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.

 

[GregLocock]

Agreed, " (mass of balloon x acceleration) = (spring constant x 5 inches). " is wrong, that would give the peak acceleration, which is no real help.

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.

 

[zdas04]

More importantly you're taking the exam in 5 days. You've learned everything you can learn from this 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

http://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

 

[JStephen]

Well, I just wasted an inordinate amount of time on the problem, and hope you don't do the same.

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).

 

[bshadel]

Sorry everyone, I should have included that the launch angle is ZERO degrees.

Bryan.

 

[bshadel]

JStephen,

Yea, even with a zero degree launch angle, I got somewhere on the order of 10^3 springs.

Bryan.

 

[bvanhiel]

Bryan,

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

 

[bshadel]

Here is the exact problem description:

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.

 

[JStephen]

Ahh, that vastly simplifies the problem. So drop time is 1.763 seconds, horizontal velocity must be 170.2 ft/second. Energy is then 2,025 ft-lbs, which requires 9,723 lbs peak force and that gives me 3,890 springs. So, either I've made a major goof or the problem-writing people have.

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.

 

[bvanhiel]

I still don't get what the 1 second has to do with anything. I think you are correct to solve this problem with energy.

-b

 

[bshadel]

JStephen, got the same answer as you using energy methods. I'm thinking it's a faulty solution presented in the book. Wonder how many others are like this?

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).

 

[tlee123]

Yes bvanhiel, lets clean up the units. I’ll put everything in terms of ft, lbs, and seconds.

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

 

[bvanhiel]

Agreed

-b

 

[johnvergel]

Hey there folk!

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.

 

[bvanhiel]

Virgilio,

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

 

[HVACctrl]

bshadel,
Have you tried engineerboards.com? That board is filled with people studying for the PE test.

Ed

http://www.engineerboards.com

 

[GBor]

I think the answer is 10, but I haven't taken the PE exam in its new format (I'm feeling my age), so I would recommend you pick the nearest answer if you get in a difficult situation.

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

 

[zekeman]

As stated, the problem cannot be solved since you are bound by the launch velocity (which everyone got), the distance of launch (5 inches) and the launch time, 1 second. On the face of it it is unsolvable. To do it you can only spec two (not three)of these variables.
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.

 

[JStephen]

I take the 1 second to be exactly what it says: "approximately 1 second to launch", not "accelerates for 1 second", which is different (and contradicts the rest of the problem). If you're shooting, your rounds per minute have nothing whatever to do with muzzle velocity. Large cannons might take a minute or more to fire a round, but the round sure doesn't spend a minute traveling leisurely down the barrel.

 

[dvd]

I think that a viable solution would be to take water balloons and catapult materials to the exam hall and quickly build a prototype and start launching balloons. (Remember, though, that you only have an hour.) It should help to relieve stress for all of your fellow examinees.