Physics 8 Wednesday, November 20, 2013

Transcription

1 Physics 8 Wednesday, November 20, 2013 I plan next time to use Statics & Strength of Materials for Architecture & Building Construction by Onouye & Kane for these few weeks supplemental topics. Used copies < $10. I bought a few copies from Amazon. me if you d like to evaluate one or more chapters for extra credit. Quiz today: one problem from HW10. No quiz next week. Quiz #11 will be on Weds. 12/4. Richard Farley (ARCH 435 prof Structures) will probably visit us the week after Thanksgiving. We covered trusses & beams very quickly, to give you a hint of where physics fits in with structures. You will see these topics in much more depth if you take ARCH 435 as a senior. Last two topics for the term: gravity & oscillations. Physics 9 will start by reviewing oscillations.

2 Remember: the deflection of a beam is inversely proportional to I x = y 2 da For instance, for a cantilever with concentrated load at the end, max = PL3 3EI x For rectangular cross-section, I x = bh 3 /12. For this ruler, h 3b. By what factor will the ruler s deflection increase if I put the ruler on the flat (like a plank) vs. upright (like a joist)? (A) stay same (B) 2 (C) 3 (D) 9 (E) 27 (F) 81

3 Remember: the deflection of a beam is inversely proportional to I x = y 2 da Where does the square come from in I x = y 2 da? Why does the beam s resistance to bending go like y 2 da? Why not y or y 3 or y 4 or y? If you imagine fibers running along the length of a wooden beam, Why does the contribution of each little fiber go like the square of that fiber s distance y from the neutral surface? (Discuss!)

4 Remember: the deflection of a beam is inversely proportional to I x = y 2 da Where does the square come from in I x = y 2 da? Why does the beam s resistance to bending go like y 2 da? Why not y or y 3 or y 4 or y? If you imagine fibers running along the length of a wooden beam, Why does the contribution of each little fiber go like the square of that fiber s distance y from the neutral surface? (Discuss!) One factor of y comes from the fact that the farther fibers are strained (stretched or compressed) more in proportion to y. By Hooke s Law, stress strain. So the farther fibers exert proportionally more force-per-unit-area (stress).

5 Remember: the deflection of a beam is inversely proportional to I x = y 2 da Where does the square come from in I x = y 2 da? Why does the beam s resistance to bending go like y 2 da? Why not y or y 3 or y 4 or y? If you imagine fibers running along the length of a wooden beam, Why does the contribution of each little fiber go like the square of that fiber s distance y from the neutral surface? (Discuss!) One factor of y comes from the fact that the farther fibers are strained (stretched or compressed) more in proportion to y. By Hooke s Law, stress strain. So the farther fibers exert proportionally more force-per-unit-area (stress). The second factor of y comes from the fact that torque is τ = r F. Each fiber s force is multiplied by that fiber s lever arm. Each fiber s lever arm is y, the distance above/below the neutral surface.

6 Remember: the deflection of a beam is inversely proportional to I x = y 2 da For a beam with rectangular cross-section I x = bh3 12 If I instead put the beam on the flat, I would get I y = hb3 12 The ratio of these two is the square of the aspect ratio: I x I y = bh3 hb 3 = ( ) h 2 b So the ruler s resistance to bending is 3 2 = 9 times as large when it is upright vs. on the flat, since aspect ratio 3

7 For a cantilever with concentrated load (force) P at the end, max = PL3 3EI x If I make two beams of the same dimensions, one from aluminum and one from steel, and I apply the same load to both, the aluminum beam will deflect about 3 as much. Which factor in the above equation is responsible for that change? (A) I x is 3 larger for steel (B) I x is 3 larger for aluminum (C) E is 3 larger for steel (D) E is 3 larger for aluminum (E) P is 3 larger for steel (F) P is 3 larger for aluminum

8 For a cantilever with concentrated load at the end, max = PL3 3EI x If I double the length of this cantilever, how will max change? (A) stay same (B) double (C) triple (D) quadruple (E) 8 (F) 16

9 For a cantilever with concentrated load at the end, max = PL3 3EI x If I double the length of this cantilever, how will max change? (A) stay same (B) double (C) triple (D) quadruple (E) 8 (F) 16 By the way, for a cantilever with uniform load w per unit length, max = wl4 8EI x

10 Which is more likely to fail by buckling: a tall, thin column or a short, thick column? (A) A tall, thin column is more susceptible to buckling (B) A short, thick column is more susceptible to buckling (C) It makes no difference what the ratio of height/thickness is

11 If you think of the individual members of a truss as columns, which ones can potentially fail by buckling: the tension members or the compression members? (A) Only the tension members can potentially fail by buckling (B) Only the compression members can potentially fail by buckling (C) Both kinds can potentially fail by buckling

12 Here (in the next few slides) is what I want you to know about gravity. The rest of Chapter 13 was for your broader education. F = Gm 1m 2 r 2 where F points along the axis connecting m 1 to m N m2 G = kg 2 is a universal constant the same on Earth, on Mars, in distant galaxies, etc. That is a remarkable fact.

13 F = Gm 1m 2 r 2 11 N m2 G = kg 2 where F points along the axis connecting m 1 to m 2. Question: What is the magnitude of the gravitational force exerted by Earth on an object of mass 1 kg which is located 6400 km from Earth s center? (A) 1.0 Newton (B) 2.5 Newtons (C) 9.8 Newtons (D) Newtons (E) Newtons

14 F = Gm 1m 2 r 2 where F points along the axis connecting m 1 to m N m2 G = kg 2 is a universal constant the same on Earth, on Mars, in distant galaxies, etc. That is a remarkable fact. g = 9.8 m/s 2 = GM e R 2 e shows that an apple falling onto Newton s head results from the same force that governs the motion of the Moon around Earth, Earth around the Sun, etc. That too is amazing.

15 F = Gm 1m 2 r 2 11 N m2 G = kg 2 Question: What is the acceleration due to gravity at a distance of one Earth radius above Earth s surface? (A) (B) (C) GM e R 2 e GM e 2R 2 e = 9.8 m/s 2 = 4.9 m/s 2 GM e = 2.45 m/s2 (2R e ) 2

16 F = Gm 1m 2 r 2 11 N m2 G = kg 2 R e = 6400 km = m M e = kg Question: What is the acceleration due to gravity at a distance of 600 km above Earth s surface, where the Hubble Space Telescope is located? (A) (B) (C) GM e R 2 e = 9.8 m/s 2 GM e = 9.8 m/s2 (R e m) 2 GM e (R e = 8.2 m/s2 m) 2

17 F = Gm 1m 2 r 2 11 N m2 G = kg 2 R e = 6400 km = m M e = kg Is the Hubble Space Telescope moving in a circle? If so, what force provides the necessary centripetal acceleration? (A) No, it is not moving in a circle. Its motion is in a straight line. (B) Yes, it is moving in a circle, but it is not accelerating, so no force is required to maintain its state of motion. (C) Yes, it is moving in a circle. The centripetal acceleration is provided by the gravitational force exerted by Earth on the HST. This force has magnitude g = 9.8 m/s 2. (D) Yes, it is moving in a circle. The centripetal acceleration is provided by the gravitational force exerted by Earth on the HST. This force has magnitude GM e /(R e km) 2 = 8.2 m/s 2.

18 F = Gm 1m 2 r 2 11 N m2 G = kg 2 R e = 6400 km = m M e = kg How can we determine the speed at which the Hubble Space Telescope circles the earth? (A) (B) mv 2 (R e km) = GM e m (R e km) 2 v 2 = 8.2 m/s2 (R e km) (C) The speed of the HST equals Earth s escape velocity. (D) A and B are both true. (E) A, B, and C are all true.

19 About how long did it take the Space Shuttle go circle around Earth once? Assume it was at the same altitude as the HST. (A) (B) (C) (D) (E) (2π)(R e km) 7600 m/s (2π)(R e km) 7600 m/s (2π)(R e km) 7600 m/s (2π)(R e km) 7600 m/s (2π)(R e km) 7600 m/s = 97 seconds = 97 minutes = 5800 minutes = 24 hours = 27.3 days

20 How long does it take the Moon to travel around Earth? Do we have enough information to figure out how far away the Moon is from Earth s center? (A) 24 hours (B) 24 hours (C) 7 days (D) 7 days (E) 27.3 days (F) 27.3 days (G) days (H) days no yes no yes no yes no yes

21 Important stuff about gravity, continued For an orbit, gravity provides the centripetal force, so mω 2 R = mv 2 R = GMm R 2 In general angular momentum (mvr ) is constant, but we ll mainly study circular orbits, where R = constant, thus v = constant. So how far away is the Moon?

22 Things to know about gravity, continued (This page is somewhat less important.) Gravitational potential energy for a system consisting of objects 1 and 2 is U = Gm 1m 2 (note the sign) r which 0 as r. The objects are bound if K + U < 0. If K + U 0, they escape each other. They just barely escape if K + U = mv escape 2 = GmM R in which case K 0 when R.

23 Important stuff about gravity, continued (This page is also somewhat less important.) For a central force that goes like F 1/R 2, the forces from a uniform spherical shell add (if you re outside the shell) up to one force directed from the center of the shell. So a rigid sphere attracts you as if it were a point mass. That s why you re able to treat Earth s gravity as if it were due to a point mass located at Earth s center. If you re inside the spherical shell, the sum of the forces adds up to zero. So if you dig a deep enough hole (e.g km), you ll see Earth s gravity decrease. If you re a distance R from Earth s center, only the part of Earth s mass located at r < R attracts you. That s it

24 HW help this week: Zoey=Weds(2n36), Bill=Thurs(3w2) me if you want to read 1 or more Onouye/Kane chapters Physics 8, Fall 2013, Homework Quiz #10 ( ). 10 minutes. Work alone. Closed book, one page of handwritten notes OK. A box of mass M b = 20 kg is suspended from the right end of a horizontal rod of mass m r = 10 kg. The left end of the rod is affixed to a wall by a pin. A wire connects the right end of the rod to the wall directly above the pin, making an angle of 37 with the rod, as shown in the figure. (Note triangle.) (a) Draw an extended free-body diagram for the horizontal rod, showing each force acting on the rod and its line of action. (b) Find the tension in the wire. (Use correct units for tension!) (c) Determine the horizontal and vertical components of the reaction force that the pivot exerts on the rod. (Again, be sure to give your answer with correct units.)