θ = 0.330rad θ = 2.10rad

Transcription

1 Chapter What are (a) the x component and (b) the y component of a vector a in the xy plane if its directions is 250 degrees counterclockwise from the positive directions of the x axis nd its magnitude is 7.3m. (a) The x component is (b) The y component is a x = 7.3 cos250 = a y = 7.3 sin 250 = Express the following angles in radians: (a) 20, (b) 50, (c) 100. Convert the following anlges to degrees: (d) rad, (e) 2.10 rad, (f) 7.70 rad. The conversion factor we need comes from the relationship 2π radians = 2π radians θ = 20 = 0.349rad 2π radians θ = 50 = 0.873rad 2π radians θ = 100 = 1.745rad θ = 0.330rad θ = 2.10rad 2π radians = π radians = θ = 7.70rad 2π radians = = = 81.2 Note that in the last relationship, the angle became larger than 360, so to get the angle in the traditional form, we subtracted A ship sets out to sail to a point 120 km due north. An unexpected storm blows the ship to a point 100km due east of its starting point. (a) How far and (b) in what direction must it now sail to reach its original destination.

2 Destination 120km d θ 100km Origination point. The distance and angle that we need to sail is d = (100km) 2 + (120km) 2 =156.2m tanθ = θ = tan 1 ( ) θ = N of W Note: The angle that I have calculated is the complement of the angle in answer section. 3.8 A person walks in the following pattern: 3.1 km north, then 2.4 km west, and finally 5.2 km south. (a) Sketch the vector diagram that represents theis motioin. (b) Howw far and (c) in what direction would a bird fly in a straight line from the same starting point to the same final point. We can see from the picture that we can find the bird s flight by adding the vectors together. This is most easily done using vector notation. d 1 = 0 î ĵ d 2 = 2.4 î + 0 ĵ d 3 = 0 î ĵ d = d 1 + d 2 + d 3 d = 2.4 î 2.1 ĵ d = = 3.19m θ = tan 1 ( ) = S of W

3 2.4km N 5.2 km 3.1km W θ E d bird flight S 3.10 A car is driven east for a distance of 50 km, then north for 30 km, and then in a directions 30 degrees east of north for 25 km. Sketch the vector diagram and determine (a) the magnitude and (b)the angle of the car s total displacement from its starting point. y R 25km 30km 50km x

4 (a) We begin by writing the three vectors in component form and proceed to add them and compute the sum in component, magnitude and direction form. a = 50kmi ˆ + 0 ˆ j b = 0i ˆ + 30km ˆ j c = 25sin30 km i ˆ + 25cos30 km ˆ j R = ( sin30 )km i ˆ + ( cos30 )km ˆ j R = 62.5km i ˆ km ˆ j R = = 81.08km tan θ = θ = tan 1 ( ) = (a) In unit-vector notation, what is the sum a + b if a = (4.0m) i ˆ + (3.0m) ˆ j and b = ( 13.0m) i ˆ + (7.0m) ˆ j? What are (b) magnitude and (c) direction of a + b? a + b = (4 13)ˆ i + (3 + 7) ˆ j = ( 9m) ˆ i + (10m) ˆ j b) Magnitude: a + b = ( 9m) 2 + (10m) 2 =13.45m c) When we compute the angle, we need to pay attention to which quadrant the vector lies in. Here, the x component is negative and the y component is positive. θ a +b = tan 1 ( 10 9 ) = Note: Your calculator may report the angle as degrees. You will need to add 180 to shift to the correct quadrant What is the sum of the follwoing four vecrtors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the angle in degrees, and (d) the angle in radians? E : 6.00 m at rad G : 4.00 m at rad We can write out each of these vectors in vector notation. F : 5.00 m at H : 6.00 m at 210.0

5 E : 6.00 m at rad E = 6cos(0.9rad) î + 6sin(0.9rad) ĵ = 3.73 î ĵ G : 4.00 m at rad G = 4cos(1.2rad) î + 4sin(1.2rad) ĵ = 1.45 î ĵ F : 5.00 m at F = 5 cos( 75.0 ) î + 5 sin( 75.0 ) ĵ =1.29 î ĵ H : 6.00 m at H = 6cos( ) î + 6sin( ) ĵ = 5.19 î ĵ R = ( )î + ( ) ĵ = 1.28î ĵ R = = 6.73m θ = tan 1 ( ) = = rad 3.29 Three vectors are given by a = 3.0î ĵ 2.0 ˆk, b = 1.0î 4.0 ĵ ˆk and c = 2.0î ĵ +1.0 ˆk. Find (a) a ( b c) (b) a ( b + c) and (c) a ( b + c). In each case, we need to compute the value inside the parentheses first. We begin with (a). We compute the cross product first. b c i ˆ ˆ j k ˆ = det = i ˆ ( ) ˆ j ( ) + k ˆ ( ( 4.0) 2.0) = 8.0ˆ i ˆ j + 6.0k ˆ Now that we have the cross product, we can compute the dot product. a ( b c) = (3.0 î ĵ 2.0 ˆk) ( 8.0î = ( ) + ( ) + ( ) = ĵ ˆk) Now we move to (b). Since it does not involve a cross product, its easier.

6 b = 1.0î 4.0 ĵ ˆk c = 2.0î ĵ +1.0 ˆk b + c = ( )î + ( ) ĵ + ( ) ˆk = 1.0î + ( 2.0) ĵ ˆk a ( b + c) = (3.0 î ĵ 2.0 ˆk) (1.0î + ( 2.0) ĵ ˆk) = ( ) + ( ) + ( ) = 9.0 We finally consider (c). We can use the result from (b) for the sum of the vectors. b + c = 1.0î a ( b + c) = det + ( 2.0) ĵ ˆk î ĵ ˆk = î ( ( 2.0) ( 2.0)) ĵ( ( 2.0) (1.0)) + ˆk(3.0 ( 2.0) ) = 5.0î 11.0 ĵ 9.0 ˆk 3.37 Rock faults are ruptures along which opposite faces of rock have slid past each other. In Fig. 3-28, point A and B coincide before the rock in the foreground slid down to the right. The net displacement AB is along the plane of the fault. The horizontal component of AB is the strike slip AC The component of AB that is directly down the plane of the fault is the dip-slip AD. (a) What is the magnitude of the net displacement AB if the strikeslip is 22m and the dip-slip is 17m. (if the plane of the fault is inclined 52 degrees to the horizontal, what is their vertical component of AB (a) AB = (AC) 2 + (AD) 2 = (22m) 2 + (17m) 2 = 27.8m (b) Vertical component AD sin52 =17msin52 =13.4m 3.43 For the vectors in Fig. 3-35, with a=4, b=3, and c=5, calculate (a) a b, (b) a c, and (c) b c One way of solving this problem is to write the three vectors in unit vector notation. a = 4.0î ĵ b = 0.0î ĵ c = 4.0 î ĵ

7 Notice that we can write the components of the c vector since we can see the lengths of the sides of the triangle that it is the hypotenuse of... but we need to remember that the directions of both the x and y components are negative. Now we can compute the three dot products a = 4.0î ĵ b = 0.0î ĵ c = 4.0î ĵ a b = = 0 a c = = 16 b c = = A fire ant, searching for hot sauce in a picnic area, goes through three displacements along level ground. d 1 fro 0.4m southwest (that is, at 45 degrees from directly south and from directly west), d 2 for 0.5m due east, d 3 for 0.6m at 60 degrees north of east. Let the positive x direction be east and the positive y direction by north. What are (a) the x component and (b) the y component d 1 (c) the x component and (d) the y component d 2 (e) the x component and (f) the y component d 3. What are (g) the x component, (h) the y component (i) the magnitude, and (j) direction of the ant s net displacement? If the ant is to return directly to the starting point (k) how far and (l) in what direction should it move? y d d1 d3 x d2 The components for each vector can be calculated using the angles and picture.

8 d 1 = 0.4cos45i ˆ 0.4sin 45 ˆ j = ˆ i ˆ j d 2 = 0.5ˆ i + 0.0ˆ j d 3 = 0.6cos60ˆ i + 0.6sin60 ˆ j = 0.3ˆ i ˆ j Now that we have computed the components of the three legs of the trip, we can find the net displacement. d 1 = ˆ i ˆ j d 2 = 0.5i ˆ + 0.0ˆ j d 3 = 0.3ˆ i ˆ j d = ( ) i ˆ + ( ) ˆ j = i ˆ ˆ j The magnitude of the displacement is d = = m θ = tan 1 ( ) = N of E 3.75 You need to draw this one on the picture graphically