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Chapter 1: Problem 70

A standard \(1.000 \mathrm{kg}\) mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 2.50 in. The density of the steel is \(7.70 \mathrm{g} / \mathrm{cm}^{3} .\) How many inches long must the section of bar be?

Short Answer

Expert verified
The length of the section of the bar must be approximately 2.94 inches long.

Step by step solution

01

- Compute the volume of steel

Since mass, \(m\), is equal to density, \(p\), times volume, \(V\) (that is \(m = pV\)), we should first convert the given mass from kilograms to grams to match the unit of density. Thus the volume of steel needed is \(V =\frac{m}{p} = \frac{1000g}{7.7g/cm^3} \approx 129.87cm^3\).
02

- Formulating the equation for the volume of a triangular prism

The volume of a triangular prism can be calculated by multiplying the area, \(A\), of the base in \(cm^2\), by the height, \(h\), (which is the length of the bar we are looking for) in \(cm\). Since the base is an equilateral triangle, its area can be calculated using the formula \(A = \frac{\sqrt{3}}{4}a^2\), where \(a\) is the length of one side. First, convert \(a = 2.5 inches\) to \(cm\) (remember 1 inch is approximately 2.54 cm), then we get \(a \approx 6.35 cm\). So the area \(A \approx \frac{\sqrt{3}}{4}(6.35cm)^2 \approx 17.36 cm^2\), which is constant. Thus the equation for the volume of the bar is given as: \(V=A*h = \frac{\sqrt{3}}{4}a^2*h\).
03

- Solve for h

With the volume we got earlier, we now substitute values into the volume equation, solve for \(h (the length)\): \(129.87 cm^3 = 17.36cm^2 * h\). Solving this, gives \(h \approx 7.48cm\).
04

- Convert the length from cm to inches

Finally, convert \(h = 7.48 cm\) back to inches to solve the problem (with \(1 inch \approx 2.54 cm\)), which gives the length in inches, \(h \approx 7.48cm / 2.54 \approx 2.94 inches\).

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