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F = MA 2010 #10 (Two blocks on top of each other, one is being pulled) 
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#1
Jan2813, 03:29 PM

P: 112

1. The problem statement, all variables and given/known data
See: http://www.aapt.org/physicsteam/2010...aSolutions.pdf #10 EDIT: There are several constants. 2. Relevant equations F = ma Ff = μN 3. The attempt at a solution First: Draw a freebody diagram for block 2 (the bottom block) Up: Normal force from floor pushing on block 2 Down: weight of block 1 on block 2 Down: weight of block 2 down Right: Force F Left: Force of Friction Ff Up forces = Down forces N = m1g + m2g N = g(m1 + m2) Ff = μg(m1+m2) F_y = 0 F_x = F_net so Fnet = F  Ff m2a = F  ug(m1+m2) a = (F  ug(m1+m2)/ m2 This is the wrong answer. The answer says it should be 2m1 + m2 not m1 + m2. Why is this? 


#2
Jan2813, 04:18 PM

P: 1,100

I think your mistake is simple, look at your calculation for the normal forces. Make sure you get the masses right, mass 1 is pushing down on mass 2. So at the point where mass 2 contacts table, how much mass is pressing down? What normal force must be needed to counteract this? 


#3
Jan2813, 04:22 PM

P: 112

I don't see what you're saying. Mass 1 is pushing down on mass 2  Yes, so there is a force m1g down. Mass 2 is also pushing on the table, so there is a force m2g down. Shouldn't that be it?
So: F_down = m1g + m2g? Where does the 2nd m1g come from? Could you please explain? 


#4
Jan2813, 04:30 PM

P: 1,100

F = MA 2010 #10 (Two blocks on top of each other, one is being pulled)
If what you said was true, I could put my hand under mass 2 and have you increase mass 1 ten times and not feel a difference. If I am trying to feel the force under mass 2, what do I feel? What if you climbed on top of the whole thing and became mass 3, would that change the force at the table?



#5
Jan2813, 04:32 PM

P: 112

So is what your saying that the table has to overcome each of the three following forces:
1) mass 2 gravity force 2) mass 1 push on mass 2 3) mass 1 gravity force This gives 2m1 + m2 and thus the right answer. But I still can't picture why you need to account for both mass 1 push on mass 2 and mass 1 gravity force both. 


#6
Jan2813, 04:40 PM

P: 1,100

Consider some extreme cases to get an intuition about it. Replace mass 1 with a massive block from stonehenge. Clearly if the force at the table / mass 2 interface will increase. It would crush you if you were under it, not matter how small mass 2 is. Two surfaces, two normal forces, each depend on ALL the mass above them (because of gravity's direction) 


#7
Jan2813, 04:46 PM

P: 112

Ok, so you can kind of view it as one big system of mass m1 + m2. And then include the force between the m1 and m2 interface. That seems much more logical to me. Thanks.



#8
Jan2813, 05:00 PM

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For block 2, there are two different normal forces; one on the upper surface, and one on the lower surface. The first thing I would do is draw a freebodydiagram for Block 1. It's a fairly simple diagram: gravity, m_{1}g, points down, normal force, N_{1} points up, and friction, f_{1} points to the right. The resulting equations are straight forward: N_{1}=m_{1}gThe frictional force, f_{1}, which gives Block 1 its acceleration, is exerted on Block 1 by Block 2. Therefore, according tp Newton's 3rd Law, Block 1 exerts a force on Block 2, equal to f_{1} and to the left. This is the force you neglected and it's equal to μm_{1}g . 


#9
Jan2813, 05:18 PM

P: 112

Oh thanks so much  I didn't see that before.
So on block 2 we have: Friction from block 1 equal to μm1g on the left Friction from the ground equal to μ(m1+m2)g on the left Force F going to the right F  g(2m1 + m2) Thanks. I finally get it. Essentially all I was missing was that the friction force between one and two needs to be accounted for. THIS is where the extra m1 term comes from. 


#10
Jan2813, 06:07 PM

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Well, I glad I went ahead and replied. 


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