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Force. BOX. force - a push or pull acting on a body forces are vector quantities magnitude (size) direction (line of action) point of application measured in Newtons (N) 1 N = (1 kg) (1 m/s 2 ) English units (lb) 1 lb = (1 slug) (1 ft/s 2 ). pt of application. F. line of action.

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    1. Force BOX • force - a push or pull acting on a body • forces are vector quantities • magnitude (size) • direction (line of action) • point of application • measured in Newtons (N) • 1 N = (1 kg) (1 m/s2) • English units (lb) • 1 lb = (1 slug) (1 ft/s2) pt of application F line of action

    2. air resistance mg F (GRF) R Free-Body Diagrams A free body diagram illustrates all of the external forces acting on an object. If whole body is considered to be “the system” Examples of external forces Weight (gravity) Ground Reaction Force (GRF) Friction Fluid Resistance Examples of internal forces Joint Reaction Force Muscle Force Contact Forces

    3. NEWTON'S LAWS OF MOTIONPhilosophiae Naturalis Principia Mathematica (1686)

    4. Newton’s 1st Law of Motion • Law of Inertia • a body at rest will remain at rest and a body in motion will remain in motion and move at a constantvelocity until a non-zero resultant external force is applied to it. Inertia is the resistance of an object to motion - the amount of resistance to linear motion varies directly with the mass of the object. When an object is in motion its resistance to change in motion is determined by its velocity as well as its mass. Momentum is the mass of an object multiplied by the velocity (p = mv).

    5. Newton’s 1st Law of Motion V W W Fr Fp Ry Ry Fr = resistive force Fp = propulsive force if v = 0 andSF = 0 DYNAMIC EQUILIBRIUM if Ry = W then resultant force = 0 if v = 0 andSF = 0 STATIC EQUILIBRIUM

    6. Newton’s 2nd Law of Motion • Law of acceleration • when a non-zero resultant external force is applied to a body, the body will accelerate in the direction of this force. The acceleration is proportional to the force and inversely proportional to the body’s mass

    7. m m F2 SF a F1 where k1 = constant of proportionality

    8. If m is measured in kg and a is measured in m/s2 the SI unit for force is “newton” (N) 1 N = 1 kg x 1 m/s2 where k2=1 SF = ma

    9. Alternate Expression of LAW OF ACCELERATION "The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts." • SF = • SFt = mDv (impulse/momentum relation) • SF = m = ma

    10. If an elephant and a feather fall from the same height in the absence of air resistance then the resultant or net force acting on each object is simply their weight. Since W = mg then the acceleration they experience is SF = ma but SF = W = mg mg = ma or a = g

    11. Now consider when air resistance is present. This force would larger on the elephant simply because is a bigger object. But the weight of the elephant is also significantly larger than that of the feather. In fact, relative to weight the air resistance acting on the feather is larger than on the elephant. This affects the resultant force acting on each object such that the resultant force acting on the feather is much closer to 0 N. Thus the feather will have a much lower acceleration.

    12. This example further demonstrates the change in resultant force due to air resistance. Notice that initially air resistance due to the body falling through the air reduces the magnitude of the acceleration but it remains a downward acceleration. Eventually you reach a point where the air resistance equals your body weight. This is known as terminal speed and would be well over 100 mph for a human body. To allow you to land without hurting yourself you deploy your parachute. This greatly changes the resultant force such that the net force actually points upward such that the acceleration is upward. This successfully decreases your speed to a more manageable level for landing.

    13. Newton’s 3rd Law • “Law of action and reaction” • when one body exerts a force on another body, the second exerts an equal but opposite force back on the 1st body • “for every action there is an equal and opposite reaction”

    14. Rresistance Aresistance Rpropulsion Apropulsion Note: these forces (action and reaction) are never applied on the same body -- it takes two bodies for a pair of action/reaction forces to exist

    15. LAW OF GRAVITATION "All bodies attract one another with a force proportional to the product of their masses and inversely proportional to the square of the distance between them." • FG = G , G = 6.67X10-11 N-m2/kg2

    16. r M Mass and Weight • g depends on the distance from the earth’s center

    17. r M Earth is not a true sphere but rather an ellipsoid: it is fatter at the equator m = your mass M = earth’s mass r = radius (distance to (center of earth) G = gravitational constant Newton’s Law of Gravitation Fg = gravitational force If m = your mass then Fg = your weight (W)

    18. r M Latitude effect = g is smaller at equator larger at poles Altitude effect = g is smaller at high altitude larger at low altitude

    19. Weight • W = mg • weight is a force • weight = mass x acceleration due to gravity • Units • N = kg x m/s2 • weight in Newtons = mass in kg x 9.81 m/s2 • a 1 kg mass weighs 9.81 N

    20. The Relevance of Newton’s Laws These 4 laws allow all motion in the universe to be described and predicted as long as the relative speeds of the objects are small compared to the speed of light.

    21. FV FML Direction of motion FAP Force Platform Ground Reaction Forces Contact Forces

    22. Impulse - Momentum Relationship Contact Forces

    23. Impulse-Momentum “Impulse = change in momentum” Units Same units! Contact Forces

    24. Impulse/Momentum (SFt = mDv) • The impulse/momentum relationship describes the effects of a force over a period of time. • An impulse (SF*t) causes a mass to change its velocity • A 10 N force is applied to a 2 kg mass for 3 seconds. What is the change in velocity? Contact Forces

    25. Impulse/Momentum • How long would it take to stop a 7 kg mass that has a velocity of 10 m/s if you are capable of producing a maximum instantaneous force of 35 N? Contact Forces

    26. Impulse/Momentum • It has been determined that a force over 200 N can injure the hand. What is the shortest period of time it will take to stop a 2 kg object traveling at 75 m/s if the hand is to be protected? Contact Forces

    27. Impulse/Momentum • The braking impulse of a subject running across a force platform is -10 N-s. The propelling impulse during the same time period is 2 N-s. What is the change in velocity of the subject if she has a mass of 55 kg? Fy 0 0 .8 time Contact Forces

    28. impulse = area • If a 10 kg object is exposed to the above impulse what will be the change in velocity? Contact Forces

    29. Impulse = area under F v. t curve F Jp=positive impulse t Jn=negative impulse JNET = NET IMPULSE JNET = Jp + Jn Contact Forces

    30. Recall that the net force is SF = GRF - W and by Newton’s 2nd Law SF = ma so GRF - W = ma If GRF > W then a >0 If GRF < W then a < 0 W GRF

    31. When the jumper initiates the countermovement he/she speeds up in the negative direction this means that the body experiences a negative acceleration thus GRF < W W GRF body weight GRF 0

    32. As the jumper nears the bottom of the countermovement he/she slows down in the negative direction this means that the body experiences a positive acceleration thus GRF > W W lowest point GRF body weight GRF 0

    33. W GRF When the jumper begins the upward portion of the countermovement he/she speeds up in the positive direction this means that the body experiences a positive acceleration thus GRF > W lowest point GRF body weight 0 takeoff

    34. Total Net Impulse = Negative Net Impulse + Positive Net Impulse SJ = mDv = m(vf - vi) this can be solved to find the takeoff velocity (final velocity of countermovement phase) vtakeoff = SJ/m since vi = 0 Knowing vtakeoff allows you to compute jump height

    35. Knowing vtakeoff allows you to compute jump height vtop = 0 m/s vf2 = vi2 + 2ad where vf = vtop vi = vtakeoff vtakeoff

    36. SO - jump height is increased by increasing the total net impulse … not just the net force jump height

    37. Conservation of Momentum Momentum represents the total quantity of motion possessed by a body or system. The momentum of a system cannot be altered without an external force. Momentum = mass x velocity Total momentumbefore = total momentumafter for example - if there are 2 objects in the system m1v1 + m2v2 = m1u1 + m2u2 v = velocity before u = velocity after subscripts represent object number

    38. Example: A 100 kg astronaut is moving at a speed of 9 m/s and runs into a stationary astronaut (mass = 150 kg). Problem: What is the velocity of the astronauts after the collision?

    39. System = both astronauts External Forces = none Therefore the momentum must be conserved. Total momentum (p): pbefore = pafter = v = ?

    40. A 75 kg rugby player is moving at 2 m/s when he runs into a 100 kg player running at –1.5 m/s. Which direction will the resulting collision travel (-, 0, +)? If they collide in mid-air (0). If they collide on the ground the players will be able to exert external forces and the larger player will probably be able to exert larger external forces (+).

    41. pbig fish-before + plittle fish-before = pbig fish-after + plittle fish-after

    42. pbig fish-before + plittle fish-before = pbig fish-after + plittle fish-after

    43. Joint Reaction Force The net force acting across a joint e.g. when standing, the thigh exerts a downward force on the shank, conversely, the shank exerts an upward force on the thigh of equal magnitude NOTE: this JRF does not include the forces exerted on the joint by the muscle crossing the joint. The force that includes all of the forces crossing the joint is known as the bone-on-bone force Contact Forces

    44. Block Pulling Force Friction Force Normal Force (weight of the block) • There is an interaction between the surface of the block and the table. • The friction force always opposes motion. • The pulling force must be greater than the frictional force to move the block. Contact Forces

    45. Normal Force = force perpendicular to surface When surface is horizontal the perpendicular direction is vertical so the normal force is simply the weight of the object FN Weight q Normal Force When surface is inclined the perpendicular direction is NOT vertical so the normal force is only a component of the object’s weight.

    46. maximal static friction force impending motion Static Friction Force Dynamic Applied Force • At some point the pulling force will be great enough so that the friction force cannot prevent movement. Contact Forces

    47. Friction • The coefficient of static friction is expressed as: where ms= coefficient of static friction Fnormal = normal force Fmax = maximal static friction force • The coefficient of friction is a dimensionless number. It is unaffected by the mass of the object or the contact area. • The greater the magnitude of ms the greater the force necessary to move the object.

    48. Friction • As the block moves along the table, there still is a frictional force that resists motion. • Sliding and rolling friction are types of dynamic friction. where md= coefficient of dynamic friction N = normal force Ffriction = force resisting motion Contact Forces

    49. Friction • It has been found experimentally that md< ms. • md depends of the relative speed of the surfaces. • At speeds from 1 cm/s to several m/s, md is approximately constant. Contact Forces

    50. W W Why is it easier to pull a desk than push it? When you push you usually have a downward component of force -- so the normal force is increased and therefore the frictional force is increased. P R = W+v h v Contact Forces