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Bellringer

Bellringer. Compare and explain in complete sentences and formulas how the third Newton’s law is applied to find the resultant force. Forces: Maintaining Equilibrium or Changing Motion. Homework. FINISH 50 % OF THE PROJECT. Force.

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Bellringer

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  1. Bellringer Compare and explain in complete sentences and formulas how the third Newton’s law is applied to find the resultant force.

  2. Forces: • Maintaining Equilibrium or Changing Motion

  3. Homework FINISH 50 % OF THE PROJECT

  4. Force • Force: a push or pull acting on a body that causes or tends to cause a change in the linear motion of the body • Characteristics of a force • magnitude • direction • point of application. • line of action • Net Force:resultant force (overall effect of multiple forces acting on a body) • Example: push from side and front = at angle

  5. Force • Free body diagram - sketch that shows a defined system in isolation with all the force vectors acting on the system.

  6. Why is this impossible?

  7. Classifying Forces • Internal Force: acts within the object or system whose motion is being investigated • action / reaction forces both act on different parts of the system • tensile-internal pulling forces when the structure is under tension • compressive- internal pushing (squeezing) forces act on the ends of an internal structure • do not accelerate the body • Orientate segments, maintain structural integrity

  8. Internal Forces • Examples • Contraction of muscles • Do not accelerate the body

  9. Classifying Forces • External Force: acts on object as a result of interaction with the environment surrounding it • non-contact - occur even if objects are not touching each other • gravity, magnetic • contact - occur between objects in contact • fluid (air & water resistance) • reaction forces with another body (ground, implement) • vertical (normal) reaction force • acts perpendicular to bodies in contact • shear reaction force • acts parallel to surfaces in contact (friction)

  10. F = ma • Force may also be defined as the product of a body's mass and the acceleration of that body resulting from the application of the force. • Units of force are units of mass multiplied by units of acceleration.

  11. Units of Force • Metric system (systeme internationale -SI) • Newton (N) • the amount of force necessary to accelerate a mass of 1 kg at 1 m/s2 • English system • pound (lb) • the amount of force necessary to accelerate a mass of 1 slug at 1 ft/s2 • equal to 4.45 N

  12. Weight (external force) • Weight - the amount of gravitational force exerted on a body. wt=mag. • Since weight is a force, units of weight are units of force - either N or lb. • As the mass of a body increases, its weight increases proportionally.

  13. Weight • The factor of proportionality is the acceleration of gravity, which is -9.81m/s2 or-32 ft/s2. • The negative sign indicates that the acceleration of gravity is directed downward or toward the center of the earth.

  14. Weight • On the moon or another planet with a different gravitational acceleration, a body's weight would be different, although its mass would remain the same. • Space station example

  15. Weight • Because weight is a force, it is also characterized by magnitude, direction, and point of application. • The direction in which weight acts is always toward the center of the earth.

  16. Center of Weight (Gravity) • The point at which weight is assumed to act on a body is the body's center of gravity.

  17. Friction (external force) • Component of a contact force that acts parallel to the surface in contact • acts opposite to motion or motion tendency • reflects interaction between molecules in contact • reflects force “squeezing” surfaces together • acts at the area of contact between two surfaces

  18. Static friction: surfaces not moving relative to each other Maximum static friction: maximum amount of friction that can be generated between two static surfaces Dynamic friction: surfaces move relative to each other constant magnitude friction during motion always less than maximum static friction Friction

  19. F =  N Friction depends on Nature of materials in contact () Force squeezing bodies together (N) Known as the normal contact (reaction) force Friction is FUN

  20. Nature of materials in contact Coefficient of friction () value serves as an index of the interaction between two surfaces in contact. Coefficient of Friction ()

  21. Factors Affecting Friction • The greater the coefficient of friction, the greater the friction • The greater the normal contact force, the greater the friction

  22. Normal reaction force (NRF) AKA - normal contact force force acting perpendicular to two surfaces in contact. magnitude intentionally altered to increase or decrease the amount of friction present in a particular situation football coach on sled push (pull) upward to slide object pivot turn on ball of foot Reaction Force

  23. Friction Manipulation Examples of manipulating  and Nc shoe design

  24. Friction Manipulation • Examples of manipulating  and • Nc • shoe design • grips (gloves, tape, sprays, chalk) • skiing: decrease for speed, increase for safety • curling • your examples???

  25. Curling

  26. Bode Miller, Silver Medalist 2002

  27. Shark Skin Suits in Swimming

  28. Friction and Surface Area • Friction force is proportional to the normal contact force • Friction is not affected by the size of the surface area in contact • normal contact force distributed over the area in contact • Friction is affected by the nature of the materials in contact

  29. Friction and Surface Area • With dry friction, the amount of surface area in contact does not impact the amount of friction. Same force acting over more area Same force acting over smaller area

  30. Coefficient of Friction • Hot, soft, rough surfaces have higher coefficients of friction • Tires, concrete, etc • Cold, hard, smooth surfaces have lower coefficients of friction • Ice, marble, etc

  31. Friction • Calculate the force of friction when you slide on ice. • Given = Normal force of 1000N • C of F = 0.02

  32. Solution Calculate the force of friction when you slide on ice. • Given = Normal force of 1000N • C of F = 0.02 • F = µ * N • F = (0.02) * (1000N) • F = 20 N

  33. Force • Force: a push or pull acting on a body that causes or tends to cause a change in the linear motion of the body (an acceleration of the body) • Characteristics of a force • magnitude • direction • point of application • line of action • sense (push or pull along the line of action) Vector represented with an arrow

  34. Recall • Concept of Net External Force Must add all forces acting on an object together

  35. Free body diagram • Free body diagram - sketch that shows a defined system in isolation with all the force vectors acting on the system • defined system: the body of interest • vector: arrow to represent a force • length: size of the force • tip: indicates direction • location: point of application

  36. Free body diagram Note: link contains practice problems  • Me, at rest in front of class • sagittal plane view • What are the names of the forces? • How big are the forces? • What direction are the forces? • Where are the forces applied?

  37. Free body diagram ==>static analysis Note: link contains practice problems • Me (mass= 75 kg): at rest (a = 0) in front of class (assume gravity at -10m/s/s) • sagittal plane view • Wt & vGRF • C of M, at feet • 750N, ????? • - & + F = m a F = 0 Wt + vGRF = 0 vGRF = - Wt vGRF = - (-750) vGRF = 750 N

  38. Addition of Forces(calculating the net [resultant] force) • Net force = vector sum of all external forces acting on the object (body) • account for magnitude and direction • e.g., Add force of 100N and 200N

  39. Addition of Forces(calculating the net [resultant] force) • Net force = vector sum of all external forces acting on the object (body) • account for magnitude and direction • e.g., Add force of 100N and 200N • act in same direction??? • Act in opposite direction??? • Act orthogonal to each other??? • Act at angles to each other???

  40. Colinear forces • Forces have the same line of action • May act in same or different directions • ie tug of war teammates: 100N, 200N, 400N • show force on rope graphically

  41. Colinear forces • Forces have the same line of action • May act in same or different directions • ie tug of war teammates: 100N, 200N, 400N • tug of war opponents: 200N, 200N, 200N • show force on rope graphically

  42. Colinear forces • Forces have the same line of action • May act in same or different directions • ie tug of war teammates: 100N, 200N, 400N • tug of war opponents: 200N, 200N, 200N • calculate resultant of the two teams • show force on rope graphically • calculate algebraically

  43. Free body diagram ==>static analysis • Weightlifter (mass 80 kg) • 100 kg bar overhead • at rest (a = 0) • sagittal plane view • What are the forces? • Where are the forces applied? • How big are the forces? • What direction are the forces? • Free body diagrams • Weightlifter • bar Fig 1.19, p 42

  44. Free body diagram ==>static analysis • Weightlifter (80 kg) • 100 kg bar overhead • at rest (a = 0) • sagittal plane view • What are the forces? • Where are the forces applied? • How big are the forces? • What direction are the forces? • Wt & vGRF • CofM, at feet • 800N, ????? • - & +

  45. Free body diagram ==>static analysis • Weightlifter (80 kg) • 100 kg bar overhead • at rest (a = 0) • sagittal plane view • Wt, bar & vGRF • CofM, on hands, at feet • 800N, 1000N, ????? • - , -, + F = m a F = 0 Wt + bar + vGRF = 0 vGRF = - Wt - bar vGRF = - (-800) - (-1000) vGRF = + 1800 N

  46. Concurrent Forces • Forces do not act along same line, but do act through the same point • ie gymnast jumps up to grab bar. Coach stops swinging by applying force to front and back of torso. • 20 N posterior directed push on front of torso • 30 N anterior directed push on back of torso • 550N force from bar on gymnast’s hands • gymnast mass 50 kg

  47. Concurrent Forces • gymnast hanging from grab bar. Coach applies force to front and back of torso to stop swing. • 20 N posterior directed push on front of torso • 30 N anterior directed push on back of torso • 550N force from bar on gymnast’s hands • gymnast mass 50 kg Page 29 in book

  48. Concurrent Forces • gymnast hanging from grab bar. Coach applies force to front and back of torso to stop swing. • 20 N posterior directed push on front of torso • 30 N anterior directed push on back of torso • 550N force from bar on gymnast’s hands • gymnast mass 50 kg • What is the resultant force? • Tip to tail method (fig 1.8 & 1.9 in text) • separate algebraic summation of horizontal and negative forces • Pythagorean theorem to solve resultant magnitude • Inverse tangent to solve direction (angle)

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