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Energy & Work (Cont.)

Energy & Work (Cont.). 4 Basic Simple Machines. Levers Inclined planes Pulleys Wheel and axle. 490 N. =. 1.66. 294 N. Force w/o. =. Force w. Length of plane. 5 m. =. Height of plane. 3 m. How much easier is it to do the work (1470J) with the inclined plane than without?.

travis-allison
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Energy & Work (Cont.)

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  1. Energy & Work (Cont.)

  2. 4 Basic Simple Machines • Levers • Inclined planes • Pulleys • Wheel and axle

  3. 490 N = 1.66 294 N Force w/o = Force w Length of plane 5 m = Height of plane 3 m How much easier is it to do the work (1470J) with the inclined plane than without? Mechanical advantage = force without incline / force with incline For inclined planes you get the same number by: = 1.66 3 m 5 m

  4. Effort arm Fe Resistance arm Fr dr de Fulcrum 1st Class Lever

  5. The work done on one side of the lever must equal or exceed the work performed by the other side. • Fe*de ≥ Fr*dr Fe Effort arm Resistance arm Fr de dr Fulcrum

  6. Fr 150 kg Fe If we have a 150 kg mass and we need to life it 1 m, and we have a 1st class lever that is 3 m on the other end, what is the minimum force necessary to lift the weight? F = m*a = (150 kg)(9.8 m/s2) = 1470 kg m / s2 W = F*d = (1470 kg m / s2)(1m) = 1470 kg m2 / s2 = 1470 Joules (J) Recall : We ≥ Wr F = W/d F = (1470 kg*m2/s2)(3 m) = 490 kg*m/s2 3m de = 490 N 1m dr

  7. Fout*dout Fin = din m*a*dout (150 kg*9.8m/s2)(1 m) Fin = = din 3 m Alternative way to look at the problem: For all simple machines: Win = Wout in = effort out = resistance But W= F*d, so: Fin*din = Fout*dout If you know any 3 of these, you can solve for the 4th But Fout=m*a; = 490 kg*m/s2 = 490 N

  8. Mechanical Advantage • It is a measure of how much easier or harder it is to do work with a machine • F out / F in • In the previous example = 1470 N / 490 N = 3 • For levers, a quick way to determine mechanical advantage is: • MA = Effort arm / Resistance arm

  9. R E F out F in d e What is the mechanical advantage of the lever system? in = effort out = resistance 5 kg 15 kg 15 kg * 9.8 m / s2 = 5 kg * 9.8 m / s2

  10. 2nd Class Lever Fr Resistance arm Fe de dr Effort arm Fulcrum

  11. Fr Fe Effort arm dr de Resistance arm Fulcrum 3rd Class Lever

  12. Ulna Radius Olecranon Process Humerus

  13. Fr Fe Effort arm dr de Resistance arm Fulcrum If I am lifting a 25 pound weight with my biceps, what is the mechanical advantage, and how much force (minimum) must the biceps apply? What do we need to know? It is about 5 cm from elbow to point of attachment of muscle and the whole bone is about 33 cm Recall: mechanical advantage is effort arm/resist. arm Therefore, 5 cm/ 3 cm = 0.15 Recall We = Wr Fede = Frdr Fe = Frdr/de Fe = 25lb.*33cm / 5 cm = 165 lbs.

  14. W=F*d W=F*d W=F*d W=F*d Pulleys • Fixed pulleys only change the direction of a force • They provide no mechanical advantage

  15. W=F*d W=F*d W=F*d W=F*d Moveable pulleys supply mechanical advantage to the number of lines attached to the mass If the box weights 100 kg, how much force is needed to lift it without the pulley? F = ma = (100 kg)(9.8 m/s2) = 980 N How much force would you have to use using the pulley? For all simple machines: Win = Wout But W= F*d, so: Fin*din = Fout*dout 980N*d = F*2*d 980N / 2 = F 490 N = F F = 490 N

  16. rwheel raxle Axle Wheel The Wheel and Axle • As in all simple machine, Fd = Fd • Circumference of axle = d = 2πraxle • Circumference of wheel = d = 2πrwheel • Therefore, Faxleraxle = Fwheelrwheel • A small force on the wheel translates into a large force on the axle • A slow velocity on the axle becomes a fast velocity on the wheel

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