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 – fnd 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±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 ² 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: ²orce = (Di³erence 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 di³erence (see fgure). We must thereFore pick the e³ective 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 di³erence acting on a small area oF the hemisphere dA (see fgure). 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 DocumentRight 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 ² · A 1 = ± F 2 π r 2 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
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.

This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas at Austin.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online