2. Advanced Programming for 3D Applications CE00383-3 Bob Hobbs Staffordshire university Forces in Human Motion Lecture 4

3. Modelling forces

4. 3 Types of force Applied LoadF Applied LoadF TensionF T TensionF T Contact: F c = kx Contact: F c = kx Friction:F f =  N Friction:F f =  N Bouyancy:F b =  Vg Bouyancy:F b =  Vg Drag:F d = AC d v Drag:F d = AC d v Applied LoadF TensionF T Contact: F c = kx Friction:F f =  N Bouyancy:F b =  Vg Drag:F d = AC d v

5. 4 Applied loads Normally act on one point Normally act on one point Typically Typically –Objects hanging from ropes –Objects being pushed –Objects being pulled F

6. 5 Example 1 A mass of 25 kg is hung from a bar by a hanger. Determine the applied load? +y 25 kg F = (m*g) = (25 * -9.81) = -245.25 N

7. 6 Tension The internal force acting within a The internal force acting within a –rope –cable –wire –component in a framework T Always PULLING Always +ve

8. 7 Example 2 A mass of 25 kg is hung from a bar by a hanger. Determine the tension in the hanger? +y 25 kg F T = (m*g) = (25 * -9.81) = 245.25 N 25 kg FTFT

9. 8 Contact Whenever two bodies touch each other there is a contact force Whenever two bodies touch each other there is a contact force –Always normal to the surface –A REACTION force –Sometimes modelled using stiffness R

10. 9 Example 3 A block of mass 75 kg is placed on a floor. What is the reaction force? +y Block “pushing” on the floor Floor pushing on the block Applied force, F = mg = (75 * -9.81) = -735.75 N Reaction force = -F = 735.75 N

11. 10 Friction Another force associated with contact Another force associated with contact –Parallel with the surface –Opposes motion –F f =  N

12. 11 Example 4 What is the maximum friction force exertable when a car, of mass 2000 kg, has been fitted with tyres of  =0.3? F = mg = 2000 * -9.81 = -20.36 kN +y Reaction, N = -F = 20.36 kN Using F f =  N = 0.3 * 20.36x10 3 = 6.1 kN

13. 12  ? ? ? ? Coefficient of friction Coefficient of friction  Block starts to slide when gravity overcomes friction FfFf  = tan(  )

14. 13 Example 5 A block is placed on a sliding table. The block begins to slide when the table reaches an angle of 25 o. Determine the coefficient of friction. Using  = tan(  ) = tan(25) = 0.466

15. 14 Stiffness All objects act as springs All objects act as springs The restoring force is proportional to deflection The restoring force is proportional to deflection x Force pulling the spring Spring pulling on the object F s = k*x

16. 15 Stiffness coefficient deflection, x (m) Applied force, F (N) + + + + + + + + + + F  x F = k*x k =  F/  x (N/m)

17. 16 Example 6 A spring is subjected to an applied load of 750 N. The spring extends by 25 mm. Determine the stiffness of the spring. Unloaded spring 750 N 25 mm Using k =  F/  x = 750 / 0.025 = 30 kN/m

18. 17 Stiffness A spring under Tension PULLS A spring under Tension PULLS A spring under Compression PUSHES A spring under Compression PUSHES

19. 18 String, Cables and Ropes They can only PULL They can only PULL You may assume constant tension throughout. You may assume constant tension throughout.

20. 19 Dampers (Viscous) Body moves with velocity v Opposing force generated proportional to v I.e. F = cv where c is the damping coefficient

21. 20 Impact forces F = constant; F =  p/t Triangular form F = 2  p/t Sine wave F =  /2  p/t  p = change in momentum General Area =  p

22. 21 Compression- Tension Cycle

23. 22 Ground Reaction Force Newton's Law of Gravitation: Newton's Law of Gravitation: –any two objects with masses attract each other and the magnitude of this attracting force is proportional to the product of the masses and inversely proportional to the square of the distance. This also holds for the gravitation between the earth and an object on the earth. The gravitational force acted upon an object by the earth is called gravity or weight of the object. Newton's Law of Reaction: Newton's Law of Reaction: –there is an equal and opposite reaction to every action. In other words, the action to the ground is always accompanied by a reaction from it. This reaction force from the ground is called the ground reaction force (R). The ground reaction force is an important external force acting upon the human body in motion. We use this force as propulsion to initiate and to control the movement. Newton's Law of Acceleration: Newton's Law of Acceleration: –the external forces acting on the body causes an acceleration:

24. 23 Ground Reaction Force F = m·a [1] where F = sum of external forces (vector), m = mass of the subject (scalar), and a = acceleration (vector) of the subject's centre of mass (CM). F = R + W [2a] Here, note that weight W always acts downward. Considering only the vertical forces and their directions: F z = R z - W [2b] where F z = net vertical force acting on the body, R z = the vertical ground reaction force, and W = weight of the body. From [1] and [2b]: R z - W = ma z [2c] where a z = magnitude of the vertical acceleration. Rearranging [2c], we obtain a z = ( R z - W ) / m [3]

25. 24 Impulse and Momentum During step sequence During step sequence –Average net Force = mass * average acceleration –Acceleration = change of velocity Force over a period of time causes a change of momentum over that time. Force over a period of time causes a change of momentum over that time. –Average net force over a period of time = m(Vf – Vi) Therefore changing momentum over a short period of time generates a greater force than over a longer period of time Therefore changing momentum over a short period of time generates a greater force than over a longer period of time –When landing from a height the start and finish velocities are the same - as is the mass – whether you land stiff-legged or ‘bounce’

26. 25 Impact The first peak is called the impact peak (P1) while the second is called the propulsion peak (P2). The impact peak is associated with the impact of the foot to the ground during early foot contact phase. The propulsion peak is associated with the propulsion of the body forward. It has always been the main focus of the shoe engineers that how to design the shoe-sole to reduce the impact peak while maintaining the propulsive characteristics. The first peak is called the impact peak (P1) while the second is called the propulsion peak (P2). The impact peak is associated with the impact of the foot to the ground during early foot contact phase. The propulsion peak is associated with the propulsion of the body forward. It has always been the main focus of the shoe engineers that how to design the shoe-sole to reduce the impact peak while maintaining the propulsive characteristics.

27. 26 Reduce impact by 1.Cushioning 2.Alter gait