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Force • force - a push or pull acting on a body • forces are vector quantities – magnitude (size) – direction (line of action) – point of application BOX • measured in Newtons (N) – 1 N = (1 kg) (1 m/s2) pt of application F • English units (lb) – 1 lb = (1 slug) (1 ft/s2) line of action 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 air resistance mg FR (GRF) Contact Forces NEWTON'S LAWS OF MOTION Philosophiae Naturalis Principia Mathematica (1686) 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 constant velocity 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). Newton’s 1st Law of Motion V W W Fr Fp Ry if Ry = W then resultant force = 0 if v = 0 and SF = 0 STATIC EQUILIBRIUM Ry Fr = resistive force Fp = propulsive force if v = 0 and SF = 0 DYNAMIC EQUILIBRIUM 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 F2 SF m m a F1 a SF a 1 m SF = ma 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 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 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. 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. 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” 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 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." m1* m2 • FG = G d2 , G = 6.67X10-11 N-m2/kg2 Mass and Weight • g depends on the distance from the earth’s center r M 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 GMm 1 Fg 2 Fg 2 and Fg m r r Fg = gravitational force If m = your mass then Fg = your weight (W) r M Fg g GMm r2 GM r2 W mg Latitude effect = g is smaller at equator larger at poles Altitude effect = g is smaller at high altitude larger at low altitude 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 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. Ground Reaction Forces FV FML FAP Force Platform Contact Forces Pressure • localized effect of a force being applied to an area of a certain surface Pressure W xxxxx x x x x x P F s xxxxx A Rn if W = 100 lb, A = 500 in2 Force Area Bird’s eye view x = pt of application for a force P 100 lb 2 0.2 psi 0 . 2 lb / in 500 in 2 Special Force Application W = 110 lb *4.45 N/lb = 489.5 N What is the avg. pressure when standing on one foot? In SI Units? 10 in2 = 0.00645 m2 2 2.54 cm 10 in 2 64.5 cm2 0.006 m2 in 489.5 N P 75.9 kPa 2 0.00645 m 40 in2 = 0.0258 m2 489.5 N P 19.0 kPa 2 0.0258 m Special Force Application Pressure distributions in the foot While standing, most of the pressure is in the heel and the forefoot Recall that the net force is SF = GRF - W and by Newton’s 2nd Law SF = ma so GRF - W = ma W If GRF > W then a >0 If GRF < W then a < 0 GRF 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 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 GRF lowest point body weight GRF 0 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 W thus GRF > W GRF lowest point body weight GRF 0 takeoff If the GRF v. time curve is approximated using rectangles the height of each rectangle is the force and the width of the rectangle is the time. The area of the rectangle (product of width and height) represents the IMPULSE (the product of Force and time). W GRF lowest point body weight GRF 0 takeoff Impulse-Momentum J p “Impulse = change in momentum” F t mv f mvi Impulse Units F t N s (kg m/ s2)s kg m/ s momentum mv kgm/ s Same units! Contact Forces 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 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? F y 0 0 time .8 Contact Forces F o r c e (N) 5 4 3 2 impulse = area 1 0 0 1 2 3 4 5 Time (s) • If a 10 kg object is exposed to the above impulse what will be the change in velocity? Contact Forces Total Net Impulse = Negative Net Impulse + Positive Net Impulse SJ = mv = 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 Knowing vtakeoff allows you to compute jump height vtop = 0 m/s vf2 = vi2 + 2ad where vf = vtop vi = vtakeoff vtakeoff SO - jump height is increased by increasing the total net impulse … not just the net force jump height 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 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? System = both astronauts External Forces = none Therefore the momentum must be conserved. Total momentum (p): pbefore = m1v1 + m2v2 = 100kg(9m/s) + 50kg(0m/s) = 900 kg-m/s pafter = m1v1 + m2v2 = (m1+ m2)v = (100kg + 150kg)v = 900 kg-m/s v = 3.6 m/s 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 (+). pbig fish-before + plittle fish-before = pbig fish-after + plittle fish-after (4m kg*5 km/hr) + (m kg* 0 km/hr) = (4m kg* v) + (m kg * v) 20m kg km/hr = 5m kg * v v = 4 km/hr pbig fish-before + plittle fish-before = pbig fish-after + plittle fish-after (4m kg*0 km/hr) + (m kg* 5 km/hr) = (4m kg* v) + (m kg * v) 5m kg km/hr = 5m kg * v v = 1 km/hr 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 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 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 Weight Normal Force FN=W*cosq q When surface is inclined the perpendicular direction is NOT vertical so the normal force is only a component of the object’s weight. Friction Force maximal static friction force Static impending motion Dynamic Applied Force • At some point the pulling force will be great enough so that the friction force cannot prevent movement. Contact Forces Friction • The coefficient of static friction is expressed as: Fmax ms Fnormal 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. 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. md F friction FNormal where md= coefficient of dynamic friction N = normal force Ffriction = force resisting motion Contact Forces 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 Ff = Wsinq N = Wcosq W q Wcosq Wsinq ms = Wsinq/Wcosq = tanq An object placed on an incline has a weight that will produce both the normal force and the force that moves the object. If the incline is slowly increased, the coefficient of static friction can be calculated from q at the instant that the object begins to slide. Contact Forces 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 h R = W+v v W W Contact Forces When you pull you usually have an upward component of force -- so the normal force is reduced and therefore the frictional force is reduced. P v R = W-v h W W Contact Forces Collisions and Impacts Elastic Inelastic Coefficient of Restitution • The coefficient of restitution (e) determines the elasticity of an impact. • If e = 1 the impact is completely elastic. This means that the object contains all of the energy it had before the impact. • If e = 0 the impact is completely plastic. This means that the object contains none of the energy it had before the impact. Special Force Application Objects typically deform during an impact This is referred to as the period of deformation This is followed by a period of restitution thus the name coefficient of restitution Coefficient of Restitution • e=- • e= relative velocity after impact relative velocity before impact h bounce h drop Work • Work is a force applied over a given distance. • W = F*d • Units are N-m, 1 N-m = 1 Joule. • Positive work occurs when the force is applied in the same direction as the motion (concentric contractions). • Negative work occurs when the force is applied in the opposite direction of the motion (eccentric contractions). Special Force Application direction of motion q • Only the component of force parallel to the direction of motion is responsible for work being done. If there is no movement there is no work being done. Special Force Application Internal Work vs External Work • External work is the result of external forces such as ground reaction forces. • Internal work is the result of internal forces such as muscle forces. While running uphill there is internal work done to rotate the segments and external work done to raise the body to a new height. Energy • Energy is the capacity to do work. • Performing positive work on an object will increase its energy while performing negative work will decrease its energy. W = E • Energy is measured in Joules. Special Force Application • Kinetic Energy (KE) is the energy that an object possesses because of its velocity. KE = ½mv2 • Potential Energy (PE) is the energy that an object possesses because of its height. PE = mgh • Strain Energy (SE) is the energy that an object possesses because of its deformation: SE = ½kx2 Special Force Application Kinetic Energy • Kinetic Energy (KE) - energy due to motion KE 1 mv 2 2 unit: kg m2 / s2 J – e.g. A diver (mass = 70 kg) hits the water after a dive from the 10 m tower with a velocity of 14 m/s. How much KE does she possess? 2 6860 J KE 1 mv 2 1 (70 kg)(14 m ) s 2 2 Special Force Application Potential Energy • Potential Energy (PE) - energy due to gravity – PE = mgh = mass*gravity*height – e.g. A diver (mass = 70 kg) is 10 meters above the water. How much PE does the diver have? PE = 70*9.8*10 = 6860 J Special Force Application Strain Energy • Strain (SE) - energy due to deformation – SE = ½kx2, k = stiffness, x = deformation – this type of energy arises in compressed springs, squashed balls ready to rebound, stretched tendons inside the body, and other deformable structures – e.g. A muscle tendon with a stiffness of 70,000 N/m is stretched by 1 cm. How much strain energy does it have? SE = ½(70,000)(.01)2 = 3.5 J Special Force Application Conservation of Energy • Energy can not be created nor destroyed, it can only change forms (von Helmholtz, 1847). PE = mgh Energy is often transferred between kinetic, potential and strain. KE =.5mv2 Special Force Application KE = 0 PE = max 1 release 3 KE = 0 PE = max .5m KE = max 2 PE = 0 • If the 4 kg pendulum above is released from a height of .5 meters what is the maximum velocity? PE = 4kg(9.81m/s/s).5m = 19.62 J KE = 19.62 J = .5(4kg)v2, v = 3.13 m/s Special Force Application Ht (m) v (m/s) PE (J) KE (J) 3.0 29.4 0 0 2.5 24.5 3.1 4.9 2.0 19.6 4.4 9.8 1.5 14.7 5.4 14.7 1.0 9.8 6.3 19.6 time Special Force Application Conversion of PE to KE and vice-versa Start with no KE b/c v=0 You can use this principle of energy conservation to solve useful problems. For instance – how high would you need to raise the slide in the picture below to successfully prevent the slider from running off the end? TEinitial = TEfinal TE = PE + KE PEinitial + KEinitial = PEfinal + KEfinal m(9.8 m/s2)(4.0 m)+ ½m(8 m/s)2 = m(9.8 m/s2)(h)+ ½m(0 m/s)2 h = 7.6 m Work-Energy Relationship • If you lift a barbell into the air you are performing work on the barbell. You apply a force over a distance. • By performing work on the barbell you change the amount of PE that the barbell has. W = E PE = mgh2 PE = mgh1 Special Force Application • If you lift a 300 Newton barbell 1 meter you have a change in energy of 300 J. This is equal to the amount of work done. W = E PE1 = (300)0 = 0 J PE2 = (300)1 = 300 J PE = mgh2 PE = mgh1 Special Force Application Work-Energy Relationship W = TE = PE+ KE+ SE When you ski down a slope you begin with only PE that is converted to KE as go down the hill. Assuming friction is negligible the TE will not change. At the bottom of the hill you have only KE which means you are moving FAST. In order to slow down you have to dissipate this kinetic energy. This requires a change in TE. By the work-energy relationship, work must be performed on the skier to bring him/her to a stop. To perform work you need a force. In this case it is the friction developed between the skier and some unpacked snow at the end of the run. AAAHHHGGG – I’M BACK IN DRIVER’S ED! Three people are driving the same type of car. Driver A is traveling at 10 mph. Driver B is traveling at 20 mph. Driver C is traveling at 30 mph. What can you say about the relationship of the stopping distance between vehicles? (Is it linear or something else?) The cars have no change in PE so only KE must be changed. Thus the work necessary to bring the cars to rest is equal to the change in KE: W = KE SF * d = ½ m(v)2 So the stopping distance (d) is proportional to the square of velocity d a v2 Kinetic Energy to Strain Energy • Kinetic energy will be used to deform the elastic tissues. Strain energy can then be transformed back into kinetic energy during the pushoff phase. • At low speeds, kangaroos use a pentapedal form of locomotion (four feet and a tail). • At 6-7 km/hr the switch to hopping. • The rate of O2 consumption increases sharply with speed during pentapedal locomotion. • When they begin to hop, the rate of O2 consumption decreases with increasing speed until 18 km/hr. Why? Dawson and Taylor, 1973 Special Force Application Kinetic to Strain to Kinetic • Kangaroos have large Achilles tendons (1.5 cm in diameter and 35 cm in length). • It is possible that a large amount of strain energy is stored in the tendon during the landing and converted back into kinetic energy as the kangaroo rebounds into the air. Special Force Application Power • Power is the rate of performing work. P= W t P = F*s t P=F*s t P = F*v Special Force Application Power • Units are Joules/second or Watts. • Greater power must be developed in order to do mechanical work more quickly. • Power can be positive or negative depending on whether F and v point in same general direction (+ power) or in opposite directions (- power) Special Force Application Power • Positive power indicates that energy is being generated and negative power indicates that energy is being absorbed thus: • Positive power is associated with concentric muscular contractions, while negative power is associated with eccentric muscular contractions. Special Force Application -1 Power (Watts·kg ) Knee Power During Running 15 10 Energy Generation 5 0 Energy Absorption -5 -10 -15 -20 0 10 20 30 40 50 60 70 80 90 100 Percent of Stance Special Force Application Power Example m = 100 kg, g = 9.8 m/s2, h = 2 m W = mgh = 100 (9.8) 2 = 1960 J now add time Case 1: raise the barbell slowly -- t = 5 s P W 1960 J 392 W t 5s Case 2: raise the barbell quickly -- t = 1.5 s P W 1960 J 1307 W t 1.5 s Special Force Application