Motion in a Plane and Equilibrium of Rigid Bodies

Introduction: Welcome to the 2D World and Objects with "Size"!

In "Basic Physics," we mainly dealt with objects moving back and forth along a straight line (one-dimensional motion) and treated objects as "point masses" whose sizes could be ignored.
Starting with this chapter, "Motion in a Plane and Equilibrium of Rigid Bodies," your journey into full-fledged Senior High School Physics truly begins!

Here, we will cover two major topics:
1. Motion in a Plane (Projectile Motion, etc.): The motion of objects moving freely in a plane (two dimensions), such as a ball thrown at an angle.
2. Equilibrium of Rigid Bodies: The conditions required for an object with size (a rigid body) to remain stationary without translating or rotating.
At first, it might seem challenging with more equations appearing, but don't worry! The golden rule for 2D motion is simply: "Break it down into vertical and horizontal components." Let's master each concept step by step.

Section 1: Motion in a Plane (Velocity and Acceleration in Curvilinear Motion)

1. Position, Displacement, Velocity, and Acceleration Are All "Vectors"

The position of an object moving on a flat surface (the \(xy\)-plane) is represented by a position vector \(\vec{r} = (x, y)\), which has a direction and distance from the origin.
When an object moves from position \(\vec{r}_1\) to \(\vec{r}_2\) over a time interval \(\Delta t\), this change in position is called the displacement vector \(\Delta \vec{r}\):
\(\Delta \vec{r} = \vec{r}_2 - \vec{r}_1 = (\Delta x, \Delta y)\)

The average velocity \(\vec{\bar{v}}\) is the displacement per unit time:
\(\vec{\bar{v}} = \frac{\Delta \vec{r}}{\Delta t}\)
When we take the limit as \(\Delta t\) approaches \(0\), we get the instantaneous velocity vector \(\vec{v} = (v_x, v_y)\).
*Note: The direction of the velocity vector is always tangent to the trajectory at that instant.

Furthermore, when the velocity vector changes from \(\vec{v}_1\) to \(\vec{v}_2\) over a time interval \(\Delta t\), the acceleration vector \(\vec{a} = (a_x, a_y)\) is defined as the rate of change of velocity:
\(\vec{a} = \frac{\Delta \vec{v}}{\Delta t}\)
The direction of acceleration indicates the direction in which the velocity is changing. Unlike 1D motion, please note that even if the speed (the magnitude of velocity) remains constant, an acceleration exists whenever the direction changes.

2. The Ultimate Strategy: "Resolve into Two Perpendicular Directions"

Thinking about motion in a plane all at once can be complicated, but if you split it into the \(x\)-direction (horizontal) and the \(y\)-direction (vertical), each one simply turns into the linear motion you already mastered in Basic Physics!
Newton's equation of motion, \(\vec{F} = m\vec{a}\), also holds true independently for each component:
- \(x\)-direction: \(F_x = m a_x\)
- \(y\)-direction: \(F_y = m a_y\)
Thus, the core principle for solving 2D motion is: "Two perpendicular directions do not interfere with each other (they are independent)."

[Key Takeaway]
Whenever you see curved motion, first resolve it into the "\(x\)-axis direction" and "\(y\)-axis direction"!
Using the Pythagorean theorem, the magnitude of velocity (speed) \(v\) is given by \(v = \sqrt{{v_x}^2 + {v_y}^2}\).

[Common Pitfall]
A frequent mistake is assuming that "if the speed is constant, the acceleration must be zero"! When turning a curve, even if your speedometer reads a constant speed, the direction of travel is changing, which means an acceleration (a change in the velocity vector) is present.

Section 2: Projectile Motion (Horizontal and Oblique Projections)

The motion of an object flying through the air under the influence of gravity alone is called projectile motion. Neglecting air resistance, the only force acting on the object is the downward gravitational force, \(mg\).
Therefore, there is no force in the horizontal direction, while a constant gravitational force acts in the vertical direction at all times.
- Horizontal direction: Acceleration \(a_x = 0\) \(\rightarrow\) Uniform linear motion (constant velocity)
- Vertical direction: Acceleration \(a_y = -g\) (taking upward as positive) \(\rightarrow\) Uniformly accelerated linear motion (free fall or vertical throw)

1. Horizontal Projection

This is the motion of an object projected horizontally with an initial velocity \(v_0\) from a cliff of height \(h\).
Let the horizontal direction be the \(x\)-axis (positive direction: direction of projection) and the vertically downward direction be the \(y\)-axis (positive direction: downward).
The velocity and position at time \(t\) are given by:
Velocity components:
- Horizontal direction (uniform linear motion): \(v_x = v_0\)
- Vertical direction (free fall): \(v_y = g t\)
Position coordinates:
- Horizontal direction: \(x = v_0 t\)
- Vertical direction: \(y = \frac{1}{2} g t^2\)

Equation of the Trajectory (Proof of a Parabola):
From \(x = v_0 t\), we get \(t = \frac{x}{v_0}\). Substituting this into the equation for \(y\):
\(y = \frac{g}{2 {v_0}^2} x^2\)
This equation takes the form of a quadratic function, \(y = k x^2\), that you learn in math. That's why the trajectory forms a "parabola"!

2. Oblique Projection

This is the motion of an object launched from the ground at an angle of elevation \(\theta\) relative to the horizontal with an initial velocity \(v_0\).
Let the horizontal direction be the \(x\)-axis and the vertically upward direction be the \(y\)-axis.
First, resolve the initial velocity into two perpendicular components:
- Horizontal component of initial velocity: \(v_{0x} = v_0 \cos\theta\)
- Vertical component of initial velocity: \(v_{0y} = v_0 \sin\theta\)

The velocity and position at time \(t\) are as follows:
Velocity components:
- Horizontal direction (uniform linear motion): \(v_x = v_0 \cos\theta\)
- Vertical direction (vertical projection upward): \(v_y = v_0 \sin\theta - g t\)
Position coordinates:
- Horizontal direction: \(x = (v_0 \cos\theta) t\)
- Vertical direction: \(y = (v_0 \sin\theta) t - \frac{1}{2} g t^2\)

Top 3 Crucial Points for Oblique Projection:
- Time to reach the highest point \(t_1\) and maximum height \(h\):
At the highest point, vertical motion stops for an instant. That means \(v_y = 0\)!
From \(0 = v_0 \sin\theta - g t_1\), we find \(t_1 = \frac{v_0 \sin\theta}{g}\).
Substituting this into the equation for \(y\) yields the maximum height: \(h = \frac{{v_0}^2 \sin^2\theta}{2g}\).
- Time of flight until landing \(t_2\):
When the projectile returns to the ground, its vertical position is zero (\(y = 0\)).
\(0 = (v_0 \sin\theta) t - \frac{1}{2} g t^2 = t \left( v_0 \sin\theta - \frac{1}{2} g t \right)\)
Since \(t > 0\), we obtain \(t_2 = \frac{2 v_0 \sin\theta}{g} = 2 t_1\).
*Due to trajectory symmetry, the time taken to reach the apex is exactly equal to the time taken to fall back to the ground.
- Horizontal range \(L\):
This is the horizontal distance traveled during the total time of flight \(t_2\).
\(L = (v_0 \cos\theta) \times t_2 = (v_0 \cos\theta) \times \frac{2 v_0 \sin\theta}{g} = \frac{{v_0}^2 (2 \sin\theta \cos\theta)}{g} = \frac{{v_0}^2 \sin 2\theta}{g}\)
*Here, we use the double-angle identity: \(2 \sin\theta \cos\theta = \sin 2\theta\).

[Physics Trivia] What angle gives the maximum range?
Looking at the horizontal range formula \(L = \frac{{v_0}^2 \sin 2\theta}{g}\), \(L\) reaches its maximum when \(\sin 2\theta\) reaches its maximum value of \(1\).
This happens when \(2\theta = 90^\circ \implies \theta = 45^\circ\)!
For a given initial speed, launching at an angle of \(45^\circ\) will send the ball the furthest distance.

[In the Presence of Air Resistance (Qualitative Understanding)]
In real air, drag acts in the direction opposite to motion.
As a result, the maximum height is lower and achieved earlier than in a vacuum, and the descent is steeper than a true parabola. The trajectory loses its neat left-right symmetry, becoming skewed in the second half of the flight.

[Common Pitfall]
Some students mistakenly think that "the velocity is zero" at the highest point of an oblique projection! Only the vertical component of velocity, \(v_y\), is zero; the horizontal component, \(v_x = v_0 \cos\theta\), remains. Even at the highest point, the projectile continues moving horizontally.

Section 3: Equilibrium of Rigid Bodies

1. From Point Masses to "Rigid Bodies"

Until now, the objects we analyzed were treated as "point masses" with negligible size. For a point mass, considering only translational motion was sufficient.
However, real objects have physical dimensions. Depending on where you apply a force, an object can spin and rotate.
An idealized body that does not deform when forces are applied and possesses a definite size is called a rigid body.
For a rigid body to remain stationary and balanced, it must satisfy the condition of "no rotation" in addition to "no translation."

2. Torque (Moment of a Force)

The measure of the tendency of a force to cause an object to rotate about a specific pivot axis is called the moment of a force (or torque).
When opening a door, pushing near the handle (far from the hinges) makes it swing open easily, while pushing right near the hinges takes a lot of effort. Furthermore, pushing perpendicular to the door works best, whereas pushing at a shallow angle causes most of the effort to be wasted.
From this everyday experience, the magnitude of the moment of force \(M\) is defined as follows.

Definition of Moment of Force:
Let \(l\) be the perpendicular distance from the axis of rotation to the line of action of the force (the lever arm), and let \(F\) be the magnitude of the force:
\(M = F l\)
The unit is the \(\text{N}\cdot\text{m}\) (newton-meter).

Direction of Rotation and Sign Convention:
Conventionally, a counterclockwise tendency to rotate is taken as positive (\(+\)), and a clockwise tendency is taken as negative (\(-\)) (choosing either convention is fine as long as you remain consistent when setting up your equations).

Transmissibility of a Force along its Line of Action:
When applying a force to a rigid body, sliding the force vector along its "line of action" (the infinite line drawn along the direction of the force) does not alter the translational or rotational effects exerted on the body.

3. Conditions for Equilibrium of a Rigid Body

For a rigid body to remain completely at rest (in static equilibrium), the following two conditions must be satisfied simultaneously.

Condition 1 (Condition for no translational motion): The resultant force must be zero
The vector sum of all external forces acting on the body must be \(\vec{0}\).
In a Cartesian coordinate system, we resolve this into components:
- Sum of forces along the \(x\)-axis: \(\sum F_x = 0\)
- Sum of forces along the \(y\)-axis: \(\sum F_y = 0\)

Condition 2 (Condition for no rotational motion): The sum of the moments of force must be zero
The algebraic sum of the torques about any arbitrary axis of rotation must equal \(0\):
\(\sum M = 0\)

[Pro-Tip: Choosing the Smartest Pivot Point]
When a rigid body is in equilibrium, the sum of the moments is zero about ANY chosen pivot point.
This fact is your greatest problem-solving shortcut!
Any force whose line of action passes through the pivot axis has a lever arm of \(0\), meaning its moment is \(0\). Therefore, if you choose a pivot point where several unknown forces intersect, those unknowns will drop out of your torque equation, making the algebra remarkably simple!

4. Center of Gravity

A rigid body is composed of countless tiny particles, and gravity pulls on each one of them. The point of application of the resultant of all these individual gravitational forces is called the center of gravity (CG).
When analyzing gravity acting on a rigid body, we can treat it as though "the entire weight of the body acts at the center of gravity."

- For uniform objects with symmetric shapes (such as uniform rods, spheres, and rectangular blocks), the center of gravity coincides with the geometric center.
- If you support an object directly beneath its center of gravity, it will balance steadily without rotating.
- When an object is tilted, as long as the vertical line passing through its center of gravity falls within its base of support, the object will tend to right itself rather than tip over.

[Common Pitfall]
A frequent misconception is assuming that "if the forces are balanced (resultant force is zero), the object will definitely stay at rest"!
For example, if you pull opposite sides of a steering wheel with equal and opposite forces, the net force is zero, yet the steering wheel turns (a pair of equal, opposite, and non-collinear forces like this is called a couple). For rigid bodies, always check both force balance and torque balance!

Chapter Summary: Study Checklist

To wrap up, let's review the core takeaways from this chapter:

- Motion in a Plane: Resolve motion into \(x\) and \(y\) components. The two directions are completely independent.
- Projectile Motion: Horizontal motion is "uniform linear motion," while vertical motion is "uniformly accelerated motion (with gravitational acceleration \(g\))."
- Key Milestones in Oblique Projection: At the highest point, \(v_y = 0\); upon landing, \(y = 0\).
- Equilibrium of a Rigid Body: ① The resultant force is zero (\(\sum \vec{F} = \vec{0}\)) AND ② The sum of torques is zero (\(\sum M = 0\)).
- Calculating Torque: \(M = F l\) (Force \(\times\) lever arm). Always pick a pivot point where unknown forces intersect!

Once you understand 2D motion and rigid body equilibrium, your physics horizons expand significantly into two dimensions.
In upcoming topics such as circular motion, momentum, and simple harmonic motion, the core skills you learned here—resolving vectors and thinking along perpendicular axes—will continue to be your greatest assets. Practice standard problems steadily, and build up your confidence!