Consider an elevator accelerating upward at $5.0~\textrm{m/s}^2$. What is the apparent gravitational acceleration inside this elevator, and how much would a scale read when a man of mass 50 kg stands on it (Hint: Scales indicate normal force.).

The elevator accelerates upward, so any object inside it experiences a downward fictitious force of $ma$ where $a= 5.0 ~\textrm{m/s}^2$. Doing some simple analysis, we can conclude that objects inside the elevator experience an effective gravitational acceleration of $g+a = 14.8 ~\textrm{m/s}^2$. The man’s effective weight, then, is:

$$F_n=mg+ma=740 ~\textrm{N}$$

Now what if the elevator was accelerating downward?

Well, then the fictitious force is simply oppositely directed and we have effective gravitational acceleration $g-a = 4.8 m/s^2$. This translates to a weight of:

$$F_n=mg-ma=240 ~\textrm{N}$$

I'll pose a little problem. Now, consider this elevator that is accelerating upwards. Can you think of what the scale will read (not as an exact value, but as just a general idea or comparison to what it would "normally" read. I'm being ambiguous on purpose so I don't give anything away.)

This is one of those cases where actually having been in an elevator helps a lot. If you've ever been in an elevator that is accelerating upwards, whether that be speeding up on the way up or slowing down on the way down, you feel yourself become heavier.

Your mass isn't actually increasing since that's an intrinsic property and doesn't change so easily. You're actually feeling your weight, which is the percieved gravitational force, change. This is because you experience a downwards fictitious force that allies itself with gravity to make you heavier (if only for a little bit). The scale will also read a greater value because it needs to support that extra weight.

Now we should consider the obvious other case: elevator accelerating down.

The inverse is true if you're accelerating down. The fictitious force is instead upwards and opposed to gravity, so you feel lighter. There is an interesting bit to be said here. In free fall, you're essentially weightless because in your reference frame (where you fall with the gravitational acceleration $g$ downwards), there is a fictitious force upwards that is equal to the gravitational force exerted on you.

We have a car cresting a circular hill, which can be approximated as a portion of a circle with radius $R$. What is the maximum velocity it can have at the top, and what happens if it exceeds that velocity?

Now, the first part of this question is a bit simpler to answer. You have a single possible velocity for a given acceleration. Now, it might be tricky to figure out what the centripetal force is at the top, but remember that hills are vertical. The normal force varies since the net force has to equal the required centripetal force for circular motion at a certain velocity, and it opposes gravity. At the critical point, there is no more normal force and the entirety of the gravitational force contributes to the centripetal force. Therefore, at the top, the maximum possible centripetal acceleration is $a_c = g$. Therefore, we can write:

$$g = \dfrac{v^2}{R}$$
This simplifies to:

$$v = \sqrt{gR}$$
Technically, this is a maximum value since the normal force can vary and allow for other velocities. If you prefer, it is more accurate to write:

$$v \leq \sqrt{gR}$$
Now, this part did not require any invocation of fictitious forces, but the second part of the problem is easier if you invoke them. Think about how the fictitious centrifugal force would be directed. In the critical case, it is directed upwards and exactly equal in magnitude to the gravitational force. Exceeding this "speed limit" would make it such that the centrifugal force is greater than the gravitational, and the car would lift off the ground. This makes sense, as if you travel over a bump in the road too fast you feel the car become airborne momentarily.

We have a car cresting a circular hill, which can be approximated as a portion of a circle with radius $R$. What forces are causing circular motion, and what happens if it goes too fast? Consider only the point where the car is at the top.

At the top of the hill, the radius of the hill is vertical, so we are only concerned with forces in the vertical direction. The only forces that can be in this direction are the gravitational and normal forces, and they are oppositely directed. In this scenario, $F_n < mg$ because the net force has to be downward to provide sufficient centripetal force for circular motion to occur. This is yet another case where the normal force is not equal to the weight of the object! For speeds low enough, the car will remain in contact with the ground, but there is a critical point where the normal force is zero and the entirety of the gravitational force is required to balance out the required centripetal force (or centrifugal force). This gives us the equation (it's one of the only ones you'll see here, I promise!):

$$m\dfrac{v^2}{R} = mg$$
Now, you can simplify this rather easily, but if you don't feel comfortable doing it, don't worry! You just need to know that the critical velocity where the normal force is zero is:

$$v = \sqrt{gR}$$
This is the maximum velocity at which circular motion is still possible. What happens if you go faster than this? Well, think of going over a speed bump really fast. Your car will go airborne momentarily. This happens because the fictitious centrifugal force upward is larger than gravity, which causes the car to lift off the ground from your perspective. In reality, you're really just experiencing inertia as your car fails to accelerate enough to go on its previous path and goes into a free-fall state, albeit only for a few fractions of a second.

Consider the scenario depicted in the figure below. What must the force $F$ be to hold all the blocks stationary relative to each other? Assume there is no friction between all surfaces. The red circle is a pulley, in case it wasn't clear. (It probably wasn't, considering how it looks.)

This problem is not trivial. It takes significant thought to figure out why this would even be possible in the first place, since there is no friction. Most people see $m_3$ and conclude that it is impossible for it to stay in place without any frictional forces. However, with fictitious forces things begin to make sense.

The best approach to the problem is actually to start with the problematic block, $m_3$. It can easily accelerate horizontally from a normal force applied on it by the big cart $m_1$, but it msut remain still vertically. We can actually ignore this horizontal acceleration because the system moves as a whole, and it doesn't offer us any particularly useful information. Since the block must remain at rest vertically, the tension force $T$ must obey the relation:

$$T = m_3g$$
Now, you might be really confused. For a normal Atwood's machine in this setup, the tension force is not this! However, this is the only way the third block can stay at rest. This leads us to the conclusion that this tension force also acts on the second block. Now is where we invoke the fictitious force. Since the cart accelerates to the right due to a force being exerted on it, we can assume it must have some acceleration $a$. This leads to a fictitious force to the left of magnitude $m_2a$ "applied" onto the second block, which must be balanced by the tension force. Thus, we have:

$$T = m_2a$$
This allows us to write:

$$m_2a = m_3g$$ $$a = \dfrac{m_3}{m_2}g$$
Now, we consider the force $F$. Since we have said that the system moves as a whole without any relative motion, we can treat it as one body and only consider the external forces on it (which is just $F$ in this scenario):

$$F = (m_1+m_2+m_3)a$$
Substituting in our value for $a$ gives us:

$$F = (m_1+m_2+m_3)\dfrac{m_3}{m_2}g$$
That was difficult, right? But it's also an interesting problem that raises some interesting implications. Accelerated reference frames can really mess with your usual intuitions, and postulates you held to be always true might not be true when working with one!

There is another interesting question I'd like to pose about this scenario. If all surfaces are frictionless, then how do the blocks affect the motion of the cart?

The pulley is the answer to our question. We know (or assume logically) that the pulley is attached to the cart. The pulley experiences tensional forces from either side as a consequence of Newton's third law, so the cart experiences those forces as well. This is how the blocks can affect the cart despite being unable to interact with it through friction.

We are going to consider a relatively complex arrangement of blocks, pulleys, and a cart. Everything is frictionless in this scenario. The red circle is a pulley.

We want to see if it's possible to push the cart such that none of the block slip or move relative to each other. In other words, we want the cart and blocks to remain in place as we push it. Now, normally this would seem impossible without friction, since $m_2$ cannot stop $m_3$ from sliding down because it is placed horizontally. However, the blocks are accelerating.

Since there is a net external force represented by $F$ on the system, it will tend to accelerate to the right. This means there will be a fictitious force to the left on each block. The only way block 3 can remain at rest is if the tension force is equal to its weight, so we can assume that. Next, we know that this tension force is the only real force acting in the horizontal direction on the second block. It is directed to the right, so we know that it has to balance the fictitious force on the second block. This enables us to do force balance and makes the scenario physically possible.

The result we get from calculations is that there is only one possible acceleration where the system can remain at rest relative to itself, which means by Newton's second law there is only one possible force $F$ that will enable this kind of motion. An interesting result, to be sure.

We know that all three blocks accelerate together, so we can treat them as one object. But for them to move together, they have to interact with each other through some force. However, it is not obvious how block 2 would interact with the cart. This is a tricky question and requires very careful force analysis.

See, block 2 feels a tension force from the rope, which is constant throughout the rope. The pulley, which is attached to the cart, also feels this same tension force. This allows block 2 to "interact" with the cart and affect its motion, which enables us to explain how block 2 would even affect the cart.