Practice

2. Practice A plane flies 1500 miles East and 200 miles South. What is the magnitude and direction of the plane’s final displacement? A hiker walks 80 m North, 20 m East, 50 m South, and 30 m West. What is the magnitude and direction of the hiker’s displacement?

3. Practice Problem #1 A plane flies 1500 miles East and 200 miles South. What is the magnitude and direction of the plane’s final displacement? **not drawn to scale** 1500 miles θ 200 miles Resultant - 1513 miles a2 + b2 = c2 (1500 m)2 + (200 m)2 = R2 R = √ (1500 m)2 + (200 m)2 R = 1513.275 m tan θ = opp/adj θ = tan-1 (200/1500) θ = 7.5946433°

4. Practice Problem #2 A hiker walks 80 m North, 20 m East, 50 m South, and 30 m West. What is the magnitude and direction of the hiker’s displacement? By subtracting the opposing directions from each other, we find the hiker’s displacement along the y-axis to be 30 m North, and the displacement on the x-axis to be 10 m West. a2 + b2 = c2 302 + 102 = R2 R = √900 + 100 R = 31.623 m tan θ = opp/adj θ = tan-1 (10/30) θ = 18.435°

5. Practice Problem #1 An airplane traveling 1001 m above the ocean at 125 km/h is going to drop a box of supplies to shipwrecked victims below. a. How many seconds before the plane is directly overhead should the box be dropped? b. What is the horizontal distance between the plane and the victims when the box is dropped?

6. Practice Answer #1 Horizontal vix = 125 km/h dx = ? t = ? ax = 0 m/s2 dx = vixt dx = (125 km/h)(14.3 s) dx = (0.03472 km/s)(14.3 s) dx = 0.4965 km = 496.5 m • Vertical • viy = 0 m/s • ay = -9.8 m/s2 • dy = 1001 m • t = ? • d = ½ at2 • 1001 m = ½ (9.8 m/s2)t2 • t = 14.3 s

7. Practice Problem #2 • Herman the human cannonball is launched from level ground at an angle of 30° above the horizontal with an initial velocity of 26 m/s. How far does Herman travel horizontally before reuniting with the ground?

8. Practice answer #2 Horizontal vix = 26(cos30) ax = 0 m/s2 dx = ? t = ? dx = vixt dx = 26(cos30)(2.66 s) dx = 60.0 m • Vertical • viy = 26(sin30) • ay = -9.8 m/s2 • vf = 0 m/s • t = ? • vf = viy + at • 0 = 26(sin30) + (-9.8 m/s2)t • -13 = -9.8 m/s2t • t = 1.33 s

9. Case 2: Block on an inclined plane FNormal Fperpendicular(F|) = Fgcosθ FN = F| FParallel FPerpendicular FGravity Fparallel (F||) causes sliding F|| = Fgsinθ

10. Practice Problem 1 (part 1) A 50 kg sled is pulled by a boy across a smooth, icy surface. If the boy is pulling the sled 500 N at 30° above the horizontal, what is the horizontal component of the force? 500 N 30° sled

11. Practice Problem 1 (part 2) What is the acceleration experienced by the sled? Does the vertical component of the force affect the acceleration? 500 N sled

12. Answers Ax = Acosθ Fx = (500 N)cos(30°) Fx = 430 N FNET = ma 430 N = (50 kg)a a = 8.6 m/s2 The vertical component doesn’t affect acceleration. It only causes the object to lift off the ground, rather than move it backward or forward.

13. Practice Problem 1 (part 3) Using the same sled from the previous example, what is the weight of the sled? Find the vertical component of the force pulling the sled. Would this force cause the sled to lift of the ground? Why? 500 N sled

14. Practice Problem 1 (part 4) What is the Normal Force felt by the sled? 500 N sled

15. Answers Weight is equal to Fg. Fg = mg. Since the mass of the sled is 50 kg, we can find weight by plugging in the numbers and solving for Fg. Fg = (50 kg)(9.81 m/s2) Fg = 490 N down This would not be enough to lift the sled because the force due to gravity is much greater. Ay = Asinθ Fy = 500 N sin(30°) Fy = 250 N up

16. Answers The Normal Force felt by the sled would be equal in magnitude to the Gravitational Force (Weight), but in the opposite direction (perpendicular to the surface). Therefore, FN = 490 N up

17. Practice Problem 2 (part 1) A block with a mass of 100 kg is at rest on an inclined plane with an angle of 30°. What is the weight of the block? What is the parallel force of the block? What is the perpendicular force of the block? 100 kg 30°

18. Drawing a Force Diagram Green Vector represents the force due to Gravity (Weight = mg) 100 kg Red Vector represents the Normal Force of the incline pushing up on the box 30° Blue Vectors represent the components of the weight (Perpendicular and Parallel Forces)

19. Answers The Weight of the block is equal to mass times acceleration due to gravity. W = mg W = (100 kg)(9.8 m/s2) W = 980 N F|| = Fg sinθ F|| = (980 N)sin(30°) F|| = 490 N F| = Fgcosθ F| = (980 N)cos(30°) F| = 850 N

20. Practice Problem 2 (part 2) What is the acceleration of the block as it slides down the inclined plane? What is the Normal Force felt by the block? As the angle of the inclined plane increases, what happens to the parallel and perpendicular forces? 100 kg 30°

21. Answers To find the acceleration of the block as it slides down the incline, we need to use the parallel force. F|| = ma 490 N = (100 kg)a a = 4.9 m/s2 Normal Force is equal to the Perpendicular Force, but opposite in direction. FN = 850 N

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