Trouble with PE exam sample problem 2

[bshadel]

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.

 

[tlee123]

JB,

OK, I also stand corrected. I must have never actually read your entire post. I apologize for that.

 

[bshadel]

I should add that there are multiple errors in this book...still surprises me this one got through editing.

Bryan.

 

[sreid]

Does anyone feel the need to comment on the shape of the balloon when it gets whacked by 2000 Gs?

 

[rb1957]

yes, we've decided that the balloon is frozen ...

maybe we should move the problem to the moon, that'll give us 6*the time to work with (about 10 sec)

 

[zekeman]

Boy, what a mess.
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.

 

[jistre]

Solve it using an impulse and momentum balance.

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.

 

[bvanhiel]

jistre,

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

 

[rb1957]

bvanhiel,
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.

 

[tlee123]

jistre,

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.

 

[telecomguy]

"19 springs have a spring rate of 19*6= 144 lb/ft"

Where did the 6 come from again?

<tg>

 

[tlee123]

tg,

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.

 

[FOETS]

Part of the test for these questions are to analyse the assumptions that are taken. If your assumptions are realistic than the examiner has the impression that you have some basic practical understanding of engineering rather than being a robotic number cruncher, knowing formulae verbatim.
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.

 

[telecomguy]

I tend to agree. The exact answer isn't what's important, but an engineer that doesn't have a good gut instinct what the right answer should be (order of magnitude?) will find himself in a tight spot one day.

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>

 

[JStephen]

I'm not sure which part of the test this would be on. But if it's multiple choice, they don't ask to see your work, right? In which case, it won't matter how you approach it you either get the right answer or not. And you get that answer in a minute or so and go on. With the horizontal launch, it does turn into a pretty quick problem.

I seem to recall one of these type tests having an area where you could challenge questions, specifically for this reason.

 

[bshadel]

Well, the exam is tomorrow. I'll let everyone know if I see this problem again.

Bryan.

 

[zeusfaber]

Hope it goes well.

A.

 

[HVACctrl]

bshadel,
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

http://www.engineerboards.com

 

[swearingen]

I know for a fact that you can appeal a question. One of my colleagues did it and was successful. If you truly believe there is a mistake, you need to go to your state's board immediately after the test and inquire about their appeal procedures.

 

[JStephen]

What I'm remembering is a section on the test form or booklet itself. But I don't remember if that was the Fundamentals or the Principals and Practice test- it's been too long.

 

[zekeman]

I withdraw all of my previous comments and please accept my apology . My bad.
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.