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Examples in Chapter 1

Examples in Chapter 1. Problem 1.42. Vector A has components A x =1.30 cm and A y = 2.25 cm; vector B has components B x =4.10 cm, B y =-3.75 cm. Find The components of the vector sum A + B the magnitude and direction of A + B The components of the vector B - A

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Examples in Chapter 1

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  1. Examples in Chapter 1

  2. Problem 1.42 Vector A has components Ax=1.30 cm and Ay= 2.25 cm; vector B has components Bx=4.10 cm, By=-3.75 cm. Find • The components of the vector sum A+B • the magnitude and direction of A+B • The components of the vector B-A • The magnitude and direction of B-A

  3. The components of the vector sum A+B • The sum of the x-components of A and B are equal to the x component of A+B

  4. (A+B)x (A+B)y (A+B) Find the magnitude of A+B

  5. (A+B)x (A+B)y (A+B) Find the direction of A+B q

  6. Components, magnitude and direction of B-A

  7. Problem 1.55

  8. Your book’s way • The book gives 3 formulae for C=AxB • Cx=AyBz-AzBy • Cy=AzBx-AxBz • Cz=AxBy-AyBx • Since A and B have only x- and y- components, we find Cz by • Cz=4*(-2)-5*3=-8-15=-23 • ||AxB||=23

  9. My way: First, a review? • A determinant represents a single number and is used in linear algebra

  10. A 3 x 3 determinant

  11. For a cross-product

  12. For our problem

  13. Problem 1.61 Biological tissues are typically made up of 98% water. Given that the density of water is 1 x 103 kg/m3, estimate the mass of • the heart of an adult human • a cell with a diameter of 0.5 mm. • a honey bee.

  14. Best guesses • Human heart: size of fist (cylinder 4” long with diameter of 3”) • Honey bee: 1” or 2.5 cm long cylinder with 0.25” diameter

  15. Heart Problem • 4” = 4*2.54 cm = 10.16 cm or 0.1016 m • 3” =3 * 2.54 cm = 7.62 cm • Volume of cylinder= p*(d/2)2 *L • p * (7.62/2)2 *10.16=463 cm3 or cc • Note: 0.454 kg =1 pound

  16. A cell • Assume spherical! • Volume=4*p/3*(d/2)3 Book answer differs by order of magnitude

  17. Honey bee • V=p*(d/2)2 *L=p*(.25/2)2 *1=0.049 in3 Book assumes ½ in long

  18. Problem 1.70 A sailor in a small sailboat encounters shifting winds. She sails 2.0 km east then 3.5 km southeast (-450 w.r.t. east) and then an unknown distance. Her final position is 5.8 km directly east of starting point. Find magnitude and direction of the third leg of the journey.

  19. Step 1: Draw it! 5.80 km 2.0 km 3.5 km

  20. 5.80 km 2.0 km 3.5 km Ay Ay Ay Step 2: Sketch in the details ?

  21. Step 3: Simplify 5.80 km-2.0 km=3.8km 2.0 km 3.5 km Ay Ay Ay ?

  22. Step 4: Solve 5.80 km-2.0 km=3.8km 2.0 km Ay 3.5 km Ay Ay ?=3.8-2.47=1.33 km

  23. Step 5: 5.80 km-2.0 km=3.8km 2.0 km Ay 3.5 km Ay=2.47 km Ay ?=3.8-2.47=1.33 km=Ax

  24. 2.47 km 1.33 km Step 6: q

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