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Topic Title: Mathimatical Engerineering Help!
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Created On: 10 April 2013 10:07 AM
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 10 April 2013 10:07 AM
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ceriwestcott

Posts: 1
Joined: 10 April 2013

A hammer of mass 100KG falls 4M on to a pile of mass 300kg amd drives it 80mm into the gorund.

Calculate the loss of energy on impact-

Calculate the work done by the resistance of the ground

c) Calculate the average resistance to penetration

Please help! This is very important that I figure this out but im having a lot of difficulty.
 22 April 2013 04:41 PM
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kengreen

Posts: 400
Joined: 15 April 2013

Dear Ceri,

I doubt that I am he best person to help with your problem but seeing that you have waited in vain for ten days let's see what I can do.

I suspect that the question was derived from a textbook? RULE no.1 is that you do NOT believe everything that is printed- a much wiser person than I is rumoured to have said that those who know - DO; those that don't know -TEACH; the rest go into politics.

The first question deals in Mass; it is my experience (from many centuries ago) that this is term is used only in textbooks and invites you to get entangled with a strange quantity "g".

The loss of energy on impact - to whom or to what is this energy to be lost? You are not alone in having difficulty!

Calculate the work done by the (unspecified) resistance of the Ground ?
"Average Resistance" ... ? Perhaps these last two questions are in the wrong order ?

I did not believe that I could solve these exercises in advanced Physics but I can assure you that our joint failure will not bring the
Universe to a grinding halt.

Keep trying,

Ken Green
 23 April 2013 09:19 PM
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Zuiko

Posts: 517
Joined: 14 September 2010

I would start by using the SUVAT equations of motion. Write down all you know:

s = 4 m
u = 0 m s^-1
a = 9.81 m s^-2

so, using SUVAT equation:

v = sqrt (2as)

v = 8.86 m s^-1 (this is the velocity of the hammer as it hits the target)


you can now work out the kinetic energy and momentum of the hammer at the moment of impact

E = 1/2 mv^2
E = 3924 J (this is the energy of the hammer as it hits the target)

and momentum

p = mv
p = 886 N s (this is the momentum of the hammer as it hits the target)


Now, you have to make an assumption that all of the momentum is transferred to the 300 kg mass (ie the hammer stops dead on impact). This is a fair assumption.

So you can say that the momentum of the 300 kg mass = momentum of hammer = 886 N s; so the mass's initial velocity, u =

u = p/m
u = 2.96 m s^-1 (this is the velocity of the mass as it shoots away from the hammer head)

and so its initial energy =

E = 1/2 mu^2
E = 1314 J

So the energy lost in impact = 3924J - 1314j = 2609 J

= 3 kJ (to one significant figure - remember the data is only accurate to one significant figure, so your answer cannot be more than that! An answer cannot be more accurate than the data provided. You are given 4 m as a distance. Strictly speaking, this is any distance between 3.5 m and 4.499m.)


Work is defined as force x distance

The mass comes to a stop after 0.08m so you can assume the work done is the force of the mass over 0.08m; or you can assume it is equal to the remaining energy, 1314J. Both should be the same!

so let is work it out

F = ma

so we need SUVAT to work out acceleration (ie, the mass starts at a speed of 2.96 m/s and ends with a speed of zero, so it has a negative acceleration)

u = 2.96 m s^-1
s = 0.08m

therefore:
a = - u^2 / 2s
a = - 54.76 m s^-2 (this is the acceleration of the mass)

The force of the mass is therefore its acceleration x mass:
F = ma
F = 16 428 N (this is the force of the mass)

The work done is the force x distance

Work = Fs
W = 16428 x 0.08
W = 1314

W = 1 kJ (one signficant figure)

which agrees with our previous result.


Good idea to learn the SUVAT equations off-by-heart, there are not many of them!

cheers
W


PS please double check the maths!

PPS always look to see if the numbers look correct and reasonable; off-the wall figures are normally incorrect

Edited: 23 April 2013 at 09:41 PM by Zuiko
 23 April 2013 10:00 PM
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Zuiko

Posts: 517
Joined: 14 September 2010

not sure about average resistance to penetration? any more info on what this actually means? (I'm an electrcial engineer, not a mechanical....!)

I can guess it might mean how much energy the ground can absorb per metre which in this case would be

1314 J / 0.08m = 16425 J/m

20 kJ/m (one sig figure)
 24 April 2013 09:12 AM
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kengreen

Posts: 400
Joined: 15 April 2013

Bravo Zuiko,

You make me feel like a VERY old man. A bit ashamed, maybe, but then ... ?

I too am an electrical/electronics engineer so have much for which to thank you.

Ken Green
 25 April 2013 03:05 AM
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richwin

Posts: 96
Joined: 25 July 2008

Zuiko,

I think you are right but I would have done it slightly differently majoring on energy rather than SUVAT. So:

Potential Energy of hammer is mgh i.e. 100*9.81*4=3924 J

This equals the Kinetic Energy of the hammer at impact: 0.5*m*v^2
so v=sqrt(3924/(0.5*100))=8.86 m/s (to 2 d.p.)
so momentum is mv i.e. 100*8.86=886 kg m/s

Assuming the conservation of momentum we can find the velocity of the pile from momentum/mass=velocity i.e. 886/300=2.953 (to 3 d.p.)

So Kinetic Energy after collision is 0.5*m*v^2 i.e. 0.5*300*2.953^2=1308 J

So energy lost in impact is 3924-1308=2616 J

The work done in bringing the pile to a halt will be 1308 J

Work done is force*distance. We know the distance the pile travels is 0.08 m. So the retarding force (average resistance to penetration) is work/distance i.e. 1308/0.08=16350 N

The only differences between us are rounding errors.

-------------------------
Richard Winstone MIET

All that is necessary for the triumph of evil is that good men do nothing.
Edmund Burke
Irish orator, philosopher, & politician (1729 - 1797)
 25 April 2013 10:13 AM
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cblackha

Posts: 75
Joined: 21 January 2003

Surely after the collision the hammer and mass are both moving - so there is 400kG moving after the collision
 26 April 2013 09:37 AM
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Zuiko

Posts: 517
Joined: 14 September 2010

how far is the hammer moving after the collision?

These equations are all based on simplified models in which only certain factors are considered and assumptions made. The simplifications are good for the vast majortity of problems that need to be solved.

The assumption that a hammer stops when it hits the ground is a valid one; assuming otherwise will make the model more complicated than it needs to be and provide no more accuracy.

The data provided is rough and accurate to only 1 sig figure - added accuracy in the answer is neither required nor correct!
 26 April 2013 09:55 PM
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cblackha

Posts: 75
Joined: 21 January 2003

Doesn't make the maths any more complicated and if you stick to your 1 sig figure, then

Using 300kg, work done stopping = 16350 N = 20,000 (1sf)

Using 400kg, work done stopping = 12262 N = 10,000 (1sf)

I would say that using 300kg instead of 400kg builds in quite an inaccuracy because the mass of the hammer is reasonably significant compared to mass of pile.
 27 April 2013 12:39 PM
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Zuiko

Posts: 517
Joined: 14 September 2010

Work is not measured in Netwons, forces are.

The work done stopping the (300 kg) mass in an elastic collision is 1314 J; 1 kJ to one sig figure.

if you do the same calculation for a 400 kg mass (so you assume that the collision is inelastic) then the work done is 981 J; or 1 kJ to one sig figure.


In reality, the collision is somewhere between the two, but the answer is still going to be 1 kJ.

Edited: 27 April 2013 at 01:05 PM by Zuiko
 03 May 2013 12:14 PM
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sandip

Posts: 21
Joined: 31 March 2013

Mr.Richard Winstone thanks for correct process & answer.
 07 May 2013 09:41 PM
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richwin

Posts: 96
Joined: 25 July 2008

That's OK - glad to be of help

-------------------------
Richard Winstone MIET

All that is necessary for the triumph of evil is that good men do nothing.
Edmund Burke
Irish orator, philosopher, & politician (1729 - 1797)
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