Kaplunovsky - practice 15

Kaplunovsky - practice 15 - practice 15 Due: May 8 2007,...

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practice 15 – – Due: May 8 2007, noon 1 Question 1, chap 18, sect 3. part 1 of 1 5 points The pressure exerted on the ground by a man is greatest when: 1. He stands with both feet Fat on the ground. 2. He stands on the toes of one foot. cor- rect 3. The pressure on the ground is the same, however he stands, sits or lies. 4. He sits on his butt. 5. He stands Fat on one foot. 6. He lies down on his back. Explanation: P = F A The forces exerted on the earth in di±erent cases are the same while when the man stands on the toes of one foot, the area is the smallest and the pressure is the greatest. Question 2, chap 18, sect 4. part 1 of 1 5 points The four tires of an automobile are inFated to a gauge pressure of 88100 Pa. Each tire has an area of 0 . 0212 m 2 in contact with the ground. Determine the weight of the automobile. Correct answer: 7470 . 88 N (tolerance ± 1 %). Explanation: Let W be its weight. Then each tire sup- ports W 4 , so that P = F A = W 4 A , yielding: W = 4 AP = 4 (0 . 0212 m 2 )(88100 Pa) = 7470 . 88 N . Question 3, chap 18, sect 3. part 1 of 2 5 points In 1657, Otto von Guericke, inventor of the air pump, evacuated a sphere made of two brass hemispheres. Two teams of eight horses each could pull the hemispheres apart only on some trials, and then only “with the greatest of di²culty.” (See ³gure for details.) F F P R P a If P a is the atmospheric pressure, P is the pressure inside the hemispheres and R is the radius of the hemispheres, then the force ´ required to pull the hemispheres apart is given by 1. F = π R 2 ( P a - P ) correct 2. F = 2 π R 3 3 ( P a - P ) 3. F = π R 2 2 ( P a - P ) 4. F = 4 π R 2 ( P a - P ) 5. F = 4 π R 3 3 ( P a - P ) 6. F = 2 π R 2 ( P a - P ) Explanation: Basic Concepts P F A Solution: ´orce = (Di±erence in pressure on 2 sides) × (Area). We must choose the area carefully. Each team of horses is pulling in the z direction with a force F . The hemispheres will come apart only when F is the z com- ponent of the net force on each hemisphere due to the pressure di±erence (see ³gure). We
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practice 15 – – Due: May 8 2007, noon 2 must therefore pick the eFective area which is perpendicular to the z direction. If you stand far away on the z axis and look at the hemi- sphere, you see a circle of area πR 2 . Hence, F = π R 2 ( P a - P ) . θ φ x y dA F z Hemisphere of radius R dF If we wanted to justify this more rigorously, we would have to examine the amount of force dF caused by the pressure diFerence acting on a small area of the hemisphere dA (see ±gure). We would then take the z-component of this and integrate this amount over the surface of the hemisphere (a double integral). The result is the same as our “intuitive” argument above.
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This note was uploaded on 01/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Kaplunovsky - practice 15 - practice 15 Due: May 8 2007,...

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