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A pile driver hammer of mass 150 kg falls freely through a distance of 5m to strike a pile of mass 400kg and drives it 0.075m into the ground. The hammer does not rebound when driving the pile. Determine the average resistance of the ground.

 

Thanks in advance.smiley

 Jan 16, 2017
 #1
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mgh = energy of hammer delivered to pile = 150*9.8*5 = 7350 j

Amount delivered to the pile  400kg x 9.8 x .0075=400*9.8*.0075 = 29.4 j

The difference is delivered to the ground (vibrations, ground movement, heat of friction etc) and is the resistance (?)

7350-29.4 = 7320.6 j

 

(a little unsure)

 Jan 16, 2017
 #2
avatar+2234 
+10

Time to test my genetic enhancement.

 

Velocity of hammer at impact with pileVf=2ad|29.81m/s25m=9.904568m/s

 

m1u1+m2u2=m1v1+m2v2conservation of momentum150kg(9.905m/s)=(150+400)kgmass of pile & driver(v)velocity of combined massesv=150kg(9.905m/s)(150+400)kg=2.70136m/s

 

 

 

Energies:kinetic energy=12(550kg)(02.70136m/s)2=2006.77Jpotential energy=(550kg)(9.81m/s2)(00.075m)=404.66J

 

Energy to work formula:(-F)(0.075m)=2411.43J|(F)=negative Force(-F)=2411.43J0.075m=32152.40 newtons

 

Average resistance of the ground =32,152.40 newtons

No figs. Just newtons on this ground. 

 

This concludes the test of my genetic enhancement.

I think I passed.smiley

 Jan 17, 2017
edited by GingerAle  Jan 17, 2017
 #3
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OMG! Thank you so so much Gingerale

 Jan 18, 2017

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