Why is static friction omitted here? And comparison with this problem on Newton's second Law. Solutions are all provided
The problem itself is short, but my concept question is...long. So be prepared. 1018
Note that problem 1 and problem 2 are related to my concept questions, so I am not taking advantage of asking two in one.
Problem #1
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Note it should read "to the right with an acceleration of $3.00m/s^2$
Solutions #1
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Concept Questions #1
a) In the free-body diagram for the 5kg mass, I am guessing $n_1$ is the gravitational force from the 2kg
b) Now if you look carefully in the solutions, you noticed that the static friction between the 2kg and 5kg mass is ignored.
$\sum F_x = F - \mu_k n_2 = ma$
But shouldn't it actually be?
$\sum F_x = F - (f_s + f_k) = ma$
If not, why was it ignored?
Problem #2
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Solution #2
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Concept Question #2
a) For the cart, what happened to the force exerted by the hanging mass on the cart in the free-body diagram?
b) How could the mass on the top have an acceleration and the hanging mass NOT accelerate? I ask this because they had the condition
$T = m_1 a$
$T - m_2 g = 0$
General Concept Question (both problems)
This is mostly concerned with <PERSON>'s second law for both the masses in the bottom cart/box.
How come on the "ma" side of the equation, we set it equal to $(\sum m)a$. I thought in <PERSON>'s second law, we set it equal to the single mass alone? I understand that in all the problems, they treated all the mass stacked as a single mass, but how do you do it systematically like I do? The solutions given kinda skipped a lot of trivial things, and I am not used to it.
For instance, for Problem #1, for the mass on the bottom. Shouldn't my system of equations be
$\begin{cases} \sum F_{2x} = f_s = 2a \ \sum F_{5x} = F - (f_s + f_k) = 5a \end{cases}.$
Compared with Problem 2, if I had added the force exerted by $m_2$ on M, then it would be (and i had set the Ma side to only a single mass)
For M
$\begin{cases} \sum F_{x} = F - {F_{2M}} = Ma \ \sum F_{y} = n = (M + m_1 + m_2)g \end{cases}$
For $m_1$
$\begin{cases} \sum F_{x} = T = m_1a \ \sum F_{y} = n_1 = m_1 g \end{cases}$
For $m_2$
$\begin{cases} \sum F_{x} = {F_{M2}} = m_2a \ \sum F_{y} = T = m_2 g \end{cases}$
**Note that $F_{2M}$ means the normal/contact force exerted by $m_2$ on to mass M and $F_{M2}$ is the force exerted by mass M on $m_2$
I am so sorry for making it this long and I realize normally I can only ask quick concept checks, but I feel like if I overcome this concept I can handle any Newtonian problems with forces. This is not homework, this is just self-study (it is the holidays afterall!).
Thank you so much for reading 512