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This problem is similar, but also different question from my previous question. They are both unfortunately long.

Problem/Solution #1

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Quick Concept Check

a) Could someone briefly explain to me why the block on top can accelerate, but the one hanging on the pulley does not?

Problem/Solution #2

Consider the set up below. What is the minimum force required so that the 3kg block remains on the 8kg block? Given that the coefficient of static friction between the blocks is 0.8 and the coefficient of kinetic friction between the 8kg block and the surface is 0.4.

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$\sum F_x' = F - (N' + \mu_k n) = 8a$

$\sum F_y' = n = (8g + \mu_ s N)$

$\sum F_x = N' = 3a$

$\sum F_y = \mu_s N = 3g$

Solving the systems of equations above, you should get that $F_{min} = 11g(\mu_k + \frac{1}{\mu_s}) = 177.87N$

Point of the question and why the heck is this so long

Notice how the 3kg block is actually "attached" tot he 8kg block, yet the solution here didn't include it in its free-body diagram and they even included the reaction force from the 3kg on the 8kg.

Compared with the cart problem where the hanging mass is also touching the cart, no reaction force was drawn and they even treated the three mass as a single mass. David explained this to me last night and I thought I got it, but I went to bed and started thinking about it and it became even more confusing!

Could you have solved the first problem with the cart WITHOUT taking all three mass into the cart's free body diagram? Is there another way of applying Newton's Second Law to the cart problem?

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For your concept question (a), the hanging mass cannot accelerate DOWNWARDS because then it would be moving with respect to the cart! and the sliding mass must accelerate with respect to ground so that on top of the moving cart it is stationary! –  Chris Gerig Dec 25 '11 at 5:08
    
As for question#2, the FBD does include the fact that the block is "attached", via the normal forces and friction forces! And these solutions are the natural way to solve these problems, so I don't consider them "long" since that word has relative meaning and we haven't compared these solutions to others. And you need to take all three masses into account in order to apply Newton's Law with the force F (i.e. it doesn't just push on mass M). –  Chris Gerig Dec 25 '11 at 5:12
    
On (a), aren't the hanging mass and the top mass TIED together? The hanging mass will still have motion right? I just don't see how it is NOT possible for the hanging mass to have no motion. –  jip Dec 25 '11 at 5:19
    
On Question 2, I am asking why they didn't include the gravitational force on the 8kg block. Because it is "attached" to it right? It's similiar to the cart problem with the hanging mass. They attached it. –  jip Dec 25 '11 at 5:21
    
For (a), yes the masses are tied together, they DON'T MOVE (with respect to the cart), but they both 'move to the right' with respect to the ground. For question#2, the gravitational force is absorbed in the normal force. –  Chris Gerig Dec 25 '11 at 6:34

1 Answer 1

up vote 2 down vote accepted

a) Cuold someone briefly explain to me why the block on top can accelerate, but the one hanging on the pulley does not?

Both blocks do accelerate, and in fact they have the same acceleration. You'll see that if you write out the full set of both the x and y components of Newton's second law for each system (i.e. for each block). In the given solution, they only wrote out the components that are directly needed to solve the problem.

Notice how the 3kg block is actually "attached" tot he 8kg block, yet the solution here didn't include it in its free-body diagram and they even included the reaction force from the 3kg on the 8kg.

Compared with the cart problem where the hanging mass is also touching the cart, no reaction force was drawn and they even treated the three mass as a single mass.

It's not attached. The only interactions between the 8kg and 3kg blocks are the normal force and static friction, neither of which qualifies as "attachment." This is exactly analogous to the interaction between $M$ and $m_2$ in the first problem. The reason no reaction force was drawn in the first case (by "reaction force" I assume you are referring to the reaction to the normal force acting on $m_2$) was that they didn't draw a free body diagram for the object the reaction force acts on, namely $M$. Again, the given solution omits some pieces (in this case, a free body diagram) which aren't necessary for solving the problem. If you draw a free body diagram for $M$, you will see the reaction force.

Could you have solved the first problem with the cart WITHOUT taking all three mass into the cart's FBD? Is there another way of applying Newton's Second Law to the cart problem?

If by this you are asking whether you could have solved the first problem by drawing separate free body diagrams for each of the three masses $m_1$, $m_2$, and $M$, then yes, you certainly can.

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On a), not sure what yuo mean, but they clearly set a_y = 0 –  jip Dec 25 '11 at 5:27
    
No the reaction force I am referring to is the force exerted by $m_2$ on M. –  jip Dec 25 '11 at 5:29
    
"If by this you are asking whether you could have solved the first problem by drawing separate free body diagrams for each of the three masses m1, m2, and M, then yes, you certainly can." I actually did it this, but I found no way of including $m_1$ in my $F_{horizontal} = \sum M a$ since $m_1$ only acts in the vertical direction with big M. –  jip Dec 25 '11 at 5:30
    
Meant to say $(\sum M)a$ –  jip Dec 25 '11 at 5:45
    
(1) Write out Newton's second law for all three objects. I see you've edited the question to contain your free body diagrams; now edit in the equations as well. (2) Yes, that's the force I'm talking about. It's the one labeled $F_2'$ in your diagrams. (3) As I said, include the equations. While you're at it, maybe you should show your attempt at solving the system of equations as well. –  David Z Dec 25 '11 at 6:07

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