#### GeorgianCMV

##### MCAT Studyaholic
10+ Year Member
5+ Year Member
This is from EK 1001 for physics, #254:

A man holds one end of a 30 cm spring in each hand. If he applies a 100N force to each end of the spring, by how much does he shorten the spring? k=1000N/m

The answer is 10cm...the explanation goes on to say that the length of the spring is irrelevant.

Well...I know that length is irrelevant, but wouldn't the answer be 20 cm since he's applying the force with each hand?????

#### Phlame217

##### PGY-1 IM
10+ Year Member
7+ Year Member
F=-kx so 100N=-1000N/M*(X) = .10 M; 0.10*2 = 0.20 M

I agree with you.

#### physics junkie

10+ Year Member
5+ Year Member
Every action has an equal and opposite reaction. If you think of him pushing the spring with the right hand if he doesn't apply a force with his left hand the spring isn't going to compress...it's just gonna move to the left. So the answer is 0.1.

Think of it another way. If you put 100N of force on a spring that was on the ground you would know that 0.1m is the right answer. But the ground is still exerting a force back on the spring. If it wasn't then the forces wouldn't be in equilibrium and then the object would be moving. Get it?

#### Kaustikos

##### Archerize It
Gold Donor
10+ Year Member
Every action has an equal and opposite reaction. If you think of him pushing the spring with the right hand if he doesn't apply a force with his left hand the spring isn't going to compress...it's just gonna move to the left. So the answer is 0.1.

Think of it another way. If you put 100N of force on a spring that was on the ground you would know that 0.1m is the right answer. But the ground is still exerting a force back on the spring. If it wasn't then the forces wouldn't be in equilibrium and then the object would be moving. Get it?

So then, what would happen in the case of a man applying 90N on one side and 50N on the other side?

Or 110N and 75N?

Is there a formula or something that takes into considering the two opposite forces? Because F = -kx isn't sufficient.

#### physics junkie

10+ Year Member
5+ Year Member
So then, what would happen in the case of a man applying 90N on one side and 50N on the other side?

Or 110N and 75N?

Is there a formula or something that takes into considering the two opposite forces? Because F = -kx isn't sufficient.
It is impossible to apply more force on one side than another. It would violate newton's third law.

#### physics junkie

10+ Year Member
5+ Year Member
To clarify, it would be impossible if you were standing still. You could, however, be running to the left with a force of 90N FROM the right on the spring. Then let's say you were applying 40N of force from the left to keep the spring partially compressed. Then the atmosphere would be supplying an additional 50N of force against the spring. That force would be dissipated as friction between your shoes and the ground.

#### Kaustikos

##### Archerize It
Gold Donor
10+ Year Member
NEVERMIND
I finally get it. Basically, the 100 N applied from one hand is resisting the motion that would be exhibited with the 100N applied from the right so that lets the 100N compress the spring.

God...

#### GeorgianCMV

##### MCAT Studyaholic
10+ Year Member
5+ Year Member
Geez louise. You know, physics junkie, I commend you...because I find physics THE most counterintuitive science. I can never grasp it!!!! Every time I do a conceptual question from EK1001 I'll tell myself what I think is the answer. And guess what? I'm always WRONG!

#### GeorgianCMV

##### MCAT Studyaholic
10+ Year Member
5+ Year Member
Honestly, I still don't get this!

##### No summer
Moderator Emeritus
10+ Year Member
5+ Year Member
Honestly, I still don't get this!
This problem is a subtle test of newton's third law. It is a good problem to really understand.
Think of it this way:

In the situation in the problem you are applying a force from each end of the spring. 100N from each end.

Now think of a situation with a spring affixed to a wall, and you apply a 100N force to the free end of the spring, compressing it.

What is the force of the wall on the affixed end of the spring in this situation?

Try and think about that question, and then re-approach the original problem. If you need further help, just ask. #### GeorgianCMV

##### MCAT Studyaholic
10+ Year Member
5+ Year Member

By Newton's third law, when you compress a spring against a wall, the wall is pushing back with the exact same force as you are applying (in the opposite direction). Therefore, if you are pushing with a 100N force, the wall is pushing back with a 100N force.

So, if you take the same spring and compress it with each hand 100N, then one of the hands is simply replacing the the force that the wall was giving, and it is essentially the exact same scenario. Is this right????

Seriously though, I can memorize Newton's third law and apply it to situations like this, but intuitively, this bothers me. No room for that on the MCAT though!!! #### physics junkie

10+ Year Member
5+ Year Member

By Newton's third law, when you compress a spring against a wall, the wall is pushing back with the exact same force as you are applying (in the opposite direction). Therefore, if you are pushing with a 100N force, the wall is pushing back with a 100N force.

So, if you take the same spring and compress it with each hand 100N, then one of the hands is simply replacing the the force that the wall was giving, and it is essentially the exact same scenario. Is this right????

Seriously though, I can memorize Newton's third law and apply it to situations like this, but intuitively, this bothers me. No room for that on the MCAT though!!! Yes, this is correct. If you draw a free-body diagram there is no difference between the two situations. I'm not surprised you missed this one it isn't exactly intuitive since you don't feel the force of a wall pushing back on you when you push on it, etc. Newton's 3rd law probably won't come up on the MCAT except in a "trick question" situation like this in which you would immediately be able to recognize it now that you've thought about the problem enough.

When I think about it further the difficult part to see is how if each action has an equal and opposite reaction how can you have motion? Your first instinct is that everything would be in equilibrium. http://mathforum.org/library/drmath/view/68399.html can clear this up for you.

As a side note, I really never grasped how gallileo realized that an object in motion stays in motion unless acted upon by an outside force. If you think about it for a second there is not a single observable situation where this is true in the realm of everyday experience. It was newton's similar capability for such raw abstraction that made him the greatest physicist of all time. Don't be too discouraged if you have difficulty following his logic with the third law.

#### GeorgianCMV

##### MCAT Studyaholic
10+ Year Member
5+ Year Member
Thank you so much. Yes, I agree that it is often counterintuitive. As much as I wish I could understand certain physics concepts, I have just succumbed to plain memorization in some cases. Thanks again.

Agreed. Don't sweat it GeorgianCMV. 