25. A sledgehammer is used to drive a wedge into a log to split it. When the wed

Question

25. A sledgehammer is used to drive a wedge into a log to split it. When the wedge is driven 0.20 m into the log, the log is separated a distance of 5.0 cm. A force of 1.7×104 N is needed to split the log, and the sledgehammer exerts a force of 1.1×104 N. a. What is the IMA of the wedget? b. What is the MA of the wedge? c. Calculate the efficiency of the wedge as a machine 26. A worker uses a pulley system to raise a 24.0-kg carton 16.5 m, as shown in Figure 10-14. A force of 129 N is exerted, and the rope is pulled 33.0 m. a. What is the MA of the pulley system? b. What is the efficiency of the system? 27. You exert a force of 225 N on a lever to raise a 1.25×103-N rock a distance of 13 cm. If the efficiency of the lever is 88.7 percent, how far did you move your end of the lever? 28. A winch has a crank with a 45-cm radius. A rope is wrapped around a drum with a 7.5-cm radius. One revolution of the crank turns the drum one revolution. a. What is the ideal mechanical advantage of this machine? b. If, due to friction, the machine is only 75 percent efficient how much force would have to be exerted on the handle of the crank to exert 750 N of force on the rope?

Explanation / Answer

25.

a. IMA of the wedge :-

IMA = de / dr = 2x101cm / 5cm = 4

b. MA of the wedge:-

MA = Fr / Fe = 1.7 x 104 N / 1.1 x 104 N = 1.545

c. efficiency of the wedge as a machine:-

efficiency = (MA / IMA).100% = (1.5 / 4) .100% = 37.5%

26.

A) for an ideal system the mechanical advantage is defined as =distance over which effort isapplied / distance which the load is moved

In this case, theeffort is applied over 33.0 m, and the load is moved 16.5 m,then mechanical advantage is 33/16.5 =2.

B) Efficiency is defined as = work output / work input . worker exerts 129 N over 33.0 M,

so he does 129 * 33 = 4257 J of work.

load weighs 24 kg and was raised 12.5m, so gained 24 kg * 9.8m/s^2 * 12.5 m = 2940 J of energy.

Thus, the efficiency is 2940/4257= .69 = 69%.

27.

efficiency = (W0 / Wi) . 100

                = (Frdr / Fede) .100

So , de = Frdr .100 / Fe efficiency = (1.25 x 103 N)(0.13 m)(100) / (225 N)(88.7%)

         = 0.81m

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