1. Physics

2. PHS 5043 Forces & Energy Machines Machine: Device or set of devices used to accomplish a particular task Machines are used to:  Make our work easier  Multiply force  Multiply speed  Change direction of force

3. PHS 5043 Forces & Energy Machines Machines can be:  Simple  Compound (combination of simple machines) Mechanical advantage (simple): Efficiency of a simple machine Mechanical advantage (compound): Product of mechanical advantage of each simple machine

4. PHS 5043 Forces & Energy Machines Simple machines:  Levers  Wheel & axle  Winch  Pulley  Inclined plane

5. PHS 5043 Forces & Energy Machines Levers: Rigid bar that pivots around a fulcrum

6. PHS 5043 Forces & Energy Machines Law of levers: F e l e = F r l r F e : Effort force (N) F r : Resistance force (N) l e : Effort arm (m). Distance between effort and fulcrum l r : Resistance arm (m). Distance between load and fulcrum Mechanical advantage of levers: MA = F r / F e or MA = l e / l r

7. PHS 5043 Forces & Energy Machines Class 1 levers: _F e and F r point in the same direction _l e is larger than l r _F e and F r on opposite sides of fulcrum _MA > 1

8. PHS 5043 Forces & Energy Machines Practice: You place a child (m = 30 kg) on a seesaw, 1.5 m away from the fulcrum. Where should you sit so that the seesaw is perfectly balance, if your mass is 60 kg? F e l e = F r l r l e = F r l r / F e l e = F g l r / F’ g l e = mg l r / m’g l e = (30 kg)(9.8 N/kg)(1.5 m)/(60 kg)(9.8 N/kg) l e = 0.75 m (75 cm)

9. PHS 5043 Forces & Energy Machines Practice: How far would you sit for the seesaw to have a mechanical advantage of 2? What would then be the value of the effort force? MA = l e / l r l e = MA l r l e = 2 (1.5 m) l e = 3 m MA = F r / F e F e = F r / MA F e = F g / MA F e = mg / MA F e = (30 kg)(9.8 N/kg) / 2 F e = 147 N

10. PHS 5043 Forces & Energy Machines Class 2 levers: _F e and F r point in opposite directions _l e is larger than l r _F e and F r on same sides of fulcrum _MA > 1

11. PHS 5043 Forces & Energy Machines Practice: You transport a 100 kg load of sand in a wheelbarrow. If the effort arm extends 1.5 meters away from the fulcrum, and the resistance arm is fifty centimeters long: a) How much force must you apply to lift the load? b) What is the mechanical advantage of this machine? F e l e = F r l r F e = F r l r / l e F e = F g l r / l e F e = mg l r / l e F e = (100 kg)(9.8 N/kg)(0.5 m) / (1.5 m) F e = 327 N MA = F r / F e MA = 980 N / 327 N MA = 3

12. PHS 5043 Forces & Energy Machines Class 3 levers: _F e and F r point in opposite directions _l r is larger than l e _F e and F r on same sides of fulcrum _MA < 1

13. PHS 5043 Forces & Energy Machines Practice: Suppose that your forearm measures 30 cm, and your bicep muscle is attached 4 cm from your elbow (fulcrum). a)What is the MA of your forearm? b)What force must the bicep muscle exert to lift 1 kg? MA = MA = l e / l r MA = 4 cm / 30 cm MA = 0.13 MA = F r / F e MA = F g / F e MA = mg / F e F e = mg / MA F e = (1 kg) (9.8 N/kg) / 0.13 F e = 75 N

14. PHS 5043 Forces & Energy Machines The wheel & axle: _Simple machine _Similar to first class lever _F e applied to the wheel’s circumference _F r applied on the axle’s circumference _Fulcrum is the center of the axle _l e : radius of wheel (R) _l r : radius of axle (r) _MA > 1 (MA = F r /F e = l e /l r = R/r)

15. PHS 5043 Forces & Energy Machines The winch: _Simple machine _Family of wheel & axle _Similar to first class lever _F e applied to the handle _F r applied on the cylinder _Fulcrum is the center of the cylinder _l e : length of handle (R) _l r : radius of cylinder (r) _MA > 1 (MA = F r /F e = l e /l r = R/r) (see Fig. 6.20, page 6.23)

16. PHS 5043 Forces & Energy Machines Practice: A winch has a cylinder with diameter 15 cm and handle measuring 30 cm a)What is the MA of this machine? b)What force must be exerted to lift 10 kg? MA = MA = l e / l r = R / r MA = 30 cm / 7.5 cm MA = 4 MA = F r / F e MA = F g / F e MA = mg / F e F e = mg / MA F e = (10 kg) (9.8 N/kg) / 4 F e = 24.5 N

17. PHS 5043 Forces & Energy Machines The pulley: _Simple machine _Used to change direction of force _Used to reduce friction _Suspended pulley has no MA (MA = 1) _Movable pulley’s MA = # ropes around movable pulley _Tension is the same at all points of rope _F r weight applied on the pulley’s rope _F e tension on the pulley’s rope _Fulcrum is the center of the pulley _MA > or = 1 (MA = F r / F e = F g / T)

18. PHS 5043 Forces & Energy Machines Practice: According to the pulley system below: a) What is the MA of the system? b) What mass must be suspended from the fixed pulley for the system to be in equilibrium? c)What would then be the value of the effort force? MA = MA = (MA) f (MA) m MA = 1 * 2 MA = 2 MA = F r / F e MA = F g / T MA = mg / m’g MA = m / m’ m’ = m / MA m’ = 10 kg / 2 m’ = 5 kg Fe = T T = mg T = (5 kg) (9.8 N/kg) T = Fe = 49 N

19. PHS 5043 Forces & Energy Machines The inclined plane: _Simple machine _MA = F r / F e _MA = (1 / sin θ) _MA = l / h _l: length of plane (m) _h: height of plane (m) *The smaller the angle of the plane, the greater its mechanical advantage sin θ = h / l

20. PHS 5043 Forces & Energy Machines Practice: A builder wants to install an access ramp with mechanical advantage of 8, what should be the length if it is designed to reach a level of 75 cm high? What would be the angle of said ramp? MA = l / h l = MA * h l = 8 * 0.75 m l = 6m MA = 1 / sin θ sin θ = 1 / MA sin θ = 1 / 8 θ = 7.2°

21. PHS 5043 Forces & Energy Machines Compound machines: _ Combination of simple machines _MA = Product of all MA’s _MA = F r / F e

22. PHS 5043 Forces & Energy Machines Practice: A compound machine consists of a class 2 lever (l r = 30 cm, l e = 90 cm) and a winch (R = 20 cm, r = 10 cm) designed to lift an object of mass 100 kg. a)What is the MA of this device? b)What is the value of F e at the end of the handle? MA = (MA) l *(MA) w MA = (l e /l r )(R/r) MA = (90 cm / 30 cm)(20 cm / 10 cm) MA = (3)(2) MA = 6 MA = F r / F e F e = F r / MA F e = F g / MA F e = mg / MA F e = (100 kg)(9.8 N/kg) / 6 F e = 163N