Ice skaters vs. Blocks

Question

Imagine two ice-skaters, $A$ and $B$. $A$ pushes off $B$ with a force $F_{a}$, say $160\:\mathrm{N}$. If the mass of $A$ is $40\:\mathrm{kg}$ and the mass of $B$ is $60\:\mathrm{kg}$, what is the acceleration of person $A$?

I think I understand how to solve this problem by using Newton's third law: both people will feel a force of $100\:\mathrm{N}$ in opposite directions. So, if I understand, the acceleration on $A$ is :

\begin{align} a &= \frac{F_{a}}{m_A}\\ &= \frac{160}{40}\\ &= 4 \end{align}

Correct me if I am wrong with this because I am still not certain.

However, here is where I get another question. In my physics class, my teacher explained you could solve force problems by either finding the free-body of the entire system (or, the total net force on the system) and then setting it equal to the total mass of the system.

In this case, to me at least, it seems that the only net force in the system (because there is no friction) would be $F_{a}$ (because the third-law forces are internal). So, I would do something like :

$$F_{a} = m_{A}m_{B} a$$

which would solve out to

$$160\:\mathrm{N} = 100\:\mathrm{kg} \cdot a$$

$$\frac{160\:\mathrm{N}}{100\:\mathrm{kg}} = a$$

$$1.6 = a$$

Clearly, a different acceleration than I got from before. I know I am reasoning something wrong. Either the acceleration of the system is different than the person or you cannot do the system or I am doing the system wrong.

Also, in my head, this problem is equivalent to having two blocks, on a frictionless surface, and with the same masses, where block $A$ pushes on block $B$ with $160\:\mathrm{N}$. So a picture would be the two blocks together ($A$ on left $B$ on right) and an applied force pointing (from left to right) into block $A$ .Is this equivalent? From my reasoning, the blocks move together, which means that the acceleration of any block is the acceleration of the system and vice versa.

Answer 1

I think I understand how to solve this problem by using Newton's third law, both people will feel a force of 100N in opposite directions.

No. Newton's third law says that the force B exerts on A is equal and opposite to the force A exerts on B. Therefore they will both feel a force from the other of magnitude 160N.

So, if I understand, the acceleration on A is

a = F_a/m_A

a = 160/40

a = 4,

Correct me if I am wrong with this because I am still not certain.

You are correct, providing that there is no friction force acting on the skaters. If there was, the net force on each would not be 160 N, but 160 N minus the friction force. Keep in mind that the acceleration that A and B will experience will be based on the net force exerted on each, not just the force they exert on each other. So with no friction, the acceleration of B will be 160N/60kg=2.67 m/s$^2$.

However, here is where I get another question. In my physics class, my teacher explained you could solve force problems by either finding the free-body of the entire system (or, the total net force on the system) and then setting it equal to the total mass of the system.

I'm not sure what your teacher is saying, assuming you are accurately portraying it, but if you are looking at a free body diagram (FBD) of the both people together, and you further assume zero friction from the ice, then there is no net external force acting on the system and the acceleration of its center of mass (COM) will be zero. Only the individual persons will accelerate and they will accelerate as said above.

In this case, to me at least, it seems that the only net force in the system (because there is no friction) would be F_a (because the third-law forces are internal). So, I would do something like

F_a = m_Am_Ba

Again, if the ice does not exert a friction force on the system of A and B, then there is no external force on the system. The 160 N force is an internal force as far as the system of A and B is concerned. It is an external force only as far as B and A individually are concerned. Only if $F_a$ was the net external force acting on the system, then you would have $F_{a}=(m_{A}+m_{B})a$, not $F_{a}=m_{A}m_{B}a$.

Also, in my head, this problem is equivalent to having two blocks, on a frictionless surface, and with the same masses....

I'm having a hard time following you here. If A exerts a force on B, and block A is to the left of block B, then I don't understand your statement "an applied force pointing (from left to right) into block A."

In any case, Fig 1 below is a FBD on the two block system. Note that since there is no friction there is no next external force acting on the system as a whole. The forces A and B exert on each other are internal forces. Therefore the acceleration of its COM will be zero, although individually they will accelerate.

Fig 2 and Fig 3 are FBD's of A and B alone. Note that now the forces they exert upon each other are external forces, and since the ice again exerts no force, those external forces are the net forces acting on each causing them to accelerate.

Hope this helps.

Answer 2

Your interpretation of the statement from your teacher is not correct. The vector sum of all the external forces acting on a system determines the acceleration of the center of mass, and the mass you use is the sum (not the product) of all the masses in the system. In your example with the two skaters, the net external force is zero, and the center of mass does not move. For each of the skaters (individually) the external forces would be: gravity, the normal force from the ice, and the push from the other skater. The internal forces would be the inter-atomic forces which hold the skater together. The push acts on the mass of all of his atoms to determine his acceleration.