Trouble with PE exam sample problem
Trouble with PE exam sample problem
(OP)
Hi folks,
New to the forum, I’m getting ready to take the mechanical PE exam this Friday. I’ve been studying for a better part of the year and hope for some positive results.
I’ve come across a sample problem where I don’t agree with the solution. Perhaps someone can set me straight. By the way, this problem is from “Six-Minute Solutions for Mechanical PE Exam”.
Problem:
Water balloons are launched from a catapult on the rooftop of a building 50 feet off the ground. Balloons must achieve at least 100 yards (horizontal). Catapult spring constant is 0.5 lbf/in and can be extended no more than 5 inches. Weight of each balloon is 4.5 lbf, and each one takes approximately 1 second to launch. Under these conditions, how many springs in parallel will it take to launch 100 yards?
My solution:
I solved for an initial velocity that would send the balloon 100 yards. I then used energy methods to solve for the total spring constant: 1/2kx^2 = 1/2mv^2. Unfortunately my answer was way off.
Book solution:
The acceleration of the balloon was found from the 1 second launch to the terminal velocity in 5 inches. Knowing that, they said: (mass of balloon x acceleration) = (spring constant x 5 inches).
While the units work out, I don’t agree with this simply because the spring force is not constant and varies with the deflection. This is not the only problem like this…there is another that utilizes this type of solution. Am I missing something???
Any assistance is greatly appreciated.
Thanks,
Bryan.
New to the forum, I’m getting ready to take the mechanical PE exam this Friday. I’ve been studying for a better part of the year and hope for some positive results.
I’ve come across a sample problem where I don’t agree with the solution. Perhaps someone can set me straight. By the way, this problem is from “Six-Minute Solutions for Mechanical PE Exam”.
Problem:
Water balloons are launched from a catapult on the rooftop of a building 50 feet off the ground. Balloons must achieve at least 100 yards (horizontal). Catapult spring constant is 0.5 lbf/in and can be extended no more than 5 inches. Weight of each balloon is 4.5 lbf, and each one takes approximately 1 second to launch. Under these conditions, how many springs in parallel will it take to launch 100 yards?
My solution:
I solved for an initial velocity that would send the balloon 100 yards. I then used energy methods to solve for the total spring constant: 1/2kx^2 = 1/2mv^2. Unfortunately my answer was way off.
Book solution:
The acceleration of the balloon was found from the 1 second launch to the terminal velocity in 5 inches. Knowing that, they said: (mass of balloon x acceleration) = (spring constant x 5 inches).
While the units work out, I don’t agree with this simply because the spring force is not constant and varies with the deflection. This is not the only problem like this…there is another that utilizes this type of solution. Am I missing something???
Any assistance is greatly appreciated.
Thanks,
Bryan.





RE: Trouble with PE exam sample problem
I'd do the launch with an energy method as well.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: Trouble with PE exam sample problem
Do not open this thread again or think any more about this problem until after the exam
At most you'll get one problem wrong out of a whole bunch of problems (as I recall it was 80 in the morning and 80 in the afternoon). The actual chances of a similar problem approach zero risk.
I'm sure the people in this forum will give you great approaches to solving it, but think about them after the exam.
David Simpson, PE
MuleShoe Engineering
www.muleshoe-eng.com
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
The harder I work, the luckier I seem
RE: Trouble with PE exam sample problem
Assuming I didn't make any mistakes somewhere, I come up with an optimum launch angle of about 40.25 degrees, which requires 90.425 ft/sec velocity. That equates to 571.8 ft-lbs of energy in the projectile, which requires a peak spring force of 2,745 lbs, which requires 1098 springs. That sounds awfully high, but as stated, that is a dinky little spring.
From the problem statement, it sounds like they expect you to assume a launch angle of 45 degrees, which is the optimum if the landing spot is the same elevation as the launch spot, but that isn't the case. (I get 90.96 ft/sec in that case). Anyway, the full solution is complicated enough, I'm wondering if either we aren't missing something from the problem statement (assumed angle, maybe assumed horizontal launch).
RE: Trouble with PE exam sample problem
Bryan.
RE: Trouble with PE exam sample problem
Yea, even with a zero degree launch angle, I got somewhere on the order of 10^3 springs.
Bryan.
RE: Trouble with PE exam sample problem
I'm with Greg. I'm trying to figure out why the 1 second launch number is significant. It seems to overdefine the problem. Can you post the complete text of the question? Maybe we're missing something.
Also, this problem has screwy units (yd, in, ft). Make sure you didn't just make a conversion error.
-b
RE: Trouble with PE exam sample problem
In an engineering competition, students try to launch large water balloons over a 100 yard range using catapults and slingshots. The balloons must be launched with a zero degree start angle from the roof of the engineering building, 50 feet above the ground. All springs used in the competition must have a spring constant of 0.5 lbf/in and be extended no more than 5 inches. The weight of each projectile water balloon is 4.5 lbf, and each one takes approximately 1 second to launch. Under these conditions, what is the least number of springs in parallel that will send a balloon 100 yards?
Answer choices: 1, 3, 10, 300.
RE: Trouble with PE exam sample problem
And it would not be at all surprising if the test questions supplied information that is not required for the solution- such as the 1 second in this case.
RE: Trouble with PE exam sample problem
-b
RE: Trouble with PE exam sample problem
bvanhiel, the 1 second was used calculate acceleration, as used in the author's solution: (mass of balloon x acceleration) = (spring constant x 5 inches).
RE: Trouble with PE exam sample problem
The required launch distance is 300 ft .
The spring constant of one spring is 6 lb/ft.
The mass of the water balloon is 4.5 lb/ 32.2 ft/sec^2 = .14 lb-sec^2/ft
The springs can be stretched .417 ft.
First, assume no air drag and massless springs.
Second, calculate the time for the water balloon to hit the ground (it makes no difference what the horizontal velocity is).
y=v0t+1/2at^2
v0=vertical velocity at launch=0
rearrange to solve for t
t=sqrt(2y/a)
y=50 ft a=32.2 ft/sec^2
t=1.762 sec
So now we just need to calculate what starting horizontal velocity is needed that will allow the balloon to travel 300 ft horizontally in 1.762 seconds.
v=300/1.762=170.2 ft/sec
The potential energy of the spring will be converted into kinetic energy of the balloon.
1/2 mv^2=1/2kx^2
rearrange and solve for needed spring rate k
k=mv^2/x^2 = (.14)(170.2)^2/(.417^2) = 23328 lb/ft
Each spring has a rate of 6 lb/ft so
total number of parallel springs = 23328/6 = 3888 springs
RE: Trouble with PE exam sample problem
-b
RE: Trouble with PE exam sample problem
Sometimes, thinking too much or realistic complicates the problem. In your argument that realistically, spring constant is varying with respect to deflection. You maybe right in reality, but,but if you notice most problems in Engineering especially on mechanics is not exactly computed as they should be realistically. taken into account the velocity of the balloon, isn't it that there are many factors that affects it? For example, the air resistance. But in your computation you don't need all this things or include these things. My point is this, spring constant may actually be not constant but in general or most often they are assumed to be constant. Unless otherwise specified as they say.
Hope I am right.
Virgilio B. Mendoza jr.
Mechanical Engineer
Innovatronix Inc.
RE: Trouble with PE exam sample problem
The energy answer calculated above gives you the minimum number of springs you would need with all of those factors (spring mass, air resistance, friction) negated. The number of springs required in this frictionless universe is an order of magnitude higher than the answers that the author proposes. I'm pretty sure he/she is just plain wrong.
-b
RE: Trouble with PE exam sample problem
Have you tried engineerboards.com? That board is filled with people studying for the PE test.
Ed
www.engineerboards.com
RE: Trouble with PE exam sample problem
Anyway, here's my thoughts:
You have already calculated the travel time. Now,
Initial velocity = change in distance / time = 300 feet / 1.762 seconds = 170 feet/sec (roughly) required, but this is a final velocity for your launch, so calculating the acceleration from the catapult, you have:
Acceleration = change in velocity / time = (Final velocity - initial velocity)/ time
In the case of the catapult function, final velocity = 170 ft/sec, initial velocity = 0 ft/sec and time is the time to accelerate (the mysterious 1 second), so acceleration is 170 ft/sec^2
Now go to the equation mass x acceleration = spring constant x distance. I suspect this comes from F=ma and F=kx resulting in ma=kx...just a guess, but:
0.14x170 = k (0.417)
k=57
Springs in parallel act together k = k1 + k2 + k3 + ..., and k1 =k2=k3..., so
# springs = 57/6 = 10 roughly
Right or wrong, I suspect that is the way they solved it...
Garland E. Borowski, PE
Borowski Engineering & Analytical Services, Inc.
Lower Alabama SolidWorks Users Group
RE: Trouble with PE exam sample problem
So, move on to the next problem and good luck on your exam. You have already shown that the energy approach (ignoring the launch time) is the best of a wrong lot.
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
i liked GBor's force-based answer, except that why would the balloon take 1 second to accelerate up to the average velocity ... if it does it won't reach the 100 yds in time ...
0sec 0ft/sec
1sec 170ft/sec distance covered is 85ft
1.76 sec 170ft/sec distance covered is .76*170 = 129ft
total distance covered 213ft
if 300ft =v*(1.76-t)+(v/2)*t; where t is the time taken to accelerate the balloon to speed v
and m*(v/t) = k*0.417ft, k = 6, m=0.14 slugs
v/t = 6*0.417/0.14 = 18
then 300 = 18t*(1.76-t) +(9t)*t
9t^2-32t+300 = 0
t^2-3.5t+33 = 0
t = (3.5+-sqrt(3.5^2-132))/2
which isn't real ... sigh, but then i've used only one spring.
10 springs would be k=60, v/t = 180
300 = 180t*(1.76-t)+(90t)*t
90t^2-320t+300 = 0
9t^2-32t+30 = 0
t^2-3.5t+3.3 = 0
t = (3.5+-sqrt(3.5^2-13.2))/2
maybe because of round-off, this is also not real
if 300 springs ...
k = 1800, v/t = 5400
300 = 5400t(1.76-t)+(2700t)*t
2700t^2-9600t+300 = 0
9t^2-32t+1 = 0
t^2-3.5t+0.11 = 0
t=(3.5-sqrt(3.5^2-0.44))/2
t = 0.03sec
then ...
0sec 0ft/sec
0.03sec 162ft/sec 81ft
1.76sec 162ft/sec 280ft
361ft
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
Springs (even theoretical ones) don't have the same force over the entire stroke. You may have figured out the author's mistaken assumption.
-b
RE: Trouble with PE exam sample problem
can't we say that the force the springs apply to the balloon is equal to the inertia force agained by the balloon ?
ie k*x = m*a = m*(v/t)
then we have v/t = k*x/m.
if we have 300 springs, v/t = (6*300)*0.417/0.14 = 5360
then 300ft = v*(1.76-t)+(v/2)*t
300 = 9434t-5360t^2 + 2680t^2
2680t^2-9434t+300 = 0
9t^2-31.68t+1 = 0
t^2-3.52t+0.11 = 0
t = (3.52-sqrt(3.52^2-0.44))/2 = 0.03
0sec 0ft/sec
0.03sec 172ft/sec 3ft
1.76sec 172ft/sec 297ft
total distance travelled = 300 ft
(i guess the other answer was round off ...
the problem is quite sensitive to accuracy, being based on th edifference of two small numbers)
force applied = (4.5/32.174)*(172/0.03) = 750 lbf
RE: Trouble with PE exam sample problem
Springs still don't produce the same force over the entire stroke. At .417ft deflection 300 springs with a spring constant of 6lbf/ft will produce ~750lbf. At half deflection the force will be ~375lbf. At 0 deflection the force will be 0lbf.
When you release the balloon you will get the initial acceleration, but it will quickly drop off over the 5 inches of spring deflection to 0.
-b
RE: Trouble with PE exam sample problem
The mistake you made is that you can't use the equation of motion:
x=x0+v0t+1/2at^2 unless a is constant. As bvanhiel points out, a is decreasing with distance.
RE: Trouble with PE exam sample problem
my point was that there are two phases to the travel of the balloon, a constant velocity ('cause we're assuming no losses once the balloon's on it's way) phase and an acceleration phase (from rest to the constant velocity); the point being that the constant velocity is higher than the average velocity (practically by only a little, but we were talking about accelerating for 1 sec)
but the spring energy = balloon kinetic energy calc yields an answer incompatable with the choices ...
anyone know an examiner ?
or a worked solution ?
RE: Trouble with PE exam sample problem
I'm no spring expert but the maximum force applied to the balloon, and thus the maximum acceleration, is imparted when the sling is released. As the springs compress the force applied is redeuced and so is the acceleration. The initial impulse of releasing the spring leaves the balloon traveling faster then the sling as the spring compress and the remaining 5 inches don't have an effect on the balloon.
I don't agree with that, if it's in fact what the author was going for, but it does explain the equation used in the book solution.
Now since the exam is supposed to have a practical component to it, the practical answer would be none of the above. I can't throw a water balloon 20 ft without it bursting in my hand. I don't think you could accelerate the balloon to 170 ft/sec in 5 inches an not have the thing burst. Most launchers that I have ever seen stretch out to nearly 10 ft and they would still not be able to make 100 ft launch.
RE: Trouble with PE exam sample problem
http://www
RE: Trouble with PE exam sample problem
Just because the acceleration is decreasing, it is still positive thru the entire 5 inch stroke, and therefore the velocity increases throughout that stroke. The indestructible, perfectly rigid water balloon stays in contact with the massless sling.
RE: Trouble with PE exam sample problem
I think I agree with JamesBarlow. If the initial horizontal acceleration is assumed to be constant, then the balloon will be traveling faster than the spring and the balloon gets an instantaneous burst of energy, sufficient to travel 100 yards in just under 2 seconds. Heads-up!
<tg>
RE: Trouble with PE exam sample problem
Since the balloon launches purely horizontally, all vertical motion arises from the force due to gravity. Therefore, the energy balance between the initial vertical position at 50 feet and 0 velocity and the final position at 0 feet and maximum velocity is used to determine time to fall, and hence, the horizontal velocity required to cover the 100 horizontal yards in that time.
The kinetic energy of the balloon in the horizontal can then be calculated. Since it began at 0 horizontal velocity and the only source for the kinetic energy is that energy stored in the springs, it is straightforward to calculate the number of springs required.
Conservation of energy has worked for us so far. Why would they claim that it doesn't work in this problem.
Consider me intrigued.
RE: Trouble with PE exam sample problem
The 1 second "launch time" has to do with the loading of the balloon or releasing of the latch or whatever.
And now people are saying that the balloon magically moves faster than the 3888 springs in parallel? Someone explain (with an equation, please) just how much time or distance it takes before that balloon leaves those springs in the dust. Does the ballon have its own means of propulsion? Maybe there's a pinhole in the back where water shoots out.
It's already been a long day (can you tell?).
RE: Trouble with PE exam sample problem
nothing "magical" about it (tho' there is a practical problem of getting all these springs at apply their forces to the balloon)
RE: Trouble with PE exam sample problem
I could waste an equation on this, but why? The 'magical' part of it is the nature of 'assumptions'. If you assume the initial acceleration is constant for the full 5" of travel, but you know the springs decelerate to 0, then it's obvious the balloon leaves the spring instantaneously! :oD
<tg>
RE: Trouble with PE exam sample problem
This is the quote that had me bothered:
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
"I don't agree with that, if it's in fact what the author was going for, but it does explain the equation used in the book solution."
I think we can all agree that the solution the author came up with is wrong.
So far, however, no one has been able to explain were it came from. I was simply proposing a possible assumption that the author MAY have used to arrive at their solution.
It's easy to show that the author is wrong, but the more interesting question is tring to show why they are wong, and how they go to this conclusion.
RE: Trouble with PE exam sample problem
GBor did solve the problem in a way that gave the solution the author presented.
RE: Trouble with PE exam sample problem
OK, I also stand corrected. I must have never actually read your entire post. I apologize for that.
RE: Trouble with PE exam sample problem
Bryan.
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
maybe we should move the problem to the moon, that'll give us 6*the time to work with (about 10 sec)
RE: Trouble with PE exam sample problem
It seems to me that the author of this thread has done the best job of "solving" this dilemma and 99% of the remaining comments are without any merit. If I were he, I would mercifully ask to terminate this blather and get on with your lives.
RE: Trouble with PE exam sample problem
Force * Delta t = Mass * delta V
Average force exerted by one spring is 2.5 lbf according to integral(k dx) from x=0 sec to x=5 sec. Launch takes 1 sec, so impulse imparted to the balloon by one spring is
I = F(av) * Delta t = 2.5lbf * 1 sec = 2.5 lbf-sec.
The change in velocity of the balloon imparted by one spring is
delta V = I/m = 2.5 lbf-sec/4.5 lbm = 17.87 ft/s.
Since we need delta V of 170 ft/s, impulse and momentum solutions also lead to 10 springs.
What I'm really interested in is if someone can show the disconnect between the answers generated by the summation of forces approach, the impulse/momentum approach and the answers generated by the energy balance approach, which is the one most of us seem to trust.
RE: Trouble with PE exam sample problem
The disconnect is the authors incorrect assumption that the springs will deliver their maximum force over a 1 second period.
sreid, rb1957,
Your assuming the springs are acting on the balloons directly, which is unlikely since it's a catapult. The 5 inches is a characteristic of the spring, not the launch mechanism. Springs acting with a lever arm will have plenty of time to accelerate a balloon.
-b
RE: Trouble with PE exam sample problem
i was thinking about the footprint that 3000 springs would have, and how big a 5lb water balloon would be.
jistre,
i think your assumption of a 1 sec launch is questionable,
when compared to a flight time of 1.76 sec.
zekeman,
i find it interesting (but i guess you don't) that different approaches to the problem yield way different answers.
RE: Trouble with PE exam sample problem
Interesting idea to look at an impulse analysis.
However, the average force of one spring is 1.25 lb, not 2.5 lb.
F(av)=(5*0.5)/2 = 1.25 lb
Following your argument with this change,
I = F(av) * Delta t = 1.25lbf * 1 sec = 1.25 lbf-sec.
The change in velocity of the balloon imparted by one spring is
delta V = I/m = 1.25 lbf-sec/.14 lb-sec^2/ft = 8.93 ft/s.
Since we need delta V of 170 ft/s, impulse and momentum solutions lead to 19 springs (170/8.93). 19 springs have a spring rate of 19*6= 144 lb/ft
OK, here is where the above analysis falls apart.
19 springs have a spring rate of 19*6= 114 lb/ft.
The mass of the balloon is .14 lb-sec^2/ft.
The spring is originally stretched by .417 ft.
The initial velocity of the balloon is 0 ft/sec.
Damping is 0 lb-sec/ft.
Since we know m,c, k, x(0), and v(0) we can solve the differential equation of motion giving us x as a function of t. If we do this, we find that the mass moves from x=.417 to x=0 in about .055 seconds.
This means the impulse occurs over only .055 second, not 1 second. This explains the how the impulse analysis differs with the (correct) energy analysis.
Actually, you could have a 1 second impulse if a flying monkey holds the back end of the springs and flies along with the water balloon for 1 second.
RE: Trouble with PE exam sample problem
Where did the 6 come from again?
<tg>
RE: Trouble with PE exam sample problem
Spring rate of one spring = 0.5 lb/in * (12 in/ft) = 6 lb/ft. The 0.5 lb/in was from the original post.
Also, the total 144 lb/ft in my last post was a typo. It was corrected lower in the analysis as 114 lb/ft.
RE: Trouble with PE exam sample problem
As long as you document your assumptions, and correctly use the methods of calculation full marks should be obtained.
That is unless you get a robotic examination marker that is only concerned with the correct answer using his own perceived assumptions and the poorly defined and incomplete criteria given in the question.
As can be seen everyone has his/her own set of assumptions which lead to different outcomes but all still defendable.
Looking at the verbosity of this thread i think that we would have run out of time in the examination by now.
RE: Trouble with PE exam sample problem
I think bvanhiel hit on the delimma. The problem statement says they are building catapults and sling shots (see bshadel's 3rd post). So the actual application of the spring is not defined. With this knowledge, I think the only way to solve the problem is by energy analysis. The exact design of the mechanism is left as an exercise.
<tg>
RE: Trouble with PE exam sample problem
I seem to recall one of these type tests having an area where you could challenge questions, specifically for this reason.
RE: Trouble with PE exam sample problem
Bryan.
RE: Trouble with PE exam sample problem
A.
RE: Trouble with PE exam sample problem
Better not let everyone know if you see it again- that would be against the agreement you will sign tomorrow!
I hope the best for you. Keep your head up. I thought I surely failed, but passed (ME, machine design in PM). Just keep hacking away at the hard ones and you'll get there!
Ed
www.engineerboards.com
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
Firstly, the launch time can be one second or one minute. But the solution must satisfy the energy equation which is inviolate, so the thread author's solution must be correct.
As far as the time, it appears that most people, myself included, kept thinking that tne mass was at the end of the spring which is not the case but, as pointed out by others could be at the end of a speed increasing mechanism, not unlike a geared up inertial mass, where we were all taught that the effective inertia is the inertia at the sped up gearshaft times the square of the gear ratio. Similarly, you can have an effective mass at the spring end hugely greater than the 5 lbs. In this case, that is the case and I solved the speed ratio as follows.
The solution of the ODE eqution for this spring mass system for a reflected mass, m is
x=5/12(1-cos(sqrt(k/m)t)) for and the velocity (one derivative) is
v=5/12*sqrt(k/m)sin(sqrt(k/m)t)
Now,at launch, the mass separates from the spring, so the sine term must be unity and therefore
sqrt(k/m)t=pi/2=1.57
and v=5/12*sqrt(k/m)
keeping in mind that the equations are at the spring and m is the reflected mass and v is the actual velocity at the end of the spring, not at the true mass whose velocity will be called v'
If t= 1sec
sqrt(k/m)=1.57 and
(1) k/m=1.57^2=2.46
Also at launch we know that v=170 and equating the energizies
.5*5/32*170^2=m*v^2=.5*m*(5/12)^2*k/m=.5*(5/12)^2*k and
k=26,010 lb/ft From eq 1
k/m=2.46 and
m=k/2.46=26010/2.46=10,573 slugs
Also equating KE
m*v^2=5/32*v'^2 where v' is velocity at the actual mass
The gear ratio or mechanical advantage is
v'/v=sqrt(m/(5/32)=260
This would be difficult to achieve in practice, since mass of the mechanism would be a big problem.
RE: Trouble with PE exam sample problem
i think you're saying that a single spring could generate the required force, given a big enough lever?
doesn't that mean that any of the answers would work, given different gearing ?
but you started by saying the energy solutions (saying several thousand springs are needed) are right; even tho' that isn't one of the answer choices.
how far does the balloon travel in the 1 sec "launch" ?
(bearing in mind that the ballon travels 300ft in 1.76sec)
RE: Trouble with PE exam sample problem
I haven't done the calcs but I think there is a gravitational constant in the KE that is being overlooked.
RE: Trouble with PE exam sample problem
Normally, a catapult arm doesn't give a linear relation between object distance and spring distance, and you can't solve equations of motion without knowing that relationship.
RE: Trouble with PE exam sample problem
The number of springs would be the same as previously calculated since they depend on the energy equation which remains the same for the ideal case of zero mass mechanism; and since I got 260 as the gear ratio, then the launch distance would be
5/12*260 or about 108 feet during the 1 second launch period followed by the 300 feet of horizontal travel.
The implementation of a practical solution would have to account for the energy in the launch mechanism and as JStephen pointed out, increases the number of springs.
Nobody says that a catapult will be used; with proper gearing a mechanical linear output could be achieved.
RE: Trouble with PE exam sample problem
"Normally, a catapult arm doesn't give a linear relation between object distance and spring distance, and you can't solve equations of motion without knowing that relationship."
If the mass is at the end of the boom and rotates 90 degrees from the horizontal until launch, the mass moves along the arc and indeed has a linear relationship with the spring motion, if the mechanism attached to the spring is a rack and pinion.
RE: Trouble with PE exam sample problem
Ooh! am I allowed to say that? I did sign box 3!
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
I never said that they make catapults with a rack and pinion, only a possibility and have no way of knowing nor do I care. My only point is that there is a linear relationship of the motion at the end of the catapult and the gearing that rotates the final actuator and suggested that if the end of the spring moves along the periphery of the input gear like a rack and pinion or a cable attched to an input winch one would get an overall linear relationship. Indeed launch mechanisms for planes on aircraft carriers (on which I worked on in the 50s) had this type of linearity and of course also a mechanical linear output motion.
Where catapults do not exhibit linearity the problem is solvable albeit with moderate difficulty (after writing the dynamic equations) with present computer methods.
RE: Trouble with PE exam sample problem
I tried it iteratively and got 8.745 springs, I guess 9 springs would be the answer.
You know the springs got to be reasonable, like a few, perhaps 9 is too high.
I took a single spring, deflected it the said amount at the required spring rate and launched the balloon of known mass straight horizontally. Did the balloon pass a point 100 feet horizontal and 50 feet in the air? If yes, then that spring talley is your answer. If no, then add another spring in parallel, determine the equivalent spring constant of the system and recompute. This is the iterative approach I used, by hand.
There are a lot of answers, few really give you an idea on the number of springs required. Too bad, nice problem!
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Trouble with PE exam sample problem
You just made my day. Way up in this thread I said, I've stayed out of it since then...listening, but perhaps my PE isn't as dusty as I thought
Garland E. Borowski, PE
Borowski Engineering & Analytical Services, Inc.
Lower Alabama SolidWorks Users Group
RE: Trouble with PE exam sample problem
GBor got pretty much the same answer,
but why does the energy solution give an answer in 1000s ?
Also, GBor got his answer assuming the balloon took 1 sec to accelerate up to speed. i think if this happens then the balloon doesn't travel the required distance in 1.76 sec.
from energy methods, a spring produces a velocity of ...
v = sqrt(k/m)*x = sqrt(6/(4.5/32.17))*(5/12) = 2.73 ft/sec
we need something like 300ft/1.76sec = 170 ft/sec
(it takes 1.76 sec for an object at rest to fall 50ft)
which would be (170/(5/12))^2*(4.5/32.17)/6 = 3890 springs
(like tlee got)
from force methods, a spring applies a force of 6*(5/12) = 2.5 lb, which would accelerate the balloon at 2.5/(4.5/32.17) = 18ft/sec2 ... but how long is this acceleration applied to the balloon ?
oh, there's a flash of the blindingly obvious ...
the energy solution assumes constant velocity, the force solution assumes constant acceleration
to travel 300ft horizontally in the same time as it takes to fall 50ft the acceleration required is 6g, 193 ft/sec2 which is 10 springs.
so is the spring going to accelerate the balloon or impart a delta v ?
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
the spring force is converted into an inertial force, no?
RE: Trouble with PE exam sample problem
Let's use some common sense.
A 5 inch spring, with a rate of 1/2 lbf/in (which is, incidentally, softer than a typical rubber band), exerts a maximum force of 2.5 lbf. 10 of them will exert a maximum force of 25 lbf, about the weight of a small sack of potatoes. So let's ignore how springs work and pretend that they apply that force for half their stroke, 2.5 inches.
Now, are you guys seriously proposing that the energy in letting a small sack of potatoes drop by 2.5 inches (bearing in mind that is an overestimate) will accelerate a frozen chicken to something over 100 mph?
If you ignore the 1 second piece of weird data, the energy method takes you to 3888 springs. I suppose by ignoring any one other piece of data in the question will allow a range of other answers. The energy method works because it makes no assumptions about 'average' spring forces, 'constant' accelerations, and other such unlikely concepts in the world of dynamics. It does assume an efficiency of 100%, and a frictionless massless catapult arrangement, but I don't see anybody else including those either.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
RE: Trouble with PE exam sample problem
A very interesting problem!!! In particular the above discussion illustrates how different assumptions lead to different solutions.
RE: Trouble with PE exam sample problem
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RE: Trouble with PE exam sample problem
He used x=5/12(1-cos(sqrt(k/m)t)). Setting t=0 we get x=0 which isn’t right. X should be 5/12 ft at the start. X=0 means the spring is not exerting a force (F=kx).
Here is my solution of the massless, frictionless rack and pinion based catapult.
We know from the energy based analysis that 3888 springs are required which gives a spring rate k=23328 lb/ft. We also know that these springs give up their energy to the water balloon thru a 5/12 ft stroke and will give a launch velocity of 170 ft/sec (what we want). Any other solution that gives an answer of fewer springs violates the Conservation of Energy, so I’d be very careful to promote those or your reputation may get tattered.
Before the balloon is released, let’s say that x is the motion at the end of the springs and y is the tangential motion of the mass (the water balloon). We’ll say that the radius of the pinion (where the springs push) has length r1 and the long arm welded to the pinion that holds the mass has length r2. So:
y = (r2/r1)x and
d2y/dt2 = (r2/r1) d2x/dt2
If we do a free body diagram and take the sum of the moments around the center of the pinion we get:
Fspring(r1) – m(r2)d2y/dt2 = 0
Substitute d2y/dt2 = (r2/r1) d2x/dt2
into above equation gives:
Fspring = m(r22/r12)( d2x/dt2)
From F=ma, we can see that the “effective mass” from the point of view of the springs is:
meffective= m(r22/r12)
Now the solution to the differential equation is:
x = x0(cos(k/meffective)1/2t)
When t=1 second we want x=0. We already know from the energy analysis that the velocity of the mass will be correct when the springs have moved 5/12 ft (x=0).
For the cosign term in the above equation to be zero, (k/meffective)1/2 must be equal to pi/2.
Substitute meffective= m(r22/r12) and solve for r2/r1 we get
r2/r1=(4k/(pi2m))1/2=260
RE: Trouble with PE exam sample problem
"He used x=5/12(1-cos(sqrt(k/m)t)). Setting t=0 we get x=0 which isn’t right. X should be 5/12 ft at the start. X=0 means the spring is not exerting a force (F=kx)."
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To set the record straight the differential equation I used was NOT incorrect but
mx"=5/12k-kx which references x to the start position zero where the spring is compressed at time t=0 and is quite arbitrary (just picture a spring compressed and the starting position of the system is 0). Saying that x=0 does NOT imply that the force is zero. This ODE shows that at the start there is a force of 5/12k .
The solution to this differential equation is as I showed
x=5/12(1-cos(sqrt(k/m)t)) which totally satisfies the ODE
And by the way your solution effectively says that x=-5/12 at time t=0 which is just as arbitrary but conveniently becomes a valid solution of the homogeneous equation which is correct for your proposed starting conditions in which you have x referenced to the the equilibrium position of the spring
mx"=-kx
x=-5/12cos(sqrt(k/mt))
If you look carefully you will see that the difference in the both solutions is 5/12 foot, which has only to do with the reference positions we each took.
So, in essence the both solutions are correct and not surprisingly lead to the same result.
RE: Trouble with PE exam sample problem