Physics 40 HW#3 Chapter 3 Problems from Knight you should do (but don t turn in): Ch 3 Problems: 7, 10, 15, 23, 33, 37, 40, 42, 43

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1 Physics 40 HW#3 Chapter 3 Problems from Knight you should do (but don t turn in): Ch 3 Problems: 7, 10, 15, 23, 33, 37, 40, 42, 43 7, 10, 15, 23, 33, 37, 40, 42, Visualiz e: Solve: (a) v = (10 m/s) cos(90.0 ) = 0 m/s v = (10 cm/s)sin(90.0 ) = 10 m/s (b) a y ay = (20 m/s ) cos30 = 17 m/s = (20 m/s )sin 30 = 10 m/s (c) F = (100 N)sin(36.9 ) = 60 N F = (100 N) cos(36.9 ) = 80 N y Visualize:

2 Solve: (a) Using the formulas for the magnitude and direction of a vector, we have: B = 1 4 ( 4) + (4) = 5.7, θ = tan = 45 4 (b) r = ( 2.0 cm) + ( 1.0 cm) = 2.2 cm, θ = tan = (c) v = ( 10 m/s) + ( 100 m/s) = 100 m/s, θ = tan = (d) a = (10 m/s ) + (20 m/s ) = 22 m/s, θ = tan = Visualize: Solve: (a) We have A= 4iˆ 2ˆj and B= 3iˆ+ 5. ˆj This means 4A= 16iˆ 8 ˆj and 2B= 2 6iˆ+ 10 ˆj. Thus, E = 4A+ 2 B= [16 + ( 6)] iˆ+ [( 8) + 10] ˆj = 10iˆ+ 2 ˆj. (b) Vectors A, B, and E are shown in the figure above. (c) From the E vector, E = 10 and E y = 2. Therefore, the magnitude and direction of E are So E is 10, 11 above the +-ais. E = (10) + (2) = 104 = 10, 1 1 Ey E θ = tan ( / ) = tan (2/10) = 11

3 3.23. Visualize: Refer to Figure P3.23 in your tetbook. Solve: (a) We are given that A+ B+ C = 1ˆj with A= 4, iˆ and C = 2. ˆj This means A+ C = 4iˆ 2ˆj. Thus, B= ( A+ B+ C) ( A+ C) = (1) ˆj (4iˆ 2) ˆj = 4iˆ+ 3. ˆj (b) We have ˆ ˆ B= Bi + Byj with B = 4 and B y = 3. Hence, B = ( 4) + (3) = 5.0 B 1 y 1 3 θ = tan = tan = 37 B 4 Since B has a negative -component and a positive y-component, the vector B is in the second quadrant and the angle θ made by B is measured above the -ais. Assess: Since B < B, θ < 45 as obtained above. y Visualize: Solve: We are given A= (5.0 m) iˆ and C = ( 1.0 m) kˆ Using trigonometry, B= (3.0 m) cos(45 ) iˆ (3.0 m)sin(45 ) ˆj displacement is r = A+ B+ C = (7.12 m) iˆ (2.12 m) ˆj (1.0 m) kˆ. The magnitude of r is 2 r = (7.12 m) + (2.12 m) + (1.0 m) = 7.5 m. Assess: A displacement of 7.5 m is a reasonable displacement. The total

4 3.37. Visualize: (a ) (b ) Solve: (a) The river is 100 m wide. If Mary rows due north at a constant speed of v row = 2.0 m/s, it will take her (100 m)/(2.0 m/s) = 50 s to row across. But while she s doing so, the current sweeps her boat sideways at a speed v current = 1.0 m/s. In the 50 s it takes her to cross the river, the current sweeps here a distance d =( vcurrent 50 s)= 1.0 m/s 50 s = 50 m, so she lands 50 m east of the point that was directly across the river from her when she started. (b) Mary s net displacement D net, her displacement D current due to the river s current, and her displacemnt D row due to her rowing are shown in the figure Model: We will treat the knot in the rope as a particle in static equilibrium. Visualize: Solve: Epressing the vectors in component form, we have F 1 = 30. iˆ and F ˆ ˆ 2 = 5.0sin (30 ) i + 5.0cos(30 ) j. Since we must have F 1+ F 2 + F 3 = 0 for the know to remain stationary, we can write F ˆ ˆ 3 = 2 F1 F2 = 0.50 i 4.33 j. The magnitude of F 3 is given by ( ) F 3 = ( 4.33) = 4.4 units The angle between F 3 and the negative -ais is θ = tan (4.33/0.50) = 83 below the negative -ais. Assess: The resultant vector has both components negative, and is therefore in quadrant III. Its magnitude and direction are reasonable. Note the minus sign that we have manually inserted with the force 2. 1

5 3.40. Visualize: The average velocity is the net displacement D net divided by the total time, which are marked on the graph. We also mark on the graph of the bacterium s individual displacements and the time for each. Solve: The magnitude of the net displacement is found with Pythagoreum s rule, taking the values from the graph. We have y μ μ μ The direction of this displacement is Dnet = Dnet, + Dnet, = (40 m) + ( 20 m) = 45 m. D 1 net, y 1 20 µ m θ = tan = tan = 27 Dnet, 40 µ m The total time for the displacement is the sum of the individual times, which may be found by dividing each individual distance by the bacterium s constant speed of 20 µm/s. This gives t = D /(20 μm/ s) = ( 50 μm) +( 10 μm ) /(20 μm/ s) =( 51.0 μm )/(20 μm/ s) = 2.55 s AB AB t = D /(20 μm/ s) = (10 μm)/(20 μm/ s) = 0.50 s BC BC t = D /(20 μm/ s) = ( 40 μm) +( 10 μm ) /(20 μm/ s) =( 41.0 μm )/(20 μm/ s) = 2.06 s CD t = D DE CD DE /(20 μm/ s) = ( 50 μm) + ( 50 μm ) /(20 μm/ s) = ( 70.7 μm )/(20 μm/ s) = 3.54 s The total time is therefore ttot = 2.55 s s s s = 8.65 s and the magnitude of the bacterium s net velocity is v net Dnet 45 μm = = = 5.2 μ m/ s t 8.65 s Tot

6 3.43. Visualize: Use a tilted coordinate system such that -ais is down the slope. Solve: Epressing all three forces in terms of unit vectors, we have F 1 = (3.0 N) iˆ, F ˆ 2 = + (6.0 N) j, and F ˆ ˆ 3 = (5.0 N)sin θi (5.0 N) cos θ j. (a) The component of F net parallel to the floor is ( F net ) = (3.0 N) + 0 N + (5.0 N)sin (30 ) = 0.50 N, or 0.50 N up the slope. (b) The component of F net perpendicular to the floor is ( F net ) y = 0 N + (6.0 N) (5.0 N)cos(30 ) = 1.67 N, or 1.7 N to two significant figures. (c) The magnitude of F net is Fnet = ( Fnet ) + ( Fnet ) y = ( 0.50 N) + (1.67 N) = 1.74 N, or 1.7 N to two significant figures. The angle between F net and the negative -ais is 1 ( Fnet ) y N φ = tan = tan = 73 ( Fnet ) 0.50 N F net is 73 clockwise from the -ais.

7 Use graph paper for drawing vectors by the graphical method. Make it very neat. Neatness counts. LABEL EVERYTHING. Bo final numerical answers. 1. The vectors shown have magnitudes A = 6.00m, B = 5.00m, C = 4.00m. a) Determine the magnitude and direction of 2A +B C by graphical methods. State your scale, label everything, make it neat. b) Epress A, B & C in terms of ijk unit vectors. Determine the magnitude and direction of R = 2A +B C and epress it in terms of ijk unit vectors. c) Find the magnitude and direction of a fourth vector, D, that would balance the three vectors such that 2A +B C + D = 0 d) Compare the results of the two methods. You graphical method must agree with the component method within 1 %. If it does not, redo it. 2. A radar station locates a sinking ship at range 17.3 km and bearing 136 clockwise from north. From the same station a rescue plane is at horizontal range 19.6 km, 153 clockwise from north, with elevation 2.20 km. (a) Write the position vector for the ship relative to the plane, letting ˆ j north, and ˆ k up. (b) How far apart are the plane and ship? i represent east, ˆ 3. The stoplight of weight W = 255 N is suspend by two lines, T 1 and T 2. The system is in equilibrium. Add the force vectors T 1 and T 2 graphically and find their magnitudes. Also find them using the component method. Compare the results of the two methods. You graphical method must agree with the component method within 1 %. If it does not, redo it. 4. A rabbit trying to escape a fo runs north for 8.00 m, darts northwest for 1.00 m, then drops 1.00 m down a hole into its burrow and stops. Sketch the vector displacements. What is the magnitude and direction of the net displacement of the rabbit? Find the ijk vector too. ( ) km, and is looking for her dog Benji who is at B = ( 8.00ˆ ) 5. Cindy is located at C = 26.0ˆ i ˆ j i ˆ j km. In what direction and how far should Cindy travel to reach Benji? Draw the vectors and the resultant vectors. Find the magnitude and direction of the displacement vector relative to Cindy epressed in both polar and ij vectors. What is the distance between them? 6. Three uniform spheres of mass 2.00 kg, 4.00 kg, and 6.00 kg are placed at the corners of a right triangle as shown. Calculate the resultant gravitational force on the 4.00-kg object, assuming the spheres are isolated from the rest of the Universe, by both graphical (be sure to state your scale factor!) and ij component method. The magnitude of the gravitational force between two masses m and M is given by GmM F = 2 d where d is the distance between them. G is the Universal Gravitational Constant: 2 11 Nm G = kg