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]

JStephen's explanation of the problem matches my interpretation. However, I don't know why none of the multiple choice answers are within an order of magnitude of the correct answer.

 

[rb1957]

my problem with the "1 second" is that the flight time of the doomed balloon is 1.76 seconds.

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

 

[plasgears]

Have we overlooked that the springs discharge energy at ever decreasing force along the launch path? Average energy is determined at the half way point of launch.

 

[bvanhiel]

rb1957,

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

 

[rb1957]

hey, firing water ballons isn't my speciality (tho' it does sound interesting), but ...
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

 

[bvanhiel]

rb1957,

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

 

[tlee123]

rb1957,

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.

 

[rb1957]

i agree, in that the results calc that in 0.03sec the balloon has travelled 3 ft, but this is meant to be the time that the springs transfer their energy to the balloon ... but this result looks incompatable with the 5" deflection of the spring

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 ?

 

[JamesBarlow]

I agree that the energy method is what I would have done as well but I think I can see what the author was doing.

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.

 

[unclesyd]

Here is the practical aspects of such a question.

http://www.punkinchunkin.com/images/galleries/2005/Chuck2005/index.htm

 

[tlee123]

JamesBarlow,

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.

 

[telecomguy]

Obviously, it's a frozen water balloon...

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>

 

[jistre]

I, too, believe the details of the launch are inconsequential given the constraints of the problem, assuming there are no frictional losses and the only two sources of energy are gravity and the springs.

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.

 

[tlee123]

Using the energy based solution, the 5 inch acceleration phase takes slightly less than .004 seconds. This is so small compared to the 1 second "launch time."

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?).

 

[rb1957]

the springs in parallel allow the required force to be created within the paprmeters of the problem (5" deflection, known spring stiffness) ... stiffness of n springs is n*k

nothing "magical" about it (tho' there is a practical problem of getting all these springs at apply their forces to the balloon)

 

[telecomguy]

LOL! "And now people are saying that the balloon magically moves faster than the 3888 springs in parallel?"

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! D

<tg>

 

[tlee123]

No, rb1957, the "magical" part was proposed by JamesBarlow and seconded by telecomguy.

This is the quote that had me bothered:

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.

 

[rb1957]

i reckon we'll all failed this question ... even if we're right

 

[JamesBarlow]

You forgot to add the second part of my quote in.

"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.

 

[JamesBarlow]

I stand corrected.

GBor did solve the problem in a way that gave the solution the author presented.