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Chapter 4: Problem 82

A nanowire is a (nearly) one-dimensional structure with a diameter on the order of a few nanometers. Suppose a \(100.0-\mathrm{nm}\) long nanowire made of pure silicon (density of \(\mathrm{Si}=2.33 \mathrm{~g} / \mathrm{cm}_{3}\) ) has a diameter of \(5.0 \mathrm{nm}\). This nanowire is attached at the top and hanging down vertically due to the force of gravity. a) What is the tension at the top? b) What is the tension in the middle? (Hint: Treat the nanowire as a cylinder of diameter \(5.0 \mathrm{nm}\) leneth 100 made of silican

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01

Find the volume of nanowire

We are given that the nanowire is cylindrical in shape, with a diameter of \(5.0 \,\text{nm}\) and a length of \(100.0 \,\text{nm}\). To find its volume \(V\), we can use the formula for the volume of a cylinder, \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. In this case, \(h = 100.0 \,\text{nm}\). The diameter is \(5.0 \,\text{nm}\), so the radius \(r\) is half of that, which is \(r = 2.5 \,\text{nm}\): $$V=\pi(2.5\,\text{nm})^2 (100.0\,\text{nm})$$ Now, calculate the volume.
02

Convert the volume to cubic centimeters

Before we can use the density of silicon to find the mass of the nanowire, we need to convert the volume from cubic nanometers to cubic centimeters so the units are compatible. There are \(1\,\text{cm}=10^7\,\text{nm}\), so: $$V=\pi(2.5\,\text{nm})^2 (100.0\,\text{nm})\cdot(\frac{1\,\text{cm}}{10^7\,\text{nm}})^3$$ Calculate the value of \(V\) in cubic centimeters.
03

Calculate the mass of the nanowire

Now that we have the volume of the nanowire in cubic centimeters, we can use the density (\(\rho=2.33\,\text{g/cm}^3\)) of silicon to find the mass \(m\) of the nanowire using the formula \(m = \rho V\): $$m=2.33\,\text{g/cm}^3\cdot V$$ Calculate the mass.
04

Calculate the weight of the nanowire

Now we need to find the weight of the nanowire, which is the force acting on the nanowire due to gravity. The weight \(W\) is given by the formula \(W = mg\), where \(m\) is the mass we calculated in step 3, and \(g\) is the acceleration due to gravity (approximately \(9.81\,\text{m/s}^2\) or \(9.81\times10^4\,\text{cm/s}^2\)): $$W = mg$$ Calculate the weight.
05

Find the tension at the top

The tension at the top of the nanowire is simply the weight of the entire nanowire, as it must support the weight of the whole nanowire. Since we have already calculated the weight \(W\), the tension at the top of the nanowire is: $$T_\text{top} = W$$
06

Find the tension in the middle

To find the tension in the middle of the nanowire, we need to consider the weight of half the nanowire because the tension at this point has to support only the bottom half's weight. The mass of the bottom half of the nanowire can be found by dividing the total mass by 2, and calculating the corresponding weight (using \(W_\text{half} = \frac{m}{2}g\)): $$T_\text{middle} = W_\text{half}$$ Calculate the tension in the middle. Now, we have the tension at both the top and the middle of the nanowire.

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