Ejercicios Tippens Cap 12

Chapter 12. Simple Machines

Simple Machines, Efficiency, and Mechanical Advantage

12-1. A 25 percent efficient machine performs external work of 200 J. What input work is required?

; Work in = 800 J

12-2. What is the input work of a 30 percent efficient gasoline engine if during each cycle it performs 400 J of useful work?

; Work in = 1333 J

12-3. A 60-W motor lifts a 2-kg mass to a height of 4 m in 3 s. Compute the output power.

; P = 26.1 W

12.4. What is the efficiency of the motor in Problem 12-3? What is the rate at which work is done against friction?

; E = 43.5%

Ploss = 60 W – 26.1 W; Ploss = 33.9 W

12-5. A 60 percent efficient machine lifts a 10-kg mass at a constant speed of 3 m/s. What is the required input power?

Pout = Fv = mgv = (10 kg)(9.8 m/s2)(3 m/s); Pout = 294 W

; Pin = 490 W

12-6. During the operation of a 300-hp engine, energy is lost to friction at the rate of 200 hp. What is the useful output power and what is the efficiency of the engine?

Pout = 300 hp – 200 hp = 100 hp ; ; E = 33.3%

12-7. A frictionless machine lifts a 200-lb load through a vertical distance of 10 ft. The input force moves through a distance of 300 ft. What is the ideal mechanical advantage of the machine? What is the magnitude of the input force?

= 30 ; = 6.67 lb

Applications of the Lever Principle

12-8. One edge of a 50-kg safe is lifted with a 1.2-m steel rod. What input force is required at the end of the rod if a fulcrum is placed 12 cm from the safe? (Hint: To lift one edge, a force equal to one-half the weight of the safe is required.)

W = mg = (50 kg)(9.8 m/s2) = 490 N; F = W/2 = 245 N

Fi = 54.4 N

12-9. For the nutcracker in Fig. 12-4a, the nut is located 2 cm from the fulcrum and an input force of 20-N is applied at the handles which are 10 cm from the fulcrum. What is the force applied to crack the nut?

;

Fo = 100 N

12-10. For the wheelbarrow in Fig. 12-4b,

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