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Newton's laws 2 
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#1
Aug1109, 11:15 PM

P: 38

1. The problem statement, all variables and given/known data
Three blocks are in contact with each other on a frictionless horizontal surface as shown. The masses of the blocks are m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 3.0 kg. A horizontal force F = 24 N is applied to m1. a. Find the acceleration of the three blocks. b. Find the net force on each block. c. Find the magnitudes of the contact forces between the blocks. 2. Relevant equations Not sure.. [tex]\Sigma[/tex]F = F1+F2+F3 3. The attempt at a solution a) F = a (1.0kg + 2.0 kg + 3.0 kg) 24N = a (6 kg) 24N / 6 kg = a 4 m/s^{2} = a b) F1 = 1kg (4 m/s2) = 4N F2 = 2kg (4 m/s2) = 8N F3 = 3kg (4 m/s2) = 12N c) F  F1 = 24N  4N = 20N (contact force between F1 & F2) F  F1  F2 = 24N  4N  8N = 12N (contact force between F2 & F3)  wow I feel lost after what I did.. Does this make sense? 


#2
Aug1209, 12:03 AM

P: 225

a) and b) look correct; however, contact forces isn't ringing a bell, so i can't say as much about c).



#3
Aug1209, 12:34 AM

P: 128

a) and b) are correct. As for contact force I presume the question is asking for the normal reaction, in which case c) is not correct. How would you calculate the normal reaction of the plane on the 1 kg, 2 kg and 3kg block?



#4
Aug1209, 11:42 AM

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Newton's laws 2
In part (c), the problem is asking for the contact force between each block, that is, what is the normal force (horizontal in this case) that block 1 exerts on block 2 (and vice versa, via Newton 3), and what is the normal (horizontal) contact force that block 2 exerts on block 3. So actualy, you have answered parts a, b, and c, correctly, BUT, your methodology is shaky at best. You should get used to drawing free body diagrams in problems such as this, noting all forces that act on the body and then using Newton's laws. In part (c), isolate Block 1 in a FBD. You know from Newton 2 that the sum of forces (the net force) acting on the body in the horizontal direction = ma. Thus, [tex]F  F_{21} = ma[/tex], where [tex]F_{21}[/tex] is the normal force of block 2 acting on block 1. Accordingly, [tex] 24  F_{21} = (1)(4)[/tex], from which, [tex]F_{21} = 20 N[/tex]. And now isolate block 3, and the only force acting on it in the horizontal direction is the normal force of block 2 on block 3, thus [tex] F_{23} = ma[/tex], from which [tex] F_{23} = (3)(4) = 12 N [/tex]. As a check, you can isolate block 2, or blocks 1 and 2 together, to achieve the same result. Again, your answers are correct, but I want to be sure you understand the reasoning behind the answers.



#5
Aug1209, 01:37 PM

P: 38

When I drew the blocks on a piece of paper, I added opposite arrows between the blocks. Is that what you mean by "free body diagrams"? And whats FBD?
Im not used to working with symbols but it is definitely much clearer. Thanks for the break down of the problem! :) 


#6
Aug1209, 02:05 PM

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Oh, sorry, FBD is an abbreviation for "Free Body Diagram". You should 'google' on 'Free Body Diagrams' for tons of examples. Basically the FBD shows all forces contact forces , gravity forces, etc....acting on the 'isolated' object or system of objects. It is usually best to look at the forces in the x and y directions separately, and apply Newton's laws in each direction, as necessary. In part 'a' of your problem, your FBD was the entire system of the 3 blocks. Since the only force in the horizontal direction acting on that system is 24 N (you don't look at the forces internal to that system, like the internal contact forces between the blocks), you used Newton 2 to get 24 = 6a, a=4. Now when you look at block 1 by itself, in the horizontal direction, isolating it from the rest of the blocks, you should note that the 24N force still acts on the left face of the block, pointing to the right, and the contact force between the blocks is now considered an external force to that isolated block, acting on the right face of the block, pointing left, which I think you have correctly noted. Solve for it as previously shown.



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