Dr. Crofoot RIM High School AP Physics Newton’s Laws 1st SEMESTER, Unit 2-2 Dr. Crofoot RIM High School

Newton’s Laws 10/12/2018 FN F=10N f Fg Starter / Plicker A 10 N force is horizontally applied to a 10 kg block on a friction surface. The coefficient of static friction is 0.35. What is the static friction from the surface on the block? 0.0 N 3.5 N 10 N 34 N FN F=10N f Fg

Newton’s Laws 10/15/2018 FN F=100N f Fg Starter / Plicker A 100 N force is horizontally applied to a 10 kg block on a friction surface. The coefficient of kinetic friction is 0.35. What is the kinetic friction from the surface on the block? 0.0 N 3.5 N 10 N 34 N FN F=100N f Fg

Newton’s Laws 10/12-15/17 2-Day Lab Friction Names Date Period Lab Report Name of Lab Scientific Question: Does the coefficient of static and kinetic friction stay the same on an incline. Theory: fs,max = 𝜇s FN, fk = 𝜇k FN Diagram: Label everything. Draw Free Body diagram for the block. Procedure: Reference diagram. How did you measure the frictions? Data/Results: Organized in a Table with units and sig. fig. Analysis: Talk about the test. Did coefficients change with angle? Compare 𝜇s and 𝜇k measurements. Give average and range of results. Discuss error and likely source. Suggest improvements. Friction Does the coefficient of static and kinetic friction stay the same on an incline? You will test at angles of zero and two others of your choice. Objectives Question listed above. Scale Block Incline

Newton’s Laws 10/12-15/18 2-Day Lab Scale Block Incline Friction Fnet = f + FN + Fs + Fg Fnet, x = -𝜇 mg cos(𝜙) + 0 + Fs – mg sin(𝜙) Stopped or Constant Velocity, a=0, so Fnet, x = 0 𝜇 mg cos(𝜙) = Fs – mg sin(𝜙) 𝜇 = [Fs – mg sin(𝜙)] / mg cos(𝜙) 𝜇 can be static max. or kinetic coefficient of friction. FN Fs f 𝜙𝜙 Fg Read scale and record max value just before movement, 𝜇s, max. Keep reading scale as blocks slides on incline at a constant slow rate. and record avg. value during movement, 𝜇k. Try to use same constant slow rate for each incline angle.

Newton’s Laws 10/16/2018 3.0 Kg 2.0 Kg 30° 60° Starter / Plicker A purple mass (3.0 kg) and red mass (2.0 kg) are tied together on a frictionless triangle. The base angles of the triangle are 60° and 30°. What is the tension on the string that ties them together? 16 N 23 N 31 N 49 N 3.0 Kg 2.0 Kg 30° 60°

Newton’s Laws 10/16/18 Write in Notebook Friction The magnitude of the maximum static friction is: fs,max = 𝜇s FN The magnitude of the kinetic friction is: fk = 𝜇k FN Sample Problem 6-2

Newton’s Laws 10/16/18 Home Work Friction From Chapter 6: Name Period Home Work Sheet Friction From Chapter 6: Problem 7, 13 Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s

Newton’s Laws 10/17/2018 Starter / Plicker In the drawing, what does scale B read? 6.3 N 12.5 N 15.2 N 30.5 N ? N A B 50° 50° 2.00 kg

Newton’s Laws 10/17/18 Write in Notebook Friction The magnitude of the maximum static friction is: fs,max = 𝜇s FN The magnitude of the kinetic friction is: fk = 𝜇k FN Work Problem 69 on the board. Tension on the cord. Magnitude of acceleration. mB = 2 kg mA = 4 kg 𝜇k = 0.50 frictionless 30°

Newton’s Laws 10/17/18 Write in Notebook Net Force We use vectors and angles to determine the forces on an object. Examples: 19.6 N ? N ? N A B A A B 50° 50° 2.00 kg 2.00 kg 2.00 kg

Newton’s Laws 10/17/18 Write in Notebook Net Force We use vectors and angles to determine the forces on an object. Examples: 5° B A A B 90° ? N ? N 2.00 kg 2.00 kg How do you think this applies to the design of things like bridges? Next week, you will design and build a bridge, then test it for strength.

Newton’s Laws 10/17/18 Home Work Friction From Chapter 6: Problem 65 Name Period Home Work Sheet Friction From Chapter 6: Problem 65 Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s

Newton’s Laws 10/18/2018 Starter / Plicker In the drawing, what does scale B read? 9.8 N 20 N 560 N 1120 N 1° B A ? N 2.00 kg

Newton’s Laws 10/18/18 Write in Notebook Drag Force The friction that we have been looking at is a simple function of the Normal Force, not a function of the objects velocity. An object moving through air experiences friction from the air that is a function of velocity, Drag Force. Moving slowly through air, there is no turbulence and the Drag force is proportional to velocity. D = k v Moving fast through air, there is turbulence and the Drag force is proportional to velocity squared. D = k v2

Newton’s Laws 10/18/18 Write in Notebook Drag Force, Terminal Velocity Moving fast through air, there is turbulence and the Drag force is proportional to velocity squared. D = k v2 ma = Fnet = D - Fg Fg Eventually, the drag force equals the force of gravity and the velocity stops increasing, acceleration is zero, terminal velocity. D = k v2 = Fg v = Fg /𝑘 v = 2Fg /(C ρ A ) v = 2mg /(C ρ A ) k = ½ C ρ A A = Area ρ = density C = constant

Newton’s Laws 10/18/18 Home Work Friction From Chapter 6: Problem 32 Name Period Home Work Sheet Friction From Chapter 6: Problem 32 Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s

Newton’s Laws 10/19/2018 Starter / Plicker On average, a sky diver has a terminal velocity of 60 m/s. What is his new terminal velocity if he decreases his area, A, by 50%. v = 2Fg /(C ρ A ) 30 m/s 42 m/s 85 m/s 120 m/s

Newton’s Laws 10/19/18 Write in Notebook Drag Force, Terminal Velocity Moving slow through air, there is no turbulence and the Drag force is proportional to velocity. D = k v ma = m dv/dt = Fnet = Fg - D = mg - kv dt = 𝑚𝑑𝑣 𝑚𝑔 −𝑘𝑣 = 𝑑𝑣 𝑔 −𝑘𝑣/𝑚 0 𝑡 𝑑𝑡 = 𝑣𝑜 𝑣 𝑑𝑣 𝑔 −𝑘𝑣/𝑚 t = - 𝑚 𝑘 ln(g - 𝑘𝑣 𝑚 ) Fg v vo

Newton’s Laws 10/19/18 Write in Notebook D Drag Force, Moving Slow, D = kv continued t = - 𝑚 𝑘 ln(g - 𝑘𝑣 𝑚 ) −𝑘 𝑡 𝑚 = ln(g - 𝑘𝑣 𝑚 ) - ln(g - 𝑘𝑣𝑜 𝑚 ) −𝑘 𝑡 𝑚 = ln( g − 𝑘𝑣 𝑚 g − 𝑘𝑣𝑜 𝑚 ) → 𝑒 −𝑘𝑡/𝑚 = ( g − 𝑘𝑣 𝑚 g − 𝑘𝑣𝑜 𝑚 ) 𝑒 −𝑘𝑡/𝑚 (g − 𝑘𝑣𝑜 𝑚 ) = g − 𝑘𝑣 𝑚 v = [g - 𝑒 −𝑘𝑡/𝑚 (g − 𝑘𝑣𝑜 𝑚 ) ] 𝑚 𝑘 v = 𝑚𝑔 𝑘 - 𝑒 −𝑘𝑡/𝑚 ( 𝑚𝑔 𝑘 − 𝑣𝑜) v vo Fg

v = 𝑚𝑔 𝑘 - 𝑚𝑔 𝑘 𝑒 − 𝑘𝑡 𝑚 → v(terminal) = 𝑚𝑔 𝑘 Newton’s Laws 10/19/18 Write in Notebook D Drag Force, Moving Slow, D = kv For Vo = 0, m = 50 kg, g = 9.8 m/s2, k = 50 v = 𝑚𝑔 𝑘 - 𝑚𝑔 𝑘 𝑒 − 𝑘𝑡 𝑚 → v(terminal) = 𝑚𝑔 𝑘 Fg

Newton’s Laws 10/19/18 Home Work Friction From Chapter 6: Problem 33 Name Period Home Work Sheet Friction From Chapter 6: Problem 33 Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s

Newton’s Laws 10/22/2018 Starter / Plicker What is the terminal velocity of a balloon of mass 5 grams, if the drag constant is k=0.17. Laminar motion, with no turbulence. 0.29 m/s 0.54 m/s 17 m/s 290 m/s

Newton’s Laws 10/22/18 Write in Notebook Uniform Circular Motion Recall that the magnitude of the centripetal acceleration is related to the magnitude of the velocity of an object in uniform circular motion by: a = v2/ R The direction of the acceleration vector is toward the center of the circle. The direction of the velocity vector is tangent to the circle. The direction of the radius vector is outward from the center of the circle, opposite the acceleration vector.

Newton’s Laws 10/22/18 Write in Notebook Uniform Circular Motion a = v2/ R Since F=ma: F = mv2/ R This force is required to keep the body moving in uniform circular motion. The direction of the force vector is inward toward the center of the circle, same as the acceleration vector.

Newton’s Laws 10/22/18 Write in Notebook Uniform Circular Motion F = mv2/ R We can also express this equation in angular frequency, v = 𝜔 R : F = m 𝜔 2 R Sample Problems: 6-7 and 6-10

Newton’s Laws 10/22/18 Home Work Friction From Chapter 6: Name Period Home Work Sheet Friction From Chapter 6: Problem 39, 47 Review Sample Problem 6-9 and 6-8 Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s Date Assigned Problems. Complete sentences Show math, units, sig. figs. Highlight or box answers 2.5 m/s

Newton’s Laws 10/23/2018 Starter / Plicker For people to stick to the wall of a spinning cylinder of radius R=2.1m and coefficient of static friction 𝜇k = 0.40, what is the minimum rotational velocity of the cylinder? 0.30rotations/sec 0.55 rotations/sec 7.2 rotations/sec 12 rotations/sec

Newton’s Laws 10/23/18 CRAM Sheet Work on CRAM Sheet Test Thursday and Friday