Stacked Blocks - Video Solution

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@labibaraidah4017 3 ай бұрын

Sir,I just added the friction force between the two blocks too.Because when we will push the bottom block,it will feel some trouble moving because of the friction force in its top so we will have to minus that from the given force.Am I wrong?

@lockinatti Жыл бұрын

"Any force larger than F_f_s_max overcomes the force of static friction and causes sliding to occur. The instant sliding occurs, static friction is no longer applicable" - Wikipedia. Im confused since the applied force F seems to be greater than F_f_s in your equation which would imply slipping right? I follow your logic but i don't understand how they ever move together without slipping because whatever force you apply to the bottom block will be completely apposed by friction until the force you apply becomes larger than F_s_max, at which point the bottom block will move but there will be slipping. My logic says if the bottom block is moving, there is slipping.

@levitopher Жыл бұрын

The ground is frictionless. The pushing force could be as small as possible (say, 0.5 N), and the system would accelerate (both blocks together) with some small acceleration. But in this problem, I'm trying to find the largest pushing force *before* the top block starts slipping off the bottom. That's the inequality I write around 6:45 - that's the condition for the top block to not slip. When I take the maximum value of that inequality, I'm taking the maximum pushing force before the top block starts to slip. Does that help? It's not "whatever force you apply to the bottom block", I can apply any force from 0 to F_max and the blocks will acceleration across the floor, but the top won't slip off the bottom.

@lockinatti Жыл бұрын

@@levitopherThank you for your response. I see how when I look at both blocks together as a system, even the smallest pushing force would cause acceleration. However, when I look at the equation for just the bottom block F-F_f_s = m_1a_1 and consider this from wiki on friction "Any force smaller than F_s_max attempting to slide one surface over the other is opposed by a frictional force of equal magnitude and opposite direction" I conclude that when looking at the bottom block, it will not accelerate if the pushing force is less that F_s_max and if the pushing force is more than that, it will accelerate but it will slip. Do you see my argument when looking at just the force equation for the bottom block? Where am I mistaken? thank you very much

@hannahxie7414 2 жыл бұрын

My lifesaver. THANK YOU SO MUCH. This is so clear

@blayses3116 5 ай бұрын

Hi! How did you get the +1 part?

@levitopher 5 ай бұрын

you mean around 6:11 right? Yeah so that's from algebraic simplification, in (m1/m2)*F_s + F_s, both terms have F_s in them so you can pull that out and write (m1/m2+1)*F_s. Makes sense?

@blayses3116 5 ай бұрын

@@levitopher Oh I see it now. Thank you!

@kingpusspuss1527 Жыл бұрын

How did you get (m1/m2 + 1) Fs 1(2) from the simplification on 6:14?

@simon2046 Жыл бұрын

I think it's supposed to be -1 instead bc of newton's third law

@levitopher Жыл бұрын

In my notation, F^{1(2)}=F^{2(1)}, with the negative sign from Newton's third law being taken care in that previous line.

@paulie2641 2 жыл бұрын

To make it simpler, the max force b4 m2 moves is equal to the friction force. So, the acceleration of m2 or the entire system a = (mu * m2 * g)/m2 = mu * g. Now, consider a force to move the entire system m1 & m2 with a (mu * g). Since there is no friction under m1 and (m1+m2) move together, we will get F = (m1 + m2) * a. If you substitute a with (mu * g), you will get the same answer (F) as shown in the clip.

@levitopher 2 жыл бұрын

I think that logical chain is not as simple as the application of Newton's 2nd law I use here, because it relies on your intuition about several different conditional forces and accelerations. But it's still correct of course!