/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 You cast some metal of density \... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 77

You cast some metal of density \(\rho_{\mathrm{m}}\) in a mold, but you are worried that there might be cavities within the casting. You measure the weight of the casting to be \(w\) , and the buoyant force when it is completely surrounded by water to be \(B\) . (a) Show that \(V_{0}=\) \(B /\left(\rho_{\text { water }} g\right)-w /\left(\rho_{\text { m }} g\right)\) is the total volume of any enclosed cavities. (b) If your metal is copper, the casting's weight is 156 \(\mathrm{N}\) , and the buoyant force is 20 \(\mathrm{N}\) , what is the total volume of any enclosed cavities in your casting? What fraction is this of the total volume of the casting?

Short Answer

Expert verified
The cavity volume is approximately 1.30 脳 10鈦宦� m鲁, about 63.7% of the total volume.

Step by step solution

01

Understand the problem

We need to determine the total volume of cavities in a cast metal object by using its weight and the buoyant force exerted on it when submerged in water.
02

Recall Archimedes' principle

According to Archimedes' principle, the buoyant force \(B\) on an object submerged in a fluid is equal to the weight of the fluid displaced by the object, which is calculated as \(B = V \cdot \rho_{\text{water}} \cdot g\), where \(V\) is the volume of the fluid displaced.
03

Determine the actual volume of the casting

The actual volume of the casting \(V_{m}\) can be calculated by determining the mass of the casting using its weight \(w\) and density \(\rho_{\text{m}}\): \(V_m = \frac{w}{\rho_{\text{m}}g}\).
04

Determine the volume of water displaced

The volume of water displaced by the object is \(V_{0} = \frac{B}{\rho_{\text{water}}g}\). This volume accounts for the total volume of cavities and the actual volume of the metal.
05

Express the volume of the cavities

The total volume of any enclosed cavities \(V_c\) is the difference between the volume of water displaced \(V_{0}\) and the actual volume of the casting \(V_m\): \[\text{cavity volume } V_c = V_{0} - V_m = \frac{B}{\rho_{\text{water}} g} - \frac{w}{\rho_{\text{m}} g}\]
06

Calculate the cavity volume for copper casting

Given \(w = 156 \, \text{N}\), \(B = 20 \, \text{N}\), \(\rho_{\text{copper}} = 8960 \, \text{kg/m}^3\), and \(\rho_{\text{water}} = 1000 \, \text{kg/m}^3\), use the formula: \[V_c = \left(\frac{20}{1000 \cdot 9.8}\right) - \left(\frac{156}{8960 \cdot 9.8}\right)\] Calculating gives \(V_c \approx 1.30 \times 10^{-3} \, \text{m}^3\).
07

Calculate fraction of cavity volume to total volume

The total volume of the casting \(V_0\) is \(\frac{B}{\rho_{\text{water}} g} = \frac{20}{1000 \cdot 9.8} \approx 2.04 \times 10^{-3} \, \text{m}^3\). The fraction is given by: \(\frac{V_c}{V_0} = \frac{1.30 \times 10^{-3}}{2.04 \times 10^{-3}} \approx 0.637\).

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Key Concepts

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

Buoyant Force
The concept of buoyant force is rooted in Archimedes' Principle, which states that any object submerged in a fluid is buoyed up by a force equal to the weight of the fluid that the object displaces. In our specific problem with the casting, this buoyant force, denoted by \(B\), becomes crucial in determining hidden cavities within the metal. Since the casting is entirely surrounded by water, the buoyant force can be calculated by evaluating the weight of the water displaced by the casting. This force acts upward, countering the weight of the object as reported when it is in the air. Understanding the buoyant force is key to calculating volumes and diagnosing potential imperfections in cast metal products.
Density of Metals
Density is a fundamental property of materials that describes mass per unit volume, commonly expressed in kilograms per cubic meter (kg/m鲁). In this problem, the density of the metal \(\rho_{\mathrm{m}}\) determines how much volume a certain weight of metal will occupy. For metals like copper, which is our subject here, the density is notably high at approximately 8960 kg/m鲁, indicating the metal is quite compact. As we calculate the actual volume of the metal casting, we use its weight \(w\) divided by \(\rho_{\mathrm{m}}\) and gravitational acceleration \(g\) to derive the metal's volume. This density-driven calculation allows us to distinguish the actual volume from any total measured volume that might suggest cavities.
Volume of Cavities
This refers to the volume within the casting that is not occupied by metal, potentially due to incomplete filling during the casting process or air pockets. The formula to find the volume of cavities \(V_c\) within the casting is derived from the difference between the total submerged volume \(V_0\) and the volume calculated from the metal's weight \(V_m\). By subtracting \(V_m\) from \(V_0\), we can determine how much of the casting's measured volume is actually empty space, or cavities: \[ V_c = \frac{B}{\rho_{\text{water}} g} - \frac{w}{\rho_{\text{m}} g} \] This method identifies flaws in the casting that could weaken its structural integrity. Understanding the volume of cavities helps manufacturers ensure the quality and durability of metal products.
Water Displacement
Water displacement is the process of measuring the volume of an object by observing the amount of water it displaces when submerged. In this exercise, it is a pivotal step because the displaced water volume directly relates to the buoyant force observed. Using the formula \( V_0 = \frac{B}{\rho_{\text{water}} g} \), where \(\rho_{\text{water}}\) is the density of water at 1000 kg/m鲁, we calculate \(V_0\), the total volume the casting displaces in terms of water. This calculated volume includes both the metal and any cavities within it. Accurate water displacement measurements ensure that manufacturers can properly assess the consistency and integrity of their castings.
Casting Weight
The casting weight, denoted as \(w\) in this problem, is simply the gravitational force exerted on the casting in air, expressed in Newtons. It plays a dual role by helping calculate both the actual volume of the metal and serve as a basis for identifying cavities. The formula \( V_m = \frac{w}{\rho_{\mathrm{m}} g} \) allows us to translate this weight into a perceivable volume, assuming the metal is solid with no defects. By comparing actual volume calculations with total displaced volume \(V_0\), any discrepancies suggest imperfections in the casting. This process ensures quality control, revealing crucial differences that might otherwise be missed during routine inspections. Understanding the interplay between weight, buoyancy, and volume ensures the integrity and reliability of cast metal products.

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Most popular questions from this chapter

The earth does not have a uniform density; it is most dense at its center and least dense at its surface. An approximation of its density is \(\rho(r)=A-B r,\) where \(A=12,700 \mathrm{kg} / \mathrm{m}^{3}\) and \(B=\) \(1.50 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{4}\) . Use \(R=6.37 \times 10^{6} \mathrm{m}\) for the radius of the earth approximated as a sphere. (a) Geological evidence indicates that the densities are \(13,100 \mathrm{kg} / \mathrm{m}^{3}\) and \(2,400 \mathrm{kg} / \mathrm{m}^{3}\) at the earth's center and surface, respectively. What values does the linear approximation model give for the densities at these two locations? (b) Imagine dividing the earth into concentric, spherical shells. Each shell has radius \(r\) , thickness \(d r\) , volume \(d V=4 \pi r^{2} d r,\) and mass \(d m=\rho(r) d V .\) By integrating from \(r=0\) to \(r=R,\) show that the mass of the earth in this model is \(M=\frac{4}{3} \pi R^{3}\left(A-\frac{3}{4} B R\right)\) (c) Show that the given values of \(A\) and \(B\) give the correct mass of the earth to within 0.4\(\%\) (d) We saw in Section 12.6 that a uniform spherical shell gives no contribution to \(g\) inside it. Show that \(g(r)=\frac{4}{3} \pi G r\left(A-\frac{3}{4} B r\right)\) inside the earth in this model. (e) Verify that the expression of part (d) gives \(g=0\) at the center of the earth and \(g=9.85 \mathrm{m} / \mathrm{s}^{2}\) at the surface. (f) Show that in this model \(g\) does not decrease uniformly with depth but rather has a maximum of \(4 \pi G A^{2} / 9 B=10.01 \mathrm{m} / \mathrm{s}^{2}\) at \(r=2 A / 3 B=5640 \mathrm{km} .\)

Submarines on Europa. Some scientists are eager to send a remote-controlled submarine to Jupiter's moon Europa to search for life in its oceans below an icy crust. Europa's mass has been measured to be \(4.78 \times 10^{22} \mathrm{kg}\) , its diamcter is 3130 \(\mathrm{km}\) , and it has no appreciable atmosphere. Assume that the layer of ice at the surface is not thick enough to exert substantial force on the water. If the windows of the submarine you are designing are 25.0 \(\mathrm{cm}\) square and can stand a maximum inward force of 9750 \(\mathrm{N}\) per window, what is the greatest depth to which this submarine can safely dive?

A cylindrical bucket, open at the top, is 25.0 \(\mathrm{cm}\) high and 10.0 \(\mathrm{cm}\) in diameter. A circular hole with a cross-sectional area 1.50 \(\mathrm{cm}^{2}\) is cutin the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of \(2.40 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s}\) . How high will the water in the bucket rise?

SHM of a Floating Object. An object with height \(h\) , mass \(M,\) and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\) . (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section \(13.7 ) .\)

A block of balsa wood placed in one scale pan of an equal-arm balance is exactly balanced by a \(0.0950-\mathrm{kg}\) brass mass in the other scale pan. Find the true mass of the balsa wood if its density is 150 \(\mathrm{kg} / \mathrm{m}^{3}\) . Explain why it is accurate to ignore the buoyancy in air of the brass but not the buoyancy in air of the balsa wood.
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