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Reason for inter-box force size when force is applied to one side of a line of boxes?

by InvisibleMan1
Tags: applied, boxes, force, interbox, line, size
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InvisibleMan1
#1
Jul21-10, 02:29 PM
P: 40
1. The problem statement, all variables and given/known data
Note: To make this easier, gravity, air resistance, and friction do not exist in this problem.

Given two boxes which are sitting directly next to eachother, and a force applied to the leftmost box in the rightward direction (straight at the other box), are there any rules/formulas for the amount of force which the leftmost box applies to the box it is sitting next to?

Example:


What causes the 1N force being applied to A to be smaller when A applies it to B?


2. Relevant equations



3. The attempt at a solution
I have a vague theory that it has something to do with A "absorbing" some of the force due to its mass, but I haven't been able to get any further with this.

As you can see, I can actually calculate what these values are, but the problem is that I don't understand why they are the way they are. It seems to me that if a 1N force is being applied in a rightward direction on A, then A would apply a 1N rightward force on the block to its right, which is B. However, this doesn't seem to be the case, at least not when viewed like this.

I tried attacking this with the assumption that a 1N force is actually being applied to B, but is being partially counteracted somewhere. I didn't get very far with this approach either.

I think I simply need someone to explain how this works, step by step.
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6Stang7
#2
Jul21-10, 03:16 PM
P: 213
It's because there system moves as a whole unit. IF box A applied 1N instead of some value less then one, it's acceleration would be greater than box A, which means box B would move faster than box A, but we know that can't happen since the boxes are always touching.
InvisibleMan1
#3
Jul21-10, 03:24 PM
P: 40
Quote Quote by 6Stang7 View Post
It's because there system moves as a whole unit. IF box A applied 1N instead of some value less then one, it's acceleration would be greater than box A, which means box B would move faster than box A, but we know that can't happen since the boxes are always touching.
Thanks for the info, but it is still too vague... I've been sitting here thinking of ways to use that knowledge to solve the problem, but I haven't hit on anything.

I'm looking for a reason why exactly the force decreases. What prevents the force of 1N from being applied to object B? Sure object B would accelerate too far, but that's just an effect, not the cause. Object A isn't worried about keeping object B's acceleration down, so something must be preventing object A from applying the full force.

PhanthomJay
#4
Jul21-10, 03:47 PM
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Reason for inter-box force size when force is applied to one side of a line of boxes?

Although the magnitude and direction of the forces shown in your FBD's are correct, you should show them on the sides on which they act. It may be helpful to see what is happening . On Block A, the 1 N acts on the left side of the block and points right toward the block; the 1/3 N acts on the right side of the block and points left toward the block. Then on block B, per newton 3, the 1/3 N force acts on the left side, pointing right, toward the block, and there is no force on the right side. You still get the same net force, but with a better picture. Beyond that, as 6Stang7 has noted, the 1 N force accelerates all 4 blocks, per F_net = 30a (newton 2). Since each block has the same acceleration as the entire system of blocks, then again using F_net = ma for each block, F_net on each block must be less.
InvisibleMan1
#5
Jul21-10, 03:55 PM
P: 40
Quote Quote by PhanthomJay View Post
Although the magnitude and direction of the forces shown in your FBD's are correct, you should show them on the sides on which they act. It may be helpful to see what is happening . On Block A, the 1 N acts on the left side of the block and points right toward the block; the 1/3 N acts on the right side of the block and points left toward the block. Then on block B, per newton 3, the 1/3 N force acts on the left side, pointing right, toward the block, and there is no force on the right side. You still get the same net force, but with a better picture. Beyond that, as 6Stang7 has noted, the 1 N force accelerates all 4 blocks, per F_net = 30a (newton 2). Since each block has the same acceleration as the entire system of blocks, then again using F_net = ma for each block, F_net on each block must be less.
Here is the new version of the example:


Thanks for the tip, I'll probably use this method from now on.

However, the main issue still stands: Exactly what is causing A to apply a smaller force on B? As I said in a previous post, A doesn't care if B accelerates too far. Something must be preventing A from applying the full 1N of force to B... This is where my theory about A "absorbing" some of the force comes from, but that is only a guess and I do not know of any rules to support that theory.
PhanthomJay
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Jul21-10, 05:35 PM
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Consider, if you will, those 2 blocks , with the 1 N force applied to the left block, but let's assume this time that block 2 is tight up against a wall. In this case, there is no acceleration, the blocks are at rest, and A does apply the full 1N force to B. In your case, however, there is acceleration, and due to the lower mass (a measure of its resistance to motion) of the individul blocks compared to the combined system of blocks, Newton 2 demands that the net force acting on it must be less (less mass, less resistance to motion, less force required to accelerate it at the same rate).
InvisibleMan1
#7
Jul21-10, 06:11 PM
P: 40
Quote Quote by PhanthomJay View Post
Consider, if you will, those 2 blocks , with the 1 N force applied to the left block, but let's assume this time that block 2 is tight up against a wall. In this case, there is no acceleration, the blocks are at rest, and A does apply the full 1N force to B. In your case, however, there is acceleration, and due to the lower mass (a measure of its resistance to motion) of the individul blocks compared to the combined system of blocks, Newton 2 demands that the net force acting on it must be less (less mass, less resistance to motion, less force required to accelerate it at the same rate).
Hmm, alright lets go with your wall example.


What happens the instant the wall is removed (assuming the wall completely vanishes in an instant)?
PhanthomJay
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Jul21-10, 06:43 PM
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Nothing happens in an instant, but assumimg the wall is removed almost instantaneously, then the forces between the blocks reduce almost instantaneously to your original example, and the blocks accelerate, per newton 2.
InvisibleMan1
#9
Jul21-10, 10:49 PM
P: 40
Hmmm... I still don't understand how this all works. Trying to understand collisions is giving me quite a headache... I'm going to try reading a different article on physics. Thanks for the help though.
housemartin
#10
Jul22-10, 02:26 AM
P: 87
If block A (20 kg) would apply tha same 1 N force to block B (10 kg), what force will B exert on A then? According to newtons third law - 1 N, so the net force on block A is +1N - 1N = 0N, net force on block B then will be 1N. So according to this, you push block A, but it doesnt move at all, instead block B starts to move away. Do you expect this kind of behavior?
Well I'm not good at physics ;], but what you have here is just Newton's second and third laws. I don't know either why they are truth, maybe no one knows.
And I think that collisions its quite a different story (maybe not so much). There you have force over short time, and in most problems, block A will come to block B with constant velocity (that is no force on A at the start). Anyway, good luck ;]
PhanthomJay
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Jul22-10, 10:47 AM
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Quote Quote by InvisibleMan1 View Post
Trying to understand collisions is giving me quite a headache...
As housemartin noted, this is NOT a collision problem. Collisions involve variable internal impact forces acting over short time intervals (generally milliseconds), with at least one object in motion prior to the collision, and typically with no external forces applied. This problem is quite different.
InvisibleMan1
#12
Jul23-10, 07:44 AM
P: 40
Quote Quote by PhanthomJay View Post
As housemartin noted, this is NOT a collision problem. Collisions involve variable internal impact forces acting over short time intervals (generally milliseconds), with at least one object in motion prior to the collision, and typically with no external forces applied. This problem is quite different.
The purpose behind all of my threads in this forum so far has been to understand collisions better. This is a form of collision in my opinion, which is why I posted about it (even though the two objects are touching, a force is applied to one object which causes it to "collide" with the other object and set them both in motion at an equal acceleration). Collisions have been, in general, quite difficult for me to understand. I suspect this is due to the material I was studying, which is why I am moving to a different set of physics articles.

Again, thanks for the help though.


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