Answer to Question #163591 in Math for a

Question

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.

Answer 1

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

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

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

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

$$u=\dfrac{m}{M+m}v$$

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

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

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

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

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

By the principle of conservation of energy

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

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

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

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

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

$$R=28521N$$

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

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

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

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

$$u=\dfrac{m}{M+m}v$$

Find acceleration of pile

$$0^2-u^2=2ah$$

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

Find the resistance of the ground $R$

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

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

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

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

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

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

$$R=28521N$$

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.