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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.

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Presentation on theme: "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."— Presentation transcript:

1 Examples in Chapter 1

2 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 a) The components of the vector sum A+B b) the magnitude and direction of A+B c) The components of the vector B-A d) 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 Find the magnitude of A+B (A+B) y (A+B) x (A+B)

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

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 C x =A y B z -A z B y C y =A z B x -A x B z C z =A x B y -A y B x Since A and B have only x- and y- components, we find C z by C z =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 10 3 kg/m 3, estimate the mass of a) the heart of an adult human b) a cell with a diameter of 0.5  m. c) 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 = cm or m 3” =3 * 2.54 cm = 7.62 cm Volume of cylinder=  *(d/2) 2 *L  * (7.62/2) 2 *10.16=463 cm 3 or cc Note: kg =1 pound

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

17 Honey bee V=  *(d/2) 2 *L=  *(.25/2) 2 *1=0.049 in 3 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 (-45 0 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 Step 2: Sketch in the details 5.80 km 2.0 km 3.5 km AyAy AyAy AyAy ?

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

22 Step 4: Solve 5.80 km-2.0 km=3.8km 2.0 km 3.5 km AyAy AyAy AyAy ?= =1.33 km

23 Step 5: 5.80 km-2.0 km=3.8km 2.0 km 3.5 km AyAy AyAy A y =2.47 km ?= =1.33 km=A x

24 Step 6: 1.33 km 2.47 km 


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