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Warmup

Warmup. Force. Affects on Velocity and Acceleration. Force Causes Change in Velocity. Oh! Skeeter!. Ha, ha, good one. This one time at band camp…. Force and Motion. Newton’s First Law At rest – Stays at rest (until force is applied) In motion – Stays in motion (until force is applied)

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Warmup

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  1. Warmup

  2. Force Affects on Velocity and Acceleration

  3. Force Causes Change in Velocity Oh! Skeeter! Ha, ha, good one This one time at band camp…

  4. Force and Motion Newton’s First Law • At rest – Stays at rest (until force is applied) • In motion – Stays in motion (until force is applied) • Force causes change in velocity • Force causes acceleration • Force causes a change in direction

  5. Warm-up • A blue one with a nubbin was moving at 5m/s straight down when the problem started. The difference between the bottom and the top is 3 times the height of a trans-atlantic 10m building. This nubbin sporting thing is earthbound. (include units)

  6. Types of Fundamental Force • Gravitational Force • Force we use in this section • Electromagnetic Force • Includes the contact forces we work with in this section • Nuclear Force • Weak Force (summarize)

  7. Electromagnetic Force • Contact Forces • Normal • Friction • Static: friction when the object is not in motion • Sliding: friction when the object is in motion • Tension • Spring (Use article to organize with topic oval)

  8. Force • Newton’s Third Law • Every action has an opposite and equal reaction.

  9. Force • Experiment with force sensors. • equal and opposite

  10. Key Vocab • Net force (Fnet) • Vector addition/subtraction 5 N + 7N • The resultant is the net force 5N + 7N = 12N

  11. Newton’s Second Law • Two men pull a 50-kg box with forces 19.7 N and 15.6 N in the directions shown below. Find the net force of the box. 19.7 N 15.4 N or -19.7N

  12. Force • The pound-force or simply pound (abbreviations: lb, lbf, or lbf) is a unit of force • 1N=0.225lb; 1lb=4.45N • Normal plus lift (Fnet) • Weight from force gauge. • Upward force must be greater than gravity to have upward acceleration. HW: 16,17, (pg 97), Example problem 2 pg 99 with elevator 3 times with a = 2.50, 3.00, 15.0m/s^2 and t=1.50, 3.25, 3.00s 19, 20, 22, 25 (pg 100 & 101) Demo

  13. Force • Newton’s Second Law………........... • Weight is a Force………………........... • Defined with the universal constant ‘G’. 1 lbf≈ 4.448222 N

  14. Force and Gravity

  15. Key Vocab • Normal Force • Force due to gravity and mass is referred to as a normal force or ‘the normal vector’. • Normal vector also refers to a vector that intersects at 90 degrees. (also stated as: A vector that is perpendicular to the tangent line at the interface)

  16. Newton’s Second Law • Two men pull a 50-kg box with forces 19.7 N and 15.6 N in the directions shown below. Find the resultant acceleration of the box and the direction in which the box moves. 19.7 N 15.4 N or -19.7N

  17. Practice - • A large helicopter is used to lift a heat pump to the roof of a new building. The mass of the helicopter is 7.0x10^3 kg and the mass of the heat pump is 1700 kg. • a. How much force must the air exert on the helicopter to lift the heat pump with an acceleration of 1.2 m/s^2? • b. Two chains connected to the load each can withstand 95,000 N. Can the load be safely lifted at 1.2 m/s^2?

  18. Air • Drag Force: exerted by a fluid on an object moving through the fluid. • Terminal Velocity: drag force is equal to force of gravity

  19. Friction Friction Coefficients: Table 5.1 pg 129 • Static friction Ff,s • Friction when there is no motion between the objects. • Ff,s <= usFN • Sliding friction (or kinetic friction) Ff,k • Friction when surfaces are rubbing against each other (in motion). • Ff,k = ukFN

  20. Friction Normal • A wood block on a wood plank. • m= kg • us = 0.5 • uk = 0.2 • FN = N Friction

  21. Review Elevator Example • Fnet = FE + (-Fg) • Fnet = ma (represents the force on the system as a whole) • FE = Fnet + Fg

  22. Tension • Force exerted by pulling (usually a string or rope). • For now, consider strings, ropes and pulleys (also called sheaves or blocks) to be massless and frictionless. • Provide a change in direction of force.

  23. Tension Problem • The blocks shown are placed on a smooth horizontal surface and connected by a piece of string. If a 8.8-N force is applied to the 8.8-kg block, what is the tension in the string? 6.1N

  24. Tension • Three blocks A, B, and C are connected by two massless strings passing over smooth pulleys as shown below, with the 3.4-kg block on a smooth horizontal surface. Calculate the tension in the strings connecting A and B, and B and C. 54N 48N

  25. Tension

  26. Tension Practice

  27. Sketch Problem 68N 40N A B C 68N 40N ABC

  28. Sketch Problem 68N 40N A B C 40N 68N C AB 68N 40N A BC

  29. Fnet • Fnet = sum of all forces acting on a system = ma • Fnet = Ff,k + FT + Fpush + Fgravity = ma

  30. Force • Newton’s Third Law • Every action has an opposite and equal reaction.

  31. Force • Newton’s Second Law………...........

  32. Force • Newton’s first law • When the net forces are zero, an object at rest remains at rest and an object in motion remains in motion in the same direction at the same speed. • For Fnet = 0 • velocity is constant: v1 = v2 • acceleration is zero

  33. Friction Friction Coefficients: Table 5.1 pg 129 • Static friction Ff,s • Friction when there is no motion between the objects. • Ff,s <= usFN • Sliding friction (or kinetic friction) Ff,k • Friction when surfaces are rubbing against each other (in motion). • Ff,k = ukFN

  34. Friction • Static friction Ff,s • Friction when there is no motion between the objects. • Ff,s <= usFN • Ff,s <= usmg 10N 5kg

  35. Review Elevator Example • Fnet = FE + (-Fg) • Fnet = ma (represents the force on the system as a whole) • FE = Fnet + Fg

  36. Setting up the Problem • Connected by a massless string, pulled along a surface with a coefficient of friction 100N

  37. Steps • What forces are present? • What is the value of each force? • Draw the free body diagram 100N

  38. Steps • List known variables, solve for unknown

  39. Break apart • Find the tension in each string • First, identify the forces of each part 100N

  40. Consider each section as a system • Draw a free body diagram for this system

  41. Consider each section as a system • Draw a free body diagram for this system

  42. Break apart • Draw a sketch (free body diagram) for each string

  43. Trig Identities • SOHCAHTOA • Adjacent Leg = Hypotenuse * cos • Opposite Leg = Hypotenuse * sin • Opposite Leg = Adjacent Leg * tan

  44. Force components • The magnitude of the force 1 is 87 N, of force 2 is 87 N, and of 3 is 87 N. The angles θ 1 and θ 2 are 60° each. Use the Pythagorean theorem and trig identities to find the resultant of the forces 1, 2 , and 3 .

  45. Drag Force • Terminal velocity is when an object in free fall has reached equilibrium. That is, the drag force is equal to the force of gravity. • The Fnet for a system in equilibrium is always zero.

  46. Friction with Force components • The system shown below is in equilibrium. Calculate the force of friction acting on the block A. The mass of block A is 7.10 kg and that of block B is 7.30 kg. The angle θ is 48.0°.

  47. Break Apart • The system shown below is in equilibrium: This means Fnet=0, since Fnet=ma, then ma=0 and a=0 • In order for the net force to be zero on block A (no acceleration) the tension in the rope pulling block A has to be matched by the friction between block A and the surface. The tension here must be equal to the friction resisting motion The friction here must be equal to the tension pulling A

  48. Break Apart • There is a force from gravity acting on block B. • Convert this force to its x and y components. • Using the concept of equilibrium, realize that the sum of x components is 0 θ is 48.0°. 7.30kg

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