# Trouble with Mechanics Problem

• MHB
• Tyrzone
In summary: For reference, the kinetic energy of the combined system, right after impact is:$$E_{combined} = \frac 12 (m_{driver} + m_{pile}) v_{combined}^2=m_{driver}\cdot g\cdot s$$So:$$v_{combined}^2=\frac{2\, m_{driver}\cdot g\cdot s}{m_{driver} + m_{pile}} = \frac{2 \cdot 800\text{ kg}\cdot 9.81 \frac{\text{m}}{\text{s}^2} \cdot 2\text{ m}}{800\text Tyrzone A pile driver of a mass of 800kg falls freely through a height of 2m onto a pile of mass 300kg and drives the pile 350mm into the ground. Determine the resisting force in the ground, assuming that the pile driver remains in contact with the pile and does not bounce. Firstly I thought I should work out the force and to do that I needed the acceleration so I rearranged a suvat equation which is v2=u2+2as into a= (v2-u2)/2s After that I thought that there was no initial or final velocity so they would both be 0. That then worked out to be 4m/s. I worked out force by doing 800x4=3200N and got stuck from there on. Any help from someone who knows where to go next? Thanks I would use energy considerations here...what is the initial gravitational potential energy of the pile driver/pile system? Tyrzone said: A pile driver of a mass of 800kg falls freely through a height of 2m onto a pile of mass 300kg and drives the pile 350mm into the ground. Determine the resisting force in the ground, assuming that the pile driver remains in contact with the pile and does not bounce. Firstly I thought I should work out the force and to do that I needed the acceleration so I rearranged a suvat equation which is v2=u2+2as into a= (v2-u2)/2s After that I thought that there was no initial or final velocity so they would both be 0. That then worked out to be 4m/s. I worked out force by doing 800x4=3200N and got stuck from there on. Any help from someone who knows where to go next? Thanks Hi Tyrzone! Welcome to MHB! ;) I suggest to use conservation of energy and the definition of work. The work W done by the ground is given by:$$W = F_{ground} \cdot d$$where d is the distance that the pile was driven into the ground. What is the gravitational energy of the pile driver and the pile, before and after the pile driving? I like Serena said: Hi Tyrzone! Welcome to MHB! ;) I suggest to use conservation of energy and the definition of work. The work W done by the ground is given by:$$W = F_{ground} \cdot d$$where d is the distance that the pile was driven into the ground. What is the gravitational energy of the pile driver and the pile, before and after the pile driving? Thanks for the helps. I think I got the right answer but I thought I would check here first. Would it be: 17.93kN/17930/17.93x103? I find the initial energy is: $$\displaystyle E_i=\left(800\cdot9.8\cdot2.35+300\cdot9.8\cdot0.35\right)\text{ J}=19465\text{ J}$$ Since the final energy is zero, this energy must have been dissipated during the time the ground was working to stop the system. Using the work-energy theorem and the relationship between work and the distance over which a force is exerted, we then find: $$\displaystyle \overline{F}=\frac{W}{d}=\frac{19465\text{ J}}{0.35\text{ m}}=55580\text{ N}$$ MarkFL said: I find the initial energy is: $$\displaystyle E_i=\left(800\cdot9.8\cdot2.35+300\cdot9.8\cdot0.35\right)\text{ J}=19465\text{ J}$$ Since the final energy is zero, this energy must have been dissipated during the time the ground was working to stop the system. Using the work-energy theorem and the relationship between work and the distance over which a force is exerted, we then find: $$\displaystyle \overline{F}=\frac{W}{d}=\frac{19465\text{ J}}{0.35\text{ m}}=55580\text{ N}$$ Wouldn't it be (1100x9.8x0.35)+(1/2x1100x4.55)=6275.5 due to kinetic energy once the pile has been hit 6275.5/0.35=17930N Tyrzone said: Wouldn't it be (1100x9.8x0.35)+(1/2x1100x4.55)=6275.5 due to kinetic energy once the pile has been hit 6275.5/0.35=17930N It would help if you show your workings... (Wasntme) For reference, the kinetic energy of the combined system, right after impact is:$$E_{combined} = \frac 12 (m_{driver} + m_{pile}) v_{combined}^2=m_{driver}\cdot g\cdot s $$So:$$v_{combined}^2=\frac{2\, m_{driver}\cdot g\cdot s}{m_{driver} + m_{pile}}
= \frac{2 \cdot 800\text{ kg}\cdot 9.81 \frac{\text{m}}{\text{s}^2} \cdot 2\text{ m}}{800\text{ kg} + 300 \text{ kg}} = 28.5 \frac{\text{m}^2}{\text{s}^2}


How did you get 4.55? (Wondering)

## 1. What is the "Trouble with Mechanics Problem"?

The "Trouble with Mechanics Problem" refers to a common issue encountered by mechanics and engineers in which a mechanical system does not function as intended or expected. It can involve issues such as malfunctions, failures, or inefficiencies in machines or devices.

## 2. What are some common causes of the "Trouble with Mechanics Problem"?

There are several potential causes of the "Trouble with Mechanics Problem," including design flaws, material defects, improper maintenance, and user error. Additionally, external factors such as environmental conditions or wear and tear can also contribute to the problem.

## 3. How do scientists and engineers approach solving the "Trouble with Mechanics Problem"?

Scientists and engineers use a systematic approach to troubleshoot mechanical problems. This involves identifying the issue, gathering data and information, formulating hypotheses, and testing potential solutions. This method allows for a logical and evidence-based approach to finding and fixing the problem.

## 4. Can the "Trouble with Mechanics Problem" be prevented?

While it is impossible to completely eliminate the "Trouble with Mechanics Problem," it can be minimized through proper maintenance, regular inspections, and thorough testing during the design and development stages. Additionally, continuous improvements and updates to the system can help prevent future issues.

## 5. How important is it to address the "Trouble with Mechanics Problem" in a timely manner?

It is crucial to address the "Trouble with Mechanics Problem" as soon as possible to prevent further damage or safety hazards. Ignoring the problem can lead to more significant and costly issues down the line, and it may also put individuals at risk if the malfunctioning system is used in critical situations.

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