Armadillo Problem (Newton's Second Law)

In summary: Basically, the armadillo slips over the ice while being pushed by a force in the positive direction of the y-axis. The force of the wind is working against the downward force of the mass of the armadillo, so the armadillo's velocity is negative.
  • #1
Seraph404
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0

Homework Statement



For sport, a 12 kg armadillo runs onto a large pond of level, frictionless ice with an initial velocity of 5.0 m/s along the positive direction of an x-axis. Take its initial position on the ice as being the origin. It slips over the ice while being pushed by a wind with a force of 17N in the positive direction of the y-axis. In unit-vector notation, what are the animal's (a) velocity and (b) position vector when it has slid for 3.0 seconds?

Homework Equations



Fnet=ma

Kinematic Equations

Other equations depend on the free body diagram(s).


The Attempt at a Solution



I feel like I'm on the verge of getting this one; I just need a hint. ><

If the ice is frictionless, that means the armadillo is still sliding at a constant velocty of 5.0 m/s after he slips, right?

I think it's finding the upward velocity on part (a) that I'm kind of having trouble with.

What I did first was draw a free body diagram for the armadillo at the second position where two forces (I think) are working on the animal: The force of the wind (17N) upward, and mg downward. Was that correct, or am I missing one? I used a = (w - mg)/m (from w-mg = ma), where w is equal to the force of the wind, to find the acceleration, which came out as -8.38 m/s^2, for me. I really don't know if that's right or not. Then I tried to use the kinematic equation v = v0 + at to find the upward velocity. I rearranged the equation to v - at = v0; I remember from free-fall motion that when an object reaches its max upward height, v=0; also, I used the acceleration -8.38 m/s^2. My answer came out to be 4.26 m/s for the upward velocity, which is close to the right answer (to correct sig figs, the answer should be 4.2 m/s in the positive y direction). I still feel like I missed something. Can somebody show me where I might have gone wrong?

[Edit] Nevermind; I don't think this works after all. D:

Also, for part (b), can I just use kinematics to find x and y after 3.0 seconds? (I don't want to try it yet until I know I have part (a) right)
 
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  • #2
I don't think the question is saying that the wind is blowing upwards, rather it is blowing parallel to the pond and perpendicular to the x-axis. In otherwords, think about drawing an x and y-axis on the surface of the ice.
 
  • #3
Hootenanny said:
I don't think the question is saying that the wind is blowing upwards, rather it is blowing parallel to the pond and perpendicular to the x-axis. In otherwords, think about drawing an x and y-axis on the surface of the ice.

"Pushed by a force in the positive direction of the y-axis" doesn't sound like the wind is blowing parallel to the pond. =/
Even if it is initially blowing parallel, that doesn't matter; all that matters if the force that's applied to the armadillo.

Perhaps you could elaborate?

Better yet, what should my free body diagram for position 2 look like?
 
  • #4
Seraph404 said:
"Pushed by a force in the positive direction of the y-axis" doesn't sound lilke the wind is blowing parallel to the pond. =/
It depends how you orientate your axis; the standard way would be thus;

cararra_ch01_11.jpg
 
  • #5
Hootenanny said:
It depends how you orientate your axis; the standard way would be thus;

cararra_ch01_11.jpg

Well, this is only advanced high school physics, so I'm pretty sure we're sticking with our regular coordinate system where y is vertical and x is horizantal. I know for a fact my teacher wouldn't assign a problem where our axes were different without letting us know.

This problem doesn't even call for 3-D space; just vector components in the x and y directions (our coordinate system being like I said).
 
  • #6
Seraph404 said:
For sport, a 12 kg armadillo runs onto a large pond of level, frictionless ice with an initial velocity of 5.0 m/s along the positive direction of an x-axis. Take its initial position on the ice as being the origin. It slips over the ice while being pushed by a wind with a force of 17N in the positive direction of the y-axis. In unit-vector notation, what are the animal's (a) velocity and (b) position vector when it has slid for 3.0 seconds?
Slipping over the ice implies that the armadillo is still in contact with the ice and besides, is the wind blowing up through the ice?
 
  • #7
Hootenanny said:
Slipping over the ice implies that the armadillo is still in contact with the ice and besides, is the wind blowing up through the ice?

lol. The direction of the wind doesn't matter. It exerts an upward force on the armadillo. Honestly, wind usually spirals all over the place; it doesn't have a definite direction.

Basically, the armadillo is running with an initial velocity. As soon as it hits the ice, it trips (it goes a little up and a little forward; in other words, its velocity has an x component and a y component). Part (a) is asking for the velocity vector after 3 seconds.

The correct answer is given to me: (5.0 m/s)i + (4.2 m/s)j

It's not interpreting the problem that I can't figure out; it's really just setting up my free body diagram and finding the upward velocity.
 
  • #8
I disagree with your interpretation, how about we go with my interpretation and see if we get the right answer?
 
  • #9
Hootenanny said:
I disagree with your interpretation, how about we go with my interpreation and see if we get the right answer?


All right; I'll agree to that. (:

Although... getting the right answer doesn't exactly help me. It's knowing how to work the problem. You seem do physics differently from the curriculum I'm being taught.
 
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  • #10
So, if the y-axis is horizontal we can ignore gravity and the only force acting is the 17N wind force. So all you need to do is work out the acceleration and plug it into the appropriate kinematic equation to obtain the velocity in the y direction after three seconds.
 
  • #11
Hootenanny said:
So, if the y-axis is horizontal we can ignore gravity and the only force acting is the 17N wind force. So all you need to do is work out the acceleration and plug it into the appropriate kinematic equation to obtain the velocity in the y direction after three seconds.

I get -4.25 m/s

Close to what I was getting before. Still, I should get 4.2 m/s (after rounding to correct number of significant digits).

Meh. I'll just find out the solution in school and then post it back on here...unless someone else still once to try. We really don't do Physics that way in my class. Maybe you living in the UK has something to do with it.

I appreciate your attempt to help me, though. (:
 
  • #12
Wait, how did you get a negative answer? The force is applied in the positive y direction.
 
  • #13
Hootenanny said:
Wait, how did you get a negative answer? The force is applied in the positive y direction.

It doensn't matter. I already know the answers to the problem. The way you're working it just doesn't make sense to me: y has always been vertical. Like I said, I'll post the solution I get from school, and then perhaps you can explain to me this y-being-horizantal business and how it gives you the right answer. Does that sound like a decent plan?

What I would really love someone to do is to look through my attempt and show me my errors.
 
  • #14
Seraph404 said:
What I would really love someone to do is to look through my attempt and show me my errors.
He's one:

You have a negative vertical acceleration, that means that your armadillo would be accelerating vertically downwards, i.e. through the ice!

Also, you have a negative acceleration and yet a positive final velocity, how does that one work?
 

FAQ: Armadillo Problem (Newton's Second Law)

1. What is the Armadillo Problem?

The Armadillo Problem, also known as Newton's Second Law of Motion, is a physics concept that describes the relationship between an object's mass, acceleration, and the force acting upon it. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

2. How is the Armadillo Problem related to Newton's Second Law?

The Armadillo Problem is another name for Newton's Second Law of Motion, which is one of three fundamental laws of motion developed by Sir Isaac Newton. It is named after the animal because it helps explain how objects of different sizes and masses can have the same acceleration under the influence of a force.

3. What is the formula for the Armadillo Problem?

The formula for the Armadillo Problem is F=ma, where F represents force, m represents mass, and a represents acceleration. This formula can be used to calculate any of the three variables as long as the other two are known.

4. How does the Armadillo Problem apply to real-life situations?

The Armadillo Problem can be applied to a wide range of real-life situations, such as calculating the force needed to push a heavy object, understanding how cars accelerate and decelerate, and predicting the motion of objects in space. It is also used in engineering and design to create structures and machines that can withstand certain forces.

5. What is an example of the Armadillo Problem in action?

A classic example of the Armadillo Problem is a person pushing a shopping cart. The person's force, or push, causes the cart to accelerate, and the amount of acceleration depends on the mass of the cart. If the cart is empty, it will accelerate quickly, but if it is full of heavy items, it will accelerate more slowly due to its increased mass.

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