Question #163591

Compare and contrast the use of D’Alembert’s principle with the principle of conservation of energy when solving, A pile driver hammer of mass 140 kg falls freely through a distance of 4.5 m to strike a pile of mass 390 kg and drives it 70 mm into the ground. The hammer does not rebound when driving the pile. Determine the average resistance of the ground.


1
Expert's answer
2021-02-24T07:06:06-0500

The potential energy of the pile-driver is converted into kinetic energy. Thus


potential energy=kinetic energy,\text{potential energy} = \text{kinetic energy,}mgH=mv22mgH=\dfrac{mv^2}{2}

Find velocity of pile immediately after impact using principle of conservation of momentum


mv=(M+m)umv=(M+m)u

u=mM+mvu=\dfrac{m}{M+m}v

The pile-driver and pile together have a mass (m+M)(m+M) and possess kinetic energy (M+m)u22.\dfrac{(M+m)u^2}{2}.

The change in potential energy of the driver as it moves in the pile and the pile as it moves through the ground


(M+m)g(0h)(M+m)g(0-h)


Let R=R= the average resistance of the ground. Then


work done=R×h\text{work done}=R\times h

By the principle of conservation of energy


(M+m)u22=(M+m)g(0h)+R×h\dfrac{(M+m)u^2}{2}=(M+m)g(0-h)+R\times h

R=(M+m)g+(M+m)u22hR=(M+m)g+\dfrac{(M+m)u^2}{2h}

R=(M+m)g+m2v22h(M+m)R=(M+m)g+\dfrac{m^2v^2}{2h(M+m)}

R=(M+m)g+m2gHh(M+m)R=(M+m)g+\dfrac{m^2gH}{h(M+m)}

(140kg+390kg)(9.81m/s2)+(140kg)2(9.81m/s2)(4.5m)0.07m(140kg+390kg)(140kg+390kg)(9.81m/s^2)+\dfrac{(140kg)^2(9.81m/s^2)(4.5m)}{0.07m(140kg+390kg)}

R=28521NR=28521N



The potential energy of the pile-driver is converted into kinetic energy. Thus


potential energy=kinetic energy,\text{potential energy} = \text{kinetic energy,}mgH=mv22mgH=\dfrac{mv^2}{2}v=2gHv=\sqrt{2gH}

Find velocity of pile immediately after impact using principle of conservation of momentum


mv=(M+m)umv=(M+m)u

u=mM+mvu=\dfrac{m}{M+m}v

Find acceleration of pile


02u2=2ah0^2-u^2=2ah

a=u22ha=-\dfrac{u^2}{2h}

Find the resistance of the ground RR


(M+m)gR=(M+m)a(M+m)g-R=(M+m)a

R=(M+m)(ga)R=(M+m)(g-a)

R=(M+m)(g+u22h)R=(M+m)(g+\dfrac{u^2}{2h})

R=(M+m)(g+m2v22h(M+m)2)R=(M+m)(g+\dfrac{m^2v^2}{2h(M+m)^2})

R=(M+m)g(1+m2Hh(M+m)2)R=(M+m)g(1+\dfrac{m^2H}{h(M+m)^2})

(390kg+140kg)(9.81m/s2)(1+(140kg)2(4.5m)0.07m(390kg+140kg)2)(390kg+140kg)(9.81m/s^2)(1+\dfrac{(140kg)^2(4.5m)}{0.07m(390kg+140kg)^2})

R=28521NR=28521N

The average resistance of the ground is 28.521 kN.


In comparing the two methods, it might be concluded that in this particular case there is very little to choose between them. Both methods required use of the principle of conservation of momentum to determine the value of velocity immediately after impact of the hammer. Both methods return identical answers with the same degree of accuracy.

We see that the kinetic energy of the system at impact is not conserved.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS