Physics Explained Through a Video Game/Newton's First and Third Laws

Topic 2.4 - Newton's First and Third Laws

This section will explore the remainder of Newton's Laws, providing a broader understanding of the general properties of forces and their interactions with objects.

Example 1: A Swimming Koi Fish

A player-maneuvered koi fish is slowly swimming around in a pond.

Consider the map "fish pond" by O_dot. In this, a player is able to maneuver a koi fish at the bottom of the map. As such, the fish is able to spin and move gently across the body of water. We can see that when the fish is moving in a straight path, it appears to move at a constant velocity. To examine this further, consider the free body diagram (below). It includes the forces that are acting on the fish while it is moving in a straight line.

Definitions of each of the mentioned forces:
* Let ${\vec {F}}_{\text{grav}}$ be the downwards gravitational force that is acting on the fish.
* Let ${\vec {F}}_{\text{buoyancy}}$ be the upwards force provided by the water.
* Let ${\vec {F}}_{\text{thrust}}$ be the forward force provided by the fish's body.
* Let ${\vec {F}}_{\text{drag}}$ be the backwards force provided by water against the direction of the fish's movement.
* Let all of the displayed forces have an equal magnitude.

We can add the vectors together three-dimensionally (by placing their ends next to one another in any order) to show that the net force on the fish, ${\vec {F}}_{\text{net}}$, is zero.

Using the free body diagram, we can find that the fish's ${\vec {F}}_{\text{net}}$ is $0\;{\text{N}}$. This can be done through vector addition, as visualized on the right. By using the definition of Newton's Second Law from Topic 2.3, we can solve directly for the net acceleration of the fish, ${\vec {a}}_{\text{net}}$, as shown below.

${\vec {F}}_{\text{net}}=m{\vec {a}}_{\text{net}}$ [Definition of Newton's Second Law]

${\vec {a}}_{\text{net}}={\frac {{\vec {F}}_{\text{net}}}{m}}$ [Algebra.]

${\vec {a}}_{\text{net}}={\frac {0}{m}}\;{\frac {\text{N}}{\text{kg}}}$ [Substitution. We can assume that the fish does have mass in it.]

${\vec {a}}_{\text{net}}=0\;{\frac {\text{m}}{{\text{s}}^{2}}}$ [Algebra, ${\frac {\text{N}}{\text{kg}}}\implies {\frac {\text{m}}{{\text{s}}^{2}}}$ ]

We can see that because ${\vec {F}}_{\text{net}}$is $0\;{\text{N}}$, the fish has a ${\vec {a}}_{\text{net}}$ of $0\;{\frac {\text{m}}{{\text{s}}^{2}}}$. To note, the fish would have zero net acceleration regardless of how much mass is contained within it. Therefore, we can generalize this concept further below.

Because ${\vec {a}}_{\text{net}}={\frac {{\vec {F}}_{\text{net}}}{m}}$, if there is a zero ${\vec {F}}_{\text{net}}$ and an object has any positive amount of mass, then ${\vec {a}}_{\text{net}}=0\;{\frac {\text{m}}{{\text{s}}^{2}}}$.

Essentially, this is the definition of Newton's First Law, which is really just a special case of Newton's Second Law. As a formal definition, it states that:

Newton's First Law:

Unless an object has a net external force, it will either be motionless or remain at a constant velocity if already in motion.

Thus, from Newton's First Law, we can find that if an object has mass then if it isn't accelerating, then it must have a zero net force. To note, this property can help us significantly with solving for the magnitude of a specific force provided that we know the other forces, such as seen in Example 3.

Example 2: Sliding on Sandstone

Players playing in a sandstone cave and sliding on a smooth surface.

Consider the map Sandstone Cave by Sirestyx in which players slip around on a smooth inclined surface. To note, in Topic 2.2, we discussed that for an inclined plane system with an object resting on it:

• There exists a gravitational force, ${\vec {F}}_{\text{grav}}$, that acts on the object, pointing downwards.
• There exists a normal force, ${\vec {F}}_{\text{n}}$, that acts on the object, pointing perpendicularly away from the inclined plane

In the case of Sandstone Cave, there isn't an obvious flat plane that the players are slipping on. Instead, the surface appears curved. However, we can still use most of the techniques from Topic 2.2 to investigate the interactions between the players and the cave wall. The only difference is that we need to note that the normal force that would be acting on the player will change as the slope of the surface changes.

A graph of the cosine function using radians.

More specifically, we previously had found that we could describe the normal force of an object on an inclined plane by the formula $F_{\text{n}}=mgcos(\theta )$ where $\theta$ is the angle of elevation, in degrees. Because $cos(\theta )$ decreases from $1$ to $0$ as $\theta$ increases from $0^{\circ }=0\;{\text{rad}}$ to $90^{\circ }={\frac {\pi }{2}}\;{\text{rad}}$ (graphed on the right) where:

$cos(0^{\circ })=cos(0\;{\text{rad}})=1$

$\updownarrow \;{\mathit {cos(\theta )}}\;{\text{is decreasing as}}\;\theta \;{\text{is increasing.}}$

$cos(90^{\circ })=cos({\frac {\pi }{2}}\;{\text{rad}})=0$

As such, for where the surface angle is somewhere between being horizontal ($0^{\circ }=0\;{\text{rad}}$) and vertical ($90^{\circ }={\frac {\pi }{2}}\;{\text{rad}}$), the steeper the slope is, the lesser the normal force will be. We can see this in action for when a player is rolling along the surface of the sandstone cave. For instance, in the video above, after striking monkey butler, the player ez01 slides up along some of the right-hand wall before falling back down. Several instances of ez01's position and their associated free body diagram can be considered as diagrammed on the left below.

• 
  Force vectors illustrated on a sliding player (ez01) while at different slopes of a smooth surface. Note that for visual simplicity, an orange circle was put over the player's avatar.
• 
  Force vectors illustrated on a sliding ball at a steep slope with overlaid text showing a derivation for the normal force.

From the left diagram, we can find that $F_{\text{n}}$appears to share the same length as the Y-component of $F_{\text{g}}$. We can prove this using trigonometry on the right diagram, as shown above.

With knowing that $F_{\text{n}}=F_{\text{g,y}}$ in magnitude, one way that we can describe the magnitude of the normal force is that it can be defined as:

• Being opposite of the direction of $F_{\text{g,y}}$.
• Being equal to the magnitude of $F_{\text{g,y}}$.

In other words, the normal force from the sandstone wall "pushes" back on the player just as much as the player pushes into it. To note, the horizontal component of the player's gravitational force, $F_{\text{g,x}}$ is always perpendicular to the sandstone wall's slope (see left diagram). Thus, it does not "push" the player into the wall. Rather, only the $F_{\text{g,y}}$ does.

We can generalize this behavior through Newton's Third Law, whose definition is provided below:

Newton's Third Law:

Whenever an object with mass applies a force onto a second object with mass, the second object will apply another force of an equal magnitude and opposite direction onto the first object.

From this, we can verify that there must be a $F_{\text{n}}$ of an equal magnitude but opposite direction of $F_{\text{g,y}}$.

To rationalize this, suppose Newton's Third Law wasn't the case. If we were to calculate the net force on the player in the Y-axis (which is perpendicular to the wall's slope), $F_{\text{net,y}}\neq F_{\text{g,y}}+F_{\text{n}}$. In this situation, the player, ex01 would have been either accelerating downwards through the sandstone or had been floating into the air.

However, if we consider the video clip, ex01 slides gently with the sandstone after colliding with monkey butler, never losing contact with the surface of the wall. Thus, Newton's Third Law holds.

Example 3: Sinking Container Ship

A container ship gradually sinking after striking a rock.

Continuing to build up a foundation of Newton's Third Law, suppose a container ship strikes a small rock island and subsequently begins to sink in the map "Sinking Container Ship" by O_o O_o O_o. Before then, the shipping containers are stacked vertically on top of each other.

For this exercise, assume that we're considering the period of time before the ship hits the rock and that:

• Each of the shipping containers have a mass of $9{,}000.\;{\text{kg}}$.
• The acceleration of an object due to gravity is $9.8\;{\frac {\text{m}}{{\text{s}}^{2}}}$.
• The ship is moving with a constant velocity.
• There are no significant horizontal forces acting on the shipping containers.

  

  A reference image of the container ship standing upright. This image will be used to answer the exercises below.

Exercise 1: Consider the reference image and the other information provided to draw a free body diagram for the yellow shipping container. Then, calculate the normal force acting on it.

Answer 1:

A free body diagram overlaid on top of the yellow shipping container.

To approach this problem, we first need to consider that there needs to be a gravitational force acting on the yellow shipping container that is acting directly downwards. Because the gravitational force is acting directly downwards, all of this force is pushing the yellow container into the orange container beneath it.

Therefore, by Newton's Third Law, there must exist a force of an equal magnitude but opposite direction of the gravitational force. Using the definitions of the types of forces from Topic 2.2, this would be a normal force.

To calculate the magnitude of the normal force, we can compute $F_{\text{n}}=mgcos(\theta )$. Because $\theta =0$ such that $cos(\theta )=1$, $F_{\text{n}}=mg$. Then, we can use algebra to find the normal force on the yellow container, as shown below:
$F_{\text{n, yellow}}=mg$
$F_{\text{n, yellow}}=(9{,}000.)(9.8)\;{\text{kg}}\cdot {\frac {\text{m}}{{\text{s}}^{2}}}\quad$ [Substitution of variables.]
$F_{\text{n, yellow}}=88200\;{\text{N}}\quad$ [Algebra; Definition of a Newton (N)]
$F_{\text{n, yellow}}=8.8\cdot 10^{4}\;{\text{N}}\quad$ [2 sigfigs; Using Scientific Notation]

The yellow shipping container has a normal force acting on it of $8.8\cdot 10^{4}\;{\text{N}}$ that is pointing in the upward direction.

Exercise 2: Draw a free body diagram for the orange shipping container that is underneath the yellow shipping container from Example 1. Calculate the normal force acting on this orange shipping container.

Answer 2:

A free body diagram of the forces acting on a shipping container while underneath another shipping container.

For this problem, we can consider that in Exercise 1 that there is a gravitational force that is being exerted downwards on the yellow container ($F_{\text{grav, yellow}}$). Because it is exerting downwards (into the orange container), this means that the orange container is being pushed downwards by it.

In addition, the orange container is being pushed downwards by its own gravitational force (($F_{\text{grav, orange}})$). Because of this, the orange container is pushing into the ship with both the gravitational force of itself and of the yellow container. Since the ship's deck is flat and directly under the containers, the ship is being pushed by the entirety of the downwards forces acting on the orange container.

Because of Newton's Third Law, there must be an equal magnitude and oppositely directed force that opposes the previously mentioned downwards forces on the orange container. This will be our normal force for the orange container, $F_{\text{n, orange}}$.

Since $F_{\text{n, orange}}$ is equal in magnitude to $F_{\text{grav, orange}}$ and $F_{\text{grav, orange}}$ combined, in terms of magnitude:
$F_{\text{n, orange}}=F_{\text{grav, orange}}+F_{\text{grav, orange}}$
From here, we can solve directly for the magnitude of the normal force acting on the orange container:
$F_{\text{n, orange}}=F_{\text{grav, orange}}+88200\;{\text{N}}\quad$ [Substituting the orange box's gravitational force from Exercise 1, using as many sigfigs as possible.]
$F_{\text{n, orange}}=88200+88200\;{\text{N}}\quad$ [Because the orange box has the same mass, it will have the same magnitude of gravitational force.]
$F_{\text{n, orange}}=176400\;{\text{N}}\quad$ [Addition.]
$F_{\text{n, orange}}=1.8\cdot 10^{5}\;{\text{N}}\quad$ [2 sigfigs; scientific notation.]
The orange container has a normal force with a magnitude of $1.8\cdot 10^{5}\;{\text{N}}$ that is pointing in the upward direction.

Example 4: A Swimming Koi Fish Continued (Challenging)

A koi fish interacting with a lily pad, resulting in the system to move at a constant velocity upwards.

Swapping back to the map "fish pond" by O_dot, we will gain a further understanding of Newton's Third Law. Suppose the fish continues to swim in the pond. Then, the fish begins to interact with a lily pad in front of it. For this example, we'll be looking into the force interactions between the fish and a lily pad.

Assume that (for a extended period of time):

• The fish and the lily pad stay in direct contact at a constant velocity with one another.
• The lily pad is directly in front of the fish.
• Both the fish and lily pad always stay in vertical equilibrium (they stay at the water's surface).
• The forces from Example 1 (${\vec {F}}_{\text{grav,fish}}$, ${\vec {F}}_{\text{buoyancy,fish}}$, ${\vec {F}}_{\text{thrust,fish}}$) remain the same magnitude.
• Equality of ${\vec {F}}_{\text{drag}}$ in Example 1 is not necessary in Example 2.

Why did we not account for ${\vec {F}}_{\text{drag}}$ 's equality?

To note, the drag force of water on an object is dependent on a wide variety of factors, including the velocity of the object, the surface area of the object, the shape of the object, and other factors. For simplicity and convenience, we're assuming that ${\vec {F}}_{\text{drag, lily}}$ and ${\vec {F}}_{\text{drag, fish}}$ in Example 2 to not be necessarily tied to ${\vec {F}}_{\text{drag, fish}}$ in Example 1.

Put as simply as possible, if ${\vec {F}}_{\text{drag}}\neq {\vec {F}}_{\text{drag,fish}}+{\vec {F}}_{\text{drag,lily pad}}$, then the lily pad and fish would briefly accelerate together after collision. This is because the water either wouldn't be tugging on them as much as beforehand (or would be tugging on them more).

However, the lily pad and fish thus would have a changing speed. Therefore, their respective drag forces will change over time. This would result in the lily pad and fish system to approach a constant velocity, regardless of the specific drag forces.

---

A diagram of a fish and lily pad before they interact with one another.

Exercise: Consider the two free body diagrams of the fish and the lily pad before their interaction and the other information provided to draw two free body diagrams for the fish and the lily pad while they are interacting with each other.

Answer:

By Newton's First Law, because fish and the lily pad both have a constant velocity, the system has a ${\vec {F}}_{\text{net}}$ of zero.

However, it is clear that the fish is pushing the lily pad forward by the ${\vec {F}}_{\text{thrust, fish}}$ force. This is because this is the only force on the fish (that we know currently) that is directed towards the lily pad. Therefore, there must also exist some ${\vec {F}}_{\text{lily pad on fish}}$ force of an equal magnitude and opposite direction by Newton's Third Law.

Another factor to consider is that once the fish begins to interact with the lily pad, the lily pad as previously mentioned begins to move. This would result in a drag force on the lily pad, which can be called ${\vec {F}}_{\text{drag, lily pad}}$. Since a drag force opposes the direction of movement, the drag force vector would start at the lily pad's center of mass and point backwards. In other words, ${\vec {F}}_{\text{drag, lily pad}}$ is pushing on the fish.

Because of this, we can again apply Newton's Third Law, including for there to be an equal magnitude, opposite direction force of ${\vec {F}}_{\text{drag, lily pad}}$ from the fish to the lily pad. We can call this vector ${\vec {F}}_{\text{fish on lily pad}}$.

With this, we can draw free body diagrams for the fish and the lily pad during their interaction, as diagrammed below.

A diagram of the fish and lily pad during interaction.

Question 1: A Humble Boat

A player pushing a sailboat off a dock.

Consider a modified version of "Don't Drown" by Hexigon where a small sailboat is on a deck. The boat is being pushed into the water by a player applying a leftward force on it.

Assume that:

• The sailboat and player each have a mass of $90.\;{\text{kg}}$.
• The player is pushing leftwards on the sailboat with a force of $200.\;{\text{N}}$.
• The ground is rough, resulting in a kinetic friction force when the boat and player are moving on it.
• The sailboat and player together accelerate prior to approaching a constant velocity.
• The sailboat has a uniform density apart from the sails, which have negligible mass

Part (a):

(i) Define which forces are acting on the player in the vertical direction when he and the sailboat are moving at a constant velocity.

(ii) Also, define which forces are acting on the player in the horizontal direction at this time.
(iii) Construct a free body diagram of the player at this time.

Part (b)(i): A selection of possible center of masses for the sailboat.

Part (b):

(i) Using the options on the right and the information provided above, decide where the sailboat's center of mass is located.

(ii) Construct a free body diagram of the sailboat when the player and sailboat are moving at a constant velocity.

Part (c):

Suppose that soon after the player begins pushing on the boat, the player-boat system has a maximum horizontal acceleration magnitude of $0.3\;{\frac {\text{m}}{{\text{s}}^{2}}}$. This acceleration occurs in the leftward direction.
(i) Calculate the magnitude of the force due to kinetic friction on the player-boat system when it reaches its maximum acceleration.
(ii) Calculate the magnitude of the force due to kinetic friction on the player-boat system when it is traveling at a constant leftward velocity.

Consider discussing your solutions on this article's Talk Page. On there, you can find help from others.

References