AP Physics 1- 2.2 Forces and Free-Body Diagrams- Study Notes- New Syllabus
AP Physics 1-2.2 Forces and Free-Body Diagrams – Study Notes
AP Physics 1-2.2 Forces and Free-Body Diagrams – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Force as an Interaction
- Forces and Free-Body Diagrams (FBD)
- Choosing a Coordinate System in Free-Body Diagrams
Force as an Interaction
A force is not a property of a single object, but an interaction between two objects or systems. Whenever one object pushes or pulls another, both objects experience forces.
A force is an interaction that can change the motion (velocity or direction) or shape of an object. It is a vector quantity, having both magnitude and direction.
Newton’s Third Law Connection:
For every force that object A exerts on object B, object B exerts an equal and opposite force on object A:
\( \vec{F}_{AB} = – \vec{F}_{BA} \)
Types of Forces:
- Contact Forces: e.g., friction, tension, normal force, applied push/pull.
- Non-contact Forces: e.g., gravitational force, electrostatic force, magnetic force.
Key Notes:
- A force always acts between two objects (never in isolation).
- Forces can change motion (acceleration) or shape (deformation).
- Measured in Newtons (N): 1 N = force required to accelerate 1 kg at 1 m/s².
Example :
A book resting on a table experiences a downward gravitational force (weight). The table exerts an upward normal force on the book.
▶️Answer/Explanation
Interaction: Earth pulls book downward with weight \( mg \). Book pushes Earth upward with equal magnitude. Table pushes upward on book with normal force, while book pushes downward on table.
Example :
A magnet attracts a piece of iron. The magnet exerts a force on the iron, and the iron exerts an equal and opposite force on the magnet.
▶️Answer/Explanation
Interaction: Magnetic force acts both ways. Even if the magnet is heavier, both forces are equal in magnitude and opposite in direction.
Forces and Free-Body Diagrams (FBD)
A free-body diagram (FBD) is a graphical representation of all the external forces acting on an object or system, drawn as arrows pointing away from the object.
Definition:
In an FBD, the object is represented as a point (or a simple box), and all the forces acting on it are shown as vectors. Each vector’s direction shows the direction of the force, and its length represents relative magnitude.
Steps to Draw an FBD:
- Isolate the object from its surroundings.
- Identify all external forces acting on the object.
- Represent each force with an arrow, starting from the object.
- Label each force clearly (e.g., \( F_g \), \( F_N \), \( F_f \), \( T \)).
Common Forces in FBD:
- Weight (\( F_g \)): Downward gravitational force (\( mg \)).
- Normal Force (\( F_N \)): Perpendicular force exerted by a surface.
- Tension (\( T \)): Force transmitted through ropes/strings.
- Friction (\( F_f \)): Opposes relative motion between surfaces.
- Applied Force (\( F_{app} \)): Any external push or pull.
Example :
A block resting on a horizontal surface.
▶️Answer/Explanation
Forces acting:
• Weight (\( F_g = mg \)) downward.
• Normal force (\( F_N \)) upward.
Since there is no vertical acceleration, \( F_N = F_g \).
Example :
A block being pulled on a rough surface with force \( F_{app} \).
▶️Answer/Explanation
Forces acting:
• Weight (\( F_g = mg \)) downward.
• Normal force (\( F_N \)) upward.
• Applied force (\( F_{app} \)) to the right.
• Frictional force (\( F_f \)) to the left.
Net force = \( F_{app} – F_f \) → determines acceleration.
Choosing a Coordinate System in Free-Body Diagrams
When analyzing forces, it is often convenient to choose a coordinate system aligned with the direction of acceleration or motion. This simplifies equations by reducing the need for trigonometric components.
Example of Inclined Plane:
For an object of mass \( m \) resting on a frictionless incline of angle \( \theta \):
- Choose x-axis parallel to the incline (along motion).
- Choose y-axis perpendicular to the incline.
Forces acting:
- Gravitational force: \( F_g = mg \) downward.
- Normal force: \( F_N \), perpendicular to surface.
Resolving Weight into Components:
- Parallel to incline: \( F_{g,\parallel} = mg \sin \theta \)
- Perpendicular to incline: \( F_{g,\perp} = mg \cos \theta \)
Equations of Motion:
Net force along incline: \( F_{net} = mg \sin \theta \)
Acceleration: \( a = g \sin \theta \)
Example :
A 10 kg block slides down a frictionless incline at angle \( \theta = 30^\circ \). Find its acceleration.
▶️Answer/Explanation
Weight = \( mg = 10 \times 9.8 = 98 \, \text{N} \).
Component along incline: \( F_{g,\parallel} = 98 \sin 30^\circ = 49 \, \text{N} \).
Net force = 49 N, Mass = 10 kg.
\( a = \dfrac{F}{m} = \dfrac{49}{10} = 4.9 \, \text{m/s}^2 \).
Answer: The block accelerates down the incline at \( 4.9 \, \text{m/s}^2 \).