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Slide 16 of 36 Vertical Motion Use Newton’s 2nd law with vertical forces: where: FA = applied force in N or kg•m/s2 Fnet = FA + Fg Fg = weight (Fg = mg) in N or kg•m/s2 Fnet = net force (Fnet = ma) in N or kg•m/s2 MEMORIZE THIS FORMULA!! Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Newton’s Laws and Vertical Motion Newton’s Laws and Vertical Motion Slide 7 of 18 Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Specific Outcome: i. I can apply Newton’s laws of motion to solve, algebraically, linear motion problems in horizontal, vertical and inclined planes near the surface of Earth, ignoring air resistance. ii. I can calculate the resultant force, or its constituents, acting on an object by adding vector components graphically and algebraically. iii. I can analyze a graph of empirical data to infer the mathematical relationships among force, mass and acceleration. Vertical Motion Fg = gravitational force on the object FA = force applied OR force exerted OR tension force on the object When there is no acceleration, FA = -Fg When there is free-fall, FA = 0 N Vector directions (up = +, down = -) are still being used! Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H 1 There are six different situations for an object in vertical motion: Slide 5 of 18 Slide 5 of 18 Vertical Motion Vertical Motion There are six different situations for an object in vertical motion: 1) at rest, FA = -Fg (since a = 0 m/s2, Fnet = 0 N) 4) moving up with positive acceleration (since a > 0 m/s2, Fnet > 0 N) 2) moving up at constant speed, FA = -Fg (since a = 0 m/s2, Fnet = 0 N) 5) moving down with negative acceleration (since a < 0 m/s2, Fnet < 0 N) 3) moving down at constant speed, FA = -Fg (since a = 0 m/s2, Fnet = 0 N) 6) free fall, accelerating with gravity only (FA = 0 N, so Fnet = Fg) Vertical Motion ex. 1: Find the net force if an upward force of 100 N is applied to a 7.0 kg mass. Fg = mg = (7.0 kg)(-9.81 m/s2) FA • Fg = -68.67 N Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Slide 20 of 36 Slide 19 of 36 Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Vertical Motion ex. 2: What is the acceleration of the object in the previous example? +31.33 N Fnet a= m = 7.0 kg = +4.5 m/s2 = 4.5 m/s2 [up] Fnet = FA + Fg = (100 N) + (-68.67 N) = +31.33 N = 31 N [up] Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H 2 Slide 21 of 36 Slide 21 of 36 Vertical Motion ex. 3: Find the force needed to accelerate a 2.0 kg object upward with an acceleration of 5.0 m/s2. Fg = mg = (2.0 kg)(-9.81 m/s2) = -19.62 N Fnet = ma = (2.0 kg)(5.0 m/s2) = +10 N Fnet = FA + Fg FA = Fnet - Fg ex. 4: Will exerts a force of 450 N to jump up at 3.0 m/s2. Determine his mass. Fnet = ma FA Fg • m= FA a-g = +450 N (+3.0 m/s2) – (-9.81 m/s2) = 35 kg Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Slide 12 of 18 Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Slide 11 of 18 Fg = mg Fnet = FA + Fg becomes ma = FA + mg FA = 10 N – (-19.62 N) = 30 N [up] Vertical Motion ex. 5: A 62 kg wrestler is standing on a spring scale in an elevator. What does the scale read when: a) the elevator is accelerating up at 2.0 m/s2? Fg = mg = (62 kg)(-9.81 m/s2) = -608.22 N Fnet = ma = (62 kg)(2.0 m/s2) = +124 N FA Fnet = FA + Fg Fg FA = Fnet - Fg Vertical Motion • Vertical Motion ex. 5: A 62 kg wrestler is standing on a spring scale in an elevator. What does the scale read when: b) the elevator moves down at 2.0 m/s2? Fnet = ma = (62 kg)(-2.0 m/s2) = -124 N FA Fnet = FA + Fg Fg FA = Fnet - Fg • FA = -124 N – (-608.22 N) = 4.8 x 102 N [up] FA = 124 N – (-608.22 N) = 7.3 x 102 N [up] Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H 3 Slide 14 of 18 Slide 13 of 18 Vertical Motion ex. 5: A 62 kg wrestler is standing on a spring scale in an elevator. What does the scale read when: c) the elevator moves down at a constant speed of 2.5 m/s? Fnet = 0 N F A FA = -Fg = -(-608.22 N) Fg Vertical Motion ex. 5: A 62 kg wrestler is standing on a spring scale in an elevator. What does the scale read when: d) the elevator falls (the cable broke!)? FA = 0 N Fg • • = 6.1 x 102 N [up] Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H Dulku – Physics 20 – Unit 2 (Dynamics) – Topic H 4