{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw14 - strong(ams4684 HW14 ditmire(58216 This print-out...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
strong (ams4684) – HW14 – ditmire – (58216) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 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 di ffi culty.” 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 what is the force F required to pull the hemispheres apart? 1. F = 4 π R 2 ( P a - P ) 2. F = π R 2 ( P a - P ) correct 3. F = 4 π R 3 3 ( P a - P ) 4. F = π R 2 2 ( P a - P ) 5. F = 2 π R 2 ( P a - P ) 6. F = 2 π R 3 3 ( P a - P ) Explanation: Force = (Di ff erence in pressure on 2 sides) × (Area). We must choose the area care- fully. 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 ff erence (see figure). We must therefore pick the e ff ective 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 π R 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 di ff erence acting on a small area of the hemisphere dA (see figure). 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. 002 (part 2 of 2) 10.0 points Determine the force if P = 0 . 0977 atm and R = 0 . 23 m. Atmosphesic pressure is 1 . 013 × 10 5 Pa. Correct answer: 15 . 1903 kN. Explanation: Let : P = 0 . 0977 atm = 9897 . 01 Pa , R = 0 . 23 m , and P a = 1 . 013 × 10 5 Pa . F = π R 2 ( P a - P ) = π (0 . 23 m) 2 × (1 . 013 × 10 5 Pa - 9897 . 01 Pa) × 1 kN 1000 N = 15 . 1903 kN . 003 (part 1 of 2) 10.0 points
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
strong (ams4684) – HW14 – ditmire – (58216) 2 In a car lift, compressed air exerts a force on a piston with a radius of 1 . 88 cm. This pressure is transmitted to a second piston with a radius of 14 . 8 cm. How large a force must the compressed air exert to lift 10700 N car? Correct answer: 172 . 654 N. Explanation: Let : r 1 = 1 . 88 cm , r 2 = 14 . 8 cm , and F 2 = 10700 N . P 1 = P 2 F 1 = F 2 A 2 · A 1 = F 2 π r 2 2 · ( π r 2 1 ) = F 2 · r 2 1 r 2 2 = 10700 N · (1 . 88 cm) 2 (14 . 8 cm) 2 = 172 . 654 N . 004 (part 2 of 2) 10.0 points What pressure produces this force? Neglect the weight of the pistons.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}