The Forces present on a system drawn with FBD's

  • Thread starter Mr LoganC
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Well, to start, I am actually in my second year of doing my Honors in Physics but am helping a first year student in understanding some of the concepts in her physics class. This question came up, and I myself cannot seem to find a solution to it. I'm quite puzzled about it....

1. The problem statement, all variables and given/known data
There are 3 boxes, each with separate masses, all touch in a line.
A force of magnitude F pushes horizontally to the right against box 1 and pushes the boxes.
See diagram below:
threeboxes.gif


We need to draw a free body diagram, as worded directly from the question, of all of the individual forces acting on the system consisting of boxes 2 and 3 together. Indicate the five forces listed below by choosing from the vectors in the diagram below. If none are present, then "N" is an option.
directions.gif


i) box 2 pushing on ground
ii)the applied force pushing on block 1
iii)the applied force pushing on block 2
iv)floor pushing on box 2
v)floor pushing on box 1

2. Relevant equations
Well, we know from Newton's 3rd law that all forces have an action/reaction pair that are equal and opposite in direction. Well also know that not only is the applied horizontal force present, but also the gravitational force.


3. The attempt at a solution
At first glance it seemed quite simple and straight forward!
i) Box 2 on ground = C from gravitational force
ii) F on block 1 = B
iii) F on block 2 = N because that is not the applied force. That would be F1 on 2
iv)Floor on box 2 = A which is the normal force that the floor pushes back with. (The action/reaction pair of gravity)
v)Floor on box 1 = A. Same logic as previous one.

Needless to say, this did not seem to work. So I tried multiple other combinations taking into account that the directions say that the system consists of boxes 2 and 3 together. I thought that perhaps box 1 was not part of the FBD. I also thought that there may be some confusing bits in the fact that it states some as "boxes" and some as "blocks"

Either way, I can't seem to get it to work. What's worse is that what I thought was fairly simple Newtonian physics, is stopping me in my tracks!

Any insight or perhaps something really simple and back to basics that I am forgetting?
 
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PhanthomJay

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Yes, according to wording of the problem, I would say that block 1 is not part of the system you are looking at.
What do you get for answers when you consider the system of blocks 2 and 3 together only?
 
When I considered that, I tried: C,N,N,A,N which didn't work.
I also tried C,N,B,A,N. I choose B thinking that perhaps since block 1 was not in the equation, that F1 on 2 would be the new applied force. But unfortunately, that didn't work either.
 

PhanthomJay

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I get your first answer. Note, however, that for (i), C is correct, but it is from the Normal force of B on the ground, not the gravitational force of B on the ground.
 
Right, right, okay.
So would you think that there is the possibility of the questions answers being incorrect? Because I can't think of any other way that it'd work.
 

PhanthomJay

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Yes, they talk about system 2 and 3, but ask questions about system 1 and 2. Looks like someone made an error. What answers do you get if you look at system 1 and 2? Maybe that works. Maybe not.
 
I got C,B,N,A,A if I take into account both systems.
Which again, does not seem to work...and I'm out of ideas! Maybe I'll give the professor an email and just ask what he thinks about.
Thanks for the help!
 

PhanthomJay

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I got C,B,N,A,A if I take into account both systems.
Which again, does not seem to work...and I'm out of ideas! Maybe I'll give the professor an email and just ask what he thinks about.
Thanks for the help!
Oh i get it now, sorry. The problem asks for the free body diagram of the system 2 and 3. For (i) , although the block pushes down on the ground, that's a free body of the ground. They want the free body of the blocks. :rolleyes:
 
In that case, I would get N,B,N,A,A which also was not correct.
And in the case that neither of the last two (floor pushing on them) are part of the FBD, because the floor itself is not in the diagram, I also tried N,B,N,N,N which didn't work.

Is that what you would have gotten also?

Perhaps N,B,B,A,A??? Even though the applied force is not on block 2, in the FBD of 2, there is still a force on it pushing in the B direction.
 

PhanthomJay

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In that case, I would get N,B,N,A,A which also was not correct.
And in the case that neither of the last two (floor pushing on them) are part of the FBD, because the floor itself is not in the diagram, I also tried N,B,N,N,N which didn't work.
I'm getting lost in these letters. The floor pushing on 2 is part of the FBD of 2/3, but the floor push on 1 is not part of the FBD of 2/3. Although the floor is not part of the FBD of 2/3, its contact force IS part of it.
Is that what you would have gotten also?
No.
Perhaps N,B,B,A,A??? Even though the applied force is not on block 2, in the FBD of 2, there is still a force on it pushing in the B direction.
There is a force pushing on 2 in the B direction, but that force, as you noted earlier, F_12, the normal force of 1 on 2, is not the 'applied' force.
 
Oh right! I forgot about the whole "Only boxes 2 & 3 together" thing!

So in that case I tried N, N, N, A, N. Meaning the only force present on the diagram was the force of the floor pushing on box 2 in the vertical (A) direction.
And it worked!!! That was the solution!

So thank you very much for your help! I never would have thought of the whole forces on the ground not in the FBD detail!

Thanks again,
-LoganC
 

PhanthomJay

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Yeah, I first missed it too...good thing you got it, there are 3125 ways of combining the letters...you did it in only after a half dozen tries...which shows that you cannot get this problem correct by guessing, unless you're real real lucky:wink:
 

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