Newtons Laws Session 2 POST - Part 2 PDF

Movement Science 1: Session II Newton’s Laws Adapted with permission from Dr. Cole, George Washington University Program in Physical Therapy, Composition and Resolution of Forces  Composition of Forces A Combining t...

Movement Science 1: Session II Newton’s Laws Adapted with permission from Dr. Cole, George Washington University Program in Physical Therapy, Composition and Resolution of Forces  Composition of Forces A Combining two or more forces to determine net effect Linear forces are added together (if in line) or subtracted, if in opposite direction B Composition of Forces  Composition of Non-Parallel Forces Two or more forces acting on an object at the same time a b  Composition of Non-Parallel Forces Polygon Method Parallelogram Method Tip to tail Each vector drawn from same Draw Resultant force point. Draw sides of parallelogram Draw R a R b R b b a a Composition of Forces that are Non-Paralell  Resolution of Forces Replacing a single force with two or more components. When combined will add to the original force Usually resolved into vertical (Fy) and horizontal (Fx)Forces For our purposes every force (F) will have: –X component (Fx) –Y component (Fy) –Sometimes one of these (Fx or Fy) may be 0 Free Body Diagrams  Introduction to Freebody Diagrams 1) Identify and isolate the free body under consideration 2) Establish a coordinate reference frame 3) Draw the internal and external forces that at on the system 4) Draw the joint reaction force 5) Write the governing equations of motion –Σ = 0; –Σ = 0; –Σ = 0.  You are working with a ballerina who has with hip flexor weakness after an injury. They are working on returning to straight leg kicks and pirouettes. The weight of their upper leg is 50 N and the moment arm of the center of mass of their upper leg is 0.2m from their hip axis of rotation. The weight of their lower leg is 30 N and the moment arm of the center of mass of their lower leg is 0.50m from their hip axis of rotation. Assume that the moment arm of the hip flexors is 0.05 meters from the hip, and that the hip flexors insert into the femur at a 40 deg angle. Calculate the hip flexor force that is needed to keep the hip still at 90 degrees with the knee extended and calculate the joint reaction force at the hip. Assume the hip flexors are the only muscles creating a hip flexion torque. Follow the steps of setting up free body diagrams. Where the x-axis is horizontal: = *0.77; = *0.64; J = sqrt(Jx^2 +Jy^2)  Introduction to Freebody Diagrams 1) Identify and isolate the free body under consideration Leg Hip  Introduction to Freebody Diagrams 2) Establish a coordinate reference frame y x  Introduction to Freebody Diagrams 3) Draw the internal and external forces that at on the system y 𝐹𝑚 𝑦 =? x 𝐹𝑚 𝑥 =30N 0.05 m = 50N 0.2 m 0.5 m  Introduction to Freebody Diagrams 4) Draw the joint reaction force =? y 𝐹𝑚 𝑦 =? x 𝐹𝑚 𝑥 =30N 0.05 m = 50N 0.2 m 0.5 m  Introduction to Freebody Diagrams Torque from is zero because by definition it 5) Write the governing equations of motion goes through the axis of rotation =? Σ = 0 and = Fd y 𝐹𝑚 0 = - (0.05m) + (0.2m) + (0.5m) 𝑦 =? 0 = - (0.05m) + (0.2m) + (0.5m) x (0.05m)=25 = 500N 𝐹𝑚 𝑥 Solve for using given info, = *0.64 0.05 = (0.64) =30N m 500= (0.64) = 50N =781.25N Solve forusing given info, = (0.77) 0.2 = (0.77) m = (0.77) = 601.6N 0.5 m  Introduction to Freebody Diagrams 5) Write the governing equations of motion We know Σ = 0 0=-50N-30N+500N+ =? y 𝐹𝑚 𝑦 =? x We know Σ = 0 =-601.6N 0=601.6N+ 𝐹𝑚 𝑥 =30N 0.05 = m =733.7N = 50N 0.2 m 0.5 m Tipping vs. Slipping  Friction Friction: Resistive force that opposes motion between two surfaces in contact Coefficient of friction (μ): describes how “sticky” a surface is The force of friction is equal to the coefficient of friction times the normal force –Ffriction = μ* FNormal  Demonstration Let’s push someone around again….  Slipping Force Box will begin to slip when Fpush > Ffriction 1. Draw all forces acting on box (push, gravity, normal force, friction) 2. Since the box isn’t accelerating in the vertical direction, FN = Fg. We also know Fg = mg 3. Ffriction =μ* FN so Ffriction =μ*mg Once Fpush > μ*mg the box will begin to slide  Tipping Force Ffriction and FNormal do not create torque, since they go through the axis of rotation. Therefore, the torque created by the force we are pushing with must be greater than the torque created by gravity 1. Draw all forces acting on box 2. Calculate torque due to push about corner of box τpush = Fd τpush = Fpush *h 3. Calculate torque due to gravity about corner of box τgravity = Fd τgravity = Fg*x τgravity = mg*x 4. If τpush > τgravity the box will begin to tip over or when Fpush *h > mg*x or Fpush > Do you want a high or low coef Friction?  BOS How are you going to guard your falling patient? Impulse-Momentum  Momentum Momentum (M) –The quantity of motion possessed by an object –Any object that has both mass and velocity has a momentum –M = m*v Where m = mass and v = velocity Units = Kg*m/s or Ns –Uses Measurement in describing impact/collision between people, between objects, and between a person and an object  Conservation of Momentum “The total momentum of any given system will remain constant unless acted upon by an external force” “The momentum before a collision is equal to the momentum after a collision” Bowling: – (mass ball)(velocity rolling ball) + (mass pins)(velocity stationary pins) 3:0 6 = (mass ball)(velocity slower roll) + (mass pins)(velocity flying pins) Learn more about momentum: https://www.youtube.com/watch? v=yhTz_6NFmV0  Impulse-Momentum Relationship Impulse: The quantity of net force and the time over which the force was applied. –ΣF t The impulse applied to an object is equal to its change in momentum –ΣF t = Δmv –ΣF = Δmv/t –Therefore, the net force required to change the original momentum is just the change in momentum over time.  Impulse-Momentum: Application In terms of impulse-momentum, explain how following through with a golf drive helps improve the players shot  Impulse-Momentrum Calculations A 50 kg gymnast dismounts from the vault and sticks her landing. If she impacted the ground at a speed of –3.75 m/s, how would landing technique affect the forces she experienced in coming to a stop? –Hard landing: dissipating a large force over short time (∆t= 0.20 s) –Soft landing: dissipating a large force over long time (∆t= 0.60 s) How do airbags help reduce the severity of many automobile  Impulse Momentum: Discussion Questions 1. Your patient is a 60 year old male that has hemiplegia due to a cerebral infarction that happened 6 months ago. Due to factors unrelated to the stroke, he also has 4+ strength of the glutes and quads on the “unaffected” side. What strategies would you use to help this person perform sit to stand without assistance (or with only CGA)? 2. How does the impulse-momentum relationship influence vertical jumping performance? 3. How is landing technique related to lower extremity injury? 4. Think of one additional example of where the impulse-momentum may aid a patient (sport, pathology, all fair game) Work, Energy, Power  Work Mechanical work is the product of the force applied to an object and the displacement of the object in the direction of the force. – Work = force x distance – Unit: Joule (J) When the force is not parallel to the distance traveled, we have to use trig If the force and displacement are in the same direction, the work done is positive If the force and displacement are in opposite directions, the work done is negative What is the (mechanical) work performed during an isometric muscle contraction? Eccentric? Concentric?  Energy and Energy Transfer Energy is the capacity to do work. We are most interested in mechanical energy here (not so fair to the muscle). Kinetic Energy – “Energy in motion” – KE = ½ mv2 Potential energy – Energy due to position within a gravitational field – PE = mgh Elastic energy (strain) – EE = ½ kx2 – Where k is the stiffness and x is the amount of deformation – Eg. Muscle/tendon stretched or spring compressed Energy can be converted from one form to another  Transfer of Energy Problems How might this gymnast use concepts of energy during a vault? How does a pitcher transfer energy to throw a pitch as fast as possible?  Work-Energy Relationship The work done by the net force acting on a body is equal to the change in the body’s energy: – Work = Δ energy In other words, work is the process of transferring energy to or from an object – If thebywork a force applied over a distance. is positive, it increases the object’s energy. Example: Jumping from a crouched position. The body starts at rest (kinetic energy =0). The muscles in the legs do work by applying a force over a distance (extension of the legs), to increase the body’s velocity and kinetic energy – If the work is negative, it decreases the object’s kinetic energy Example: A box sliding along a “sticky” floor. The box is initially moving fast (high kinetic energy). Friction does work by ti o Fric applying a force over a distance, in the opposite direction the n box is moving, decreasing the speed of the box and decreasing it’s kinetic energy.  Work-Energy Problems A 60kg diver leaves the 10 m board with zero vertical velocity. Compute the potential and kinetic energies at 10 m, 5 m and 0 m above the water. How much work was performed on the diver if the water slowed the diver’s speed by 50% upon entry?  Work-Energy Problems A baseball catcher “gives” to stop a 22 m/s fastball over a distance of 0.25 m. – What was the average force needed to stop the 0.2 kg baseball? – What is the average force needed to stop the same 22 m/s fastball over 0.125 m?  Power Power is the rate at which work is done or energy is transferred within a system – Power = – Power = Force * velocity (if the force is applied to move an object at a constant velocity) – Units: Joule/sec = Watt Power is a measure of how quickly a force can move an object or how quickly energy can be delivered to or extracted from a system. Example: Mechanical power is the work done by a person to lift a weight, divided by the time it takes to lift it. – If the person has more power, they lift the weight quickly – If the person has less power, they will have to lift the weight slowly  Power problems A weight lifter is doing a clean and jerk with 100kg. He is lifting the weight from 0m to 2m. What is the power required to raise a barbel slowly, taking 5s versus lifting the weight quickly, taking 1.5s. How might reduced power affect your patient’s ability to cross the street?  Base of Support We know the box will tip if Fpush > What makes the box more stable/harder to tip over? –Increase width (base of support) –Add weight –Apply push lower down Do you want a high or low coef Friction? push > μ*mg the object will Once F begin to slide Once Fpush > the object will begin to tip Which ever condition is met first, will occur: Really sticky, will tip Really wide, will slide Push high up, will tip Push really low, will slide  Impulse-Momentum: Application In terms of impulse-momentum, explain how following through with a golf drive helps improve the players shot. Following through keeps the club in contact with the ball front a longer period of time. This increases the impulse (ΣF t) applied to the ball by increasing the time the force is applied. This leads to a greater change in the balls momentum which means a greater change in speed.  Impulse-Momentrum Calculations A 50 kg gymnast dismounts from the vault and sticks her landing. If she impacted the ground at a speed of –3.75 m/s, how would landing technique affect the forces she experienced in coming to a stop? –Hard landing: dissipating a large force over short time (∆t= 0.20 s) ΣF = Δmv/t ΣF = (50kg)*(0-3.75 m/s) / (0.2 s) ΣF = 937.5 N –Soft landing: dissipating a large force over long time (∆t= 0.60 s) ΣF = Δmv/t ΣF = (50kg)*(0-3.75 m/s) / (0.6 s) ΣF = 312.5 N How do airbags help reduce the severity of many automobile  Impulse Momentum: Discussion Questions 1. Your patient is a 60 year old male that has hemiplegia due to a cerebral infarction that happened 6 months ago. Due to factors unrelated to the stroke, he also has 4+ strength of the glutes and quads on the “unaffected” side. What strategies would you use to help this person perform sit to stand without assistance (or with only CGA)? Slow and Steady. Exert force for longer period of time 2. How does the impulse-momentum relationship influence vertical jumping performance? Deeper counter jump leads to a higher jump because it increases the contact time with the ground and therefore increases the amount of time the force was applied to the ground. This causes a larger change in momentum and also velocity 2. How is landing technique related to lower extremity injury? Longer landing time via a deep squat increases the time the decelerating force from the floor is applied. The floor applies a smaller force over a longer period of time. Smaller forces are less likely to cause injuries 4. Think of one additional example of where the impulse-momentum may aid a patient (sport, pathology, all fair game)  Work Mechanical work is the product of the force applied to an object and the displacement of the object in the direction of the force. – Work = force x distance – Unit: Joule (J) When the force is not parallel to the distance traveled, we have to use trig If the force and displacement are in the same direction, the work done is positive If the force and displacement are in opposite directions, the work done is negative What is the (mechanical) work performed during an isometric muscle contraction? (zero) Eccentric (negative)? Concentric (positive)?  Transfer of Energy Problems How might this gymnast use concepts of energy during a vault? Maximizes kinetic energy during their approach by running as fast as they can When the y plant their feet on the springboard, their kinetic energy is transferred into elastic energy, stored in the springboard The spring board release, using the elastic energy to propel the gymnast up While flying up, the gymnast is converting kinetic energy into potential How does energy a pitcher transfer energy to throw a pitch as fast as possible? During Stretch descent, potential phase: during energyphase the cocking is converted into kinetic of the throw, energy the pitcher externally rotates and extends his shoulder, stretching the pec major, latissimus dorsi and subscapularis. The stretch stores elastic energy in the muscle-tendon unit Recoil phase: As the arm moves forward to throw, the elastic energy is release, contributing to the explosive internal rotation and adduction of the shoulder  Work-Energy Problems A 60kg diver leaves the 10 m board with zero vertical velocity. Compute the potential and kinetic energies at 10 m, 5 m and 0 m above the water. How much work was performed on the diver if the water slowed the diver’s speed by 50% upon entry? – At 10m: – Velocity at entry: – KE after it slows down: – PE=mgh=60*9.8*10 = 5,880 – 5,880 J = 1/2mv2 – KE=1/2mv2 J – 5,880 = ½(60)v2 – KE=1/2(60)72 – KE=0.5mv2 = 0.5(60)(0)2=0 J – V=14 m/s – KE= 1470 J – At 5m: – Work = Δenergy – PE=mgh=60*9.8*5 = 2940 J – Velocity after water slows – Work=5888 J – 1470 J – So, KE=5,880 J – 2940 J = them down – Work = 4410 J 2940 J – 14 m/s*0.5 = 7 m/s – At 0m: – PE=mgh=60*9.8*0 = 0 J – So, KE = 5,880 J – 0 J =  Work-Energy Problems A baseball catcher “gives” to stop a 22 m/s fastball over a distance of 0.25 m. – What was the average force needed to stop the 0.2 kg baseball? – What is the average force needed to stop the same 22 m/s fastball over 0.125 m? –For d=0.25: –Work = –For d=0.125: –KE before player stops Δenergy – Work = Fd ball: –Work = 48.4 J –48.4 J = –KE=1/2mv2 F(0.125m) –KE=1/2(0.2)(22)2 – Work = Fdf –F= 387.2 N –KE=48.4 J –48.4 J = –KE after player stops ball F(0.25m) –F= 193.6 N –KE = 0 J  Power problems A weight lifter is doing a clean and jerk with 100kg. He is lifting the weight from 0m to 2m. What is the power required to raise a barbel slowly, taking 5s versus lifting the weight quickly, taking 1.5s. – So m = 100 kg, g = 9.8 m/s, h = 2 m End: mgh = 100(9.8)(2) =1960 J Start: mgh = 0 J Work = Δenergy = 1960 – 0 = 1960 J Raise the barbell slowly: 5s: Power = Work/time= 1960J/ 5s = 392 W Raise the barbell quickly: 1.5s Power = Work/time = 1960J/1.5s = 1306.7 W How might reduced power affect your patient’s ability to cross the street? If someone’s muscles lack power, they might not be able to cross the street in time allotted at crosswalks. Mechanical power is the product of force and velocity, Power = force*velocity. Even if they have enough strength, if power is low, velocity will be slow. Another way to think about it, power is work divided by time, Power =. If power is low, time required will be high.