# 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. 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

Note it should read "to the right with an acceleration of $$3.00m/s^2$$

Solutions #1 !

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

Solution #2

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 Newton'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 Newton'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

• 1b) They're looking at both blocks together (note that the $m$ is 5+2), and when you add all the forces affecting both blocks together the two $f_s$s cancel. – dfan Dec 24 '11 at 3:22

As dfan posted in a comment, the solutions to these problems write Newton's second law for a system which includes multiple physical objects. Consider the proper interpretation of Newton's second law:

\begin{align}\sum F &= ma\\\sum(\text{forces on system}) &= (\text{mass of system})(\text{acceleration of system})\end{align}

In many cases, the "system" will be a single object, but it doesn't have to be. You can write Newton's second law for any system that moves with a uniform acceleration. When you do this, it's important to remember that $\sum F$ only counts external forces acting on the system. Internal forces exerted by one piece of the system on another piece of the system (like static friction, in example #1) are omitted from the sum.

One way you can see why only internal forces are omitted is by writing Newton's second law for each part of the system individually and then adding them up. Suppose your system has two parts, A and B. Writing Newton's second law separately for each part gives you

\begin{align}\sum \vec{F}_A = \vec{F}_{BA} + \sum \vec{F}_{\text{ext},A} &= m_A \vec{a}_A \\ \sum \vec{F}_B = \vec{F}_{AB} + \sum \vec{F}_{\text{ext},B} &= m_B \vec{a}_B\end{align}

where $\vec{F}_{BA}$ represents the force exerted by B on A, and $\sum\vec{F}_{\text{ext},A}$ represents the force exerted by external sources on A. This is just saying that the total (net) force on A, $\vec{F}_A$, is equal to the force on A exerted by B plus the force on A exerted by everything other than B, which should be pretty self-explanatory. (We assume that A does not exert a net force on itself, which follows from Newton's third law.)

If you assume that $\vec{a}_A = \vec{a}_B$ (which is a prerequisite to defining A and B as a single system), you can add these equations up and get

$$\vec{F}_{BA} + \vec{F}_{AB} + \sum\vec{F}_{\text{ext},A} + \sum\vec{F}_{\text{ext},B} = (m_A + m_B)\vec{a}$$

But Newton's third law tells you that the vector sum of a force and its reaction is zero; in other words,

$$\vec{F}_{BA} + \vec{F}_{AB} = 0$$

Also, the sum of all the external forces acting on A plus all the external forces acting on B is simply the sum of all the external forces acting on the system:

$$\sum\vec{F}_{\text{ext},A} + \sum\vec{F}_{\text{ext},B} = \sum\vec{F}_{\text{ext},A+B}$$

So the sum of the two Newton's second laws becomes

$$\sum\vec{F}_{\text{ext},A+B} = m_{A+B}\vec{a}$$

which is nothing more than Newton's second law applied to the whole system.

For a system which contains more than two parts, you can apply exactly the same procedure: just write Newton's second law for each part, and add them all up, canceling out all the internal forces using Newton's third law.

• Hold on, I thought you only label external forces that exert on the body in the force diagram (as they have done). Why would the static friction from the 2.00kg block be counted as "internal" suddenly? – Hawk Dec 24 '11 at 5:13
• I am also a bit confused with your argument on >I am also a bit confused with your argument in \begin{align}\sum \vec{F}_A = \vec{F}_{BA} + \sum \vec{F}_{\text{ext},A} &= m_A \vec{a}_A \\ \sum \vec{F}_B = \vec{F}_{AB} + \sum \vec{F}_{\text{ext},B} &= m_B \vec{a}_B\end{align}. Why do the action-reaction pair forces have the same sign? A and B are two different bodies, how could they have the same sign? – Hawk Dec 24 '11 at 5:23
• The designation of a force as "internal" vs. "external" depends on what you define as "the system." If you define the small block alone to be the system, then any force that acts on the small block (like $f_s$) is considered external. Similarly for the large block. But if you define both blocks to be the system, then forces exerted by one block on the other are not external to that system. (If you're drawing a FBD, the object in the FBD is the system) – David Z Dec 24 '11 at 5:27
• Also, what do you mean by "they have the same sign"? Remember that these forces are vectors, so they don't really have a sign, they have a direction. Do you understand that the force exerted by A on B has the same magnitude and opposite direction to the force exerted by B on A? – David Z Dec 24 '11 at 5:30
• >>The designation of a force as "internal" vs. "external" depends on what you define as "the system." If you define the small block alone to be the system, then any force that acts on the small block (like fs) is considered external. Similarly for the large block. But if you define both blocks to be the system, then forces exerted by one block on the other are not external to that system. (If you're drawing a FBD, the object in the FBD is the system) Having quite a bit of trouble of putting that yellow bar to quote. Anyways so glad I asked this because this was NOT in my book – Hawk Dec 24 '11 at 5:33