# Resistance of the ground problem

• MNWO
In summary: I done it like this, can you tell me if the answer is right?Mass of the hammer(mh) x velocity*=mass of the hammer+mass of the pile (mp) v2*170x10.84=(170+410) v21842.8=580 x v2V2=1842.8/580(the velocity of the hammer and the pile together)V2=3.17 m/sA= V*-U*/2sA= 3.17*-0*/2x0.075A=66.99 m/sGR= (m1+m2)(g)+(m1+m2)(a)GR=(170+410) x(9
MNWO
Hi guys, so I've been given a physics question which is black magic to me, so I thought i will ask you for help :) Here is the question:
A pile driver hammer of mass 170kg falls freely through a distance of 6m to strike a pile of mass 410kg and drives it 75mm into the ground. The hammer does not rebound when driving the pile. Determine the average resistance of the ground.

This is what i have done so far:
m1=170kg
m2=410kg
D1=6m
D2=75mm
PE= m x g x h
PE= 170 x 9.81 x 6
PE=10006.2J
Velocity=2 x g x s
(2) x (9.81) x (6-0)=117.72 square root=10.84 m/s
KE= ½ x mass x velocity
KE=1/2 x 170 x 10.84*
KE=9987.97J

Work= change in kinetic energy + change in potential energy + work due to friction
½ ( 170kg) ( 10.84*-0) + (150kg)(9.81m/s*) (6-0)
=18816.97N without work due to friction

Now I don't know how to work out: work due to friction and how to calculate the resistance of the ground. This is what I was able to do myself and now I am stuck.

Hi and welcome to PF.
If you read the rules at the top of the General Physics Forum (but who does??) you would see that this question should be in the appropriate Forum. I see you have made a start by writing out as much as possible. I suggest you have written down more than you need and it has overwhelmed you. You started off on the right lines.

You can do all this, using Energy and Momentum considerations. 1.Speed of hammer on impact (easy). 2. Momentum is conserved so you can find the speed (then total KE) of the pair after the (instantaneous impact* of hammer on pile). Then Energy conservation says that this KE plus the gpe of the two above the final depth reached is the Work done (Force times distance). You know the distance so you can find the force.
*They will travel together through the ground after that.

I done it like this, can you tell me if the answer is right?
Mass of the hammer(mh) x velocity*=mass of the hammer+mass of the pile (mp) v2*
170x10.84=(170+410) v2
1842.8=580 x v2
V2=1842.8/580
(the velocity of the hammer and the pile together)V2=3.17 m/s
A= V*-U*/2s
A= 3.17*-0*/2x0.075
A=66.99 m/s
GR= (m1+m2)(g)+(m1+m2)(a)
GR=(170+410) x(9.81) + (170+ 410) x (66.99)
5689.8+ 38854.2
=44544N is the total amount of ground resistance

"KE= ½ x mass x velocity"
You are missing an exponent here.

"Velocity=2 x g x s"
What is "s" here? It must be the falling time. But you don't know the falling time without calculating it. Better try another approach.

"(2) x (9.81) x (6-0)=117.72 square root=10.84 m/s"
This is gibberish to me. Where did the square root come from?

MNWO said:
I done it like this, can you tell me if the answer is right?
Mass of the hammer(mh) x velocity*=mass of the hammer+mass of the pile (mp) v2*
170x10.84=(170+410) v2
1842.8=580 x v2
V2=1842.8/580
(the velocity of the hammer and the pile together)V2=3.17 m/s
A= V*-U*/2s
A= 3.17*-0*/2x0.075
A=66.99 m/s
GR= (m1+m2)(g)+(m1+m2)(a)
GR=(170+410) x(9.81) + (170+ 410) x (66.99)
5689.8+ 38854.2
=44544N is the total amount of ground resistance
Let's see. Your calculation for the hammer + mass velocity after the strike is correct, IMHO. I get almost the same result:
v3 = (170 * 10,85)/(170 + 410) = 3,18 m/s

And, concerning the negative acceleration A when sinking in the ground, you get 66,99 and I have got 67,41 m/s2, that is quite close. The equation that I used here is v2 = 2 * a * 0,075. Solving for a, I get a = 3,182/2*0,075 = 67,41 m/s2

But I don't agree with the way you calculate the ground resistance. You should delete the first term of your sum GR=(170+410) x(9.81) + (170+410) x (66.99), With the second term only, you get a result (38854 N) that is close to the one I've got by other, less simple way, and that I believe correct (39076 N)

MNWO

## 1. What is the resistance of the ground problem?

The resistance of the ground problem refers to the difficulty in achieving an accurate measurement of the electrical resistance of the ground or soil. This can be caused by various factors such as variations in soil composition, moisture levels, and depth of measurement.

## 2. Why is it important to measure the resistance of the ground?

Measuring the resistance of the ground is important in various industries, including construction, power distribution, and telecommunications. It helps in determining the safety and effectiveness of grounding systems, as well as identifying potential hazards such as stray currents and lightning strikes.

## 3. How is the resistance of the ground measured?

The resistance of the ground is typically measured using a device called a ground resistance tester or earth resistance meter. This instrument sends a small electrical current into the ground and measures the resulting voltage to calculate the resistance. The measurement is usually expressed in ohms.

## 4. What factors can affect the accuracy of ground resistance measurements?

There are several factors that can affect the accuracy of ground resistance measurements. These include the type of soil, moisture content, temperature, and the presence of rocks or other objects in the ground. It is important to take these factors into consideration when conducting measurements and to use appropriate techniques to minimize their impact.

## 5. How can the resistance of the ground problem be mitigated?

To mitigate the resistance of the ground problem, it is important to follow proper measurement techniques and use high-quality equipment. It is also recommended to take multiple measurements at different locations and times to account for any variations. In some cases, it may be necessary to improve the grounding system itself by adding more rods or using different materials to lower the overall resistance.

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