the lateral strain, \(\varepsilon_d = -u \times \varepsilon\).\[\varepsilon_d = -0.2 \times 7.39 \times 10^{-6} = -1.478 \times 10^{-6}\] The change in diameter \(\Delta d\) is given by \(\Delta d = \varepsilon_d \times d\), where \(d = 0.006 \text{ m}\). \[\Delta d = -1.478 \times 10^{-6} \times 0.006 = -8.87 \times 10^{-9} \text{ m}\]
04

Interpreting the Results

The rope's elongation is \(3.33 \times 10^{-5} \text{ m} = 0.033 \text{ mm}\). The diameter change is \(-8.87 \times 10^{-9} \text{ m} = -0.00887 \text{ mm}\), indicating a slight decrease in the diameter due to the tension.

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Key Concepts

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

Hooke's Law
Understanding Hooke's Law is essential to solving problems involving the deformation of materials under an applied force. It's a principle first proposed by 17th-century physicist Robert Hooke. Hooke's Law states that the force needed to extend or compress a spring by some distance is proportional to that distance. This relationship is commonly expressed in the formula:
  • \(F = kx\)
Here:
  • \(F\) is the force applied to the spring
  • \(k\) is the spring constant, which measures the stiffness of the spring
  • \(x\) is the distance the spring is stretched or compressed from its natural length
In the context of materials that aren't springs, such as ropes or wires, Young's Modulus comes into play. Young's Modulus \(E\) helps us understand a material's stiffness. When dealing with these materials, Hooke's Law is slightly adapted:
  • \(F = EA\varepsilon\)
Where \(\varepsilon\) is the strain (the relative change in length), \(E\) is the Young's Modulus, and \(A\) is the cross-sectional area. By rearranging this formula, you can find the strain:
  • \(\varepsilon = \frac{F}{EA}\)
This equation forms the foundation of how materials respond to mechanical stresses, crucial for engineering and mechanics.
Poisson's Ratio
Poisson's Ratio is a concept in material science that describes how a material reacts to stress in directions perpendicular to the applied force. When a material is stretched, not only does it elongate, but it also tends to get narrower in diameter, and when it is compressed, it becomes thicker. This behavior is quantified by Poisson's Ratio.
  • Poisson's Ratio \( u \) is defined as the negative ratio of transverse to axial strain.
The formula is:
  • \( u = - \frac{\varepsilon_{\text{transverse}}}{\varepsilon_{\text{axial}}} \)
Where:
  • \(\varepsilon_{\text{transverse}}\) is the strain in the direction perpendicular to the applied force
  • \(\varepsilon_{\text{axial}}\) is the strain in the direction of the applied force
Poisson's Ratio is a unitless measure, often denoted as a decimal. For example, a Poisson's Ratio of 0.2 indicates a modest reduction in diameter when a tensile stress is applied. This property helps engineers understand how materials will behave under different kinds of load.
  • In our nylon rope example, the change in the diameter can be calculated using the formula derived from Poisson's Ratio, \(\Delta d = \varepsilon_d \times d\), revealing how application of stress results in slight lateral contraction.
Strain and Elongation Calculations
To grasp the full effect of forces on materials, one must understand strain and elongation calculations. Strain is a measure of deformation representing the displacement between particles in the material body. The axial strain \(\varepsilon\) is calculated by the change in length \(\Delta L\) divided by the original length \(L\):
  • \(\varepsilon = \frac{\Delta L}{L}\)
In the exercise, the strain was calculated using the rearranged Hooke’s Law formula:
  • \(\varepsilon = \frac{F}{EA}\)
And, knowing the strain, we can compute the elongation \(\Delta L\) using:
  • \(\Delta L = \varepsilon \times L\)
For a nylon rope, as given in the exercise, the elongation is determined after calculating the strain with respect to the applied force and cross-sectional area.
  • Once you have the strain, multiplying it by the original length provides the total elongation, a crucial step in understanding how much a material stretches under stress.
In our example, a small force applied to the rope causes a measurable extension, which gives engineers insights into material performance and safety.

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Most popular questions from this chapter

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