Forces and Free-Body Diagrams
Noah Martinez
6 min read
Listen to this study note
Study Guide Overview
This study guide covers forces as vector quantities, including contact forces. It explains free-body diagrams as a tool for visualizing forces, emphasizing the representation of forces from the environment (gravitational force, normal force, tension, and applied forces) as vectors originating from the center of mass. It also highlights coordinate system selection and provides practice questions and exam tips.
#AP Physics 1: Forces and Free-Body Diagrams Study Guide 🚀
Welcome to your ultimate review for forces and free-body diagrams! Let's make sure you're fully prepped and confident for the exam. We'll break down the key concepts, highlight important connections, and get you ready to ace those questions!
#1. Forces as Interactions
#1.1. Forces as Vector Quantities 🏹
- Forces are vector quantities that describe the interactions between objects or systems. They have both magnitude and direction.
- Forces arise from the interaction of an object with another object or system. A force is always exerted on an object or system.
An object or system cannot exert a net force on itself.
#1.2. Contact Forces
- Contact forces result from the interaction of an object or system touching another object or system.
- These forces arise from the macroscopic effects of interatomic electric forces.
- Examples include:
- Friction between two surfaces
- Normal force exerted by a surface on an object resting on it
#2. Free-Body Diagrams
#2.1. Visualizing Forces
- Free-body diagrams are essential tools for visualizing the forces acting on a single object or system.
- They help determine the equations that represent a physical situation.
- By providing a clear, visual representation of all forces, they simplify the analysis.
#2.2. Forces from the Environment
- A free-body diagram depicts each of the forces exerted on the object by the environment.
- Common forces include:
- Gravitational force (weight) 🌍
- Normal force from surfaces
- Tension in ropes or strings
- Applied forces (pushes or pulls)
#2.3. Vector Representation
- Represent forces as vectors originating from the center of mass of the object (often depicted as a dot).
- The system is treated as if all its mass is located at the center of mass.
- Force vectors are drawn as arrows pointing in the direction of the force.
#2.4. Coordinate System Selection
Selecting a coordinate system with one axis parallel to the direction of acceleration simplifies the translation from a free-body diagram to algebraic equations.
Boundary Statements: On the exam, only depict forces exerted on the object, not force components. Individual forces should be drawn as straight arrows originating from the dot, without overlapping. Forces in the same direction should be drawn side by side, not on top of each other.
#Memory Aids and Quick Facts
F.R.E.E.: Force is a Result of an Environmental Exertion. This helps remember that forces are interactions and always come from the environment.
Free-body diagrams are all about showing forces acting on the object, not forces the object exerts.
#Final Exam Focus
High-Priority Topics:
- Correctly drawing and interpreting free-body diagrams.
- Identifying all forces acting on an object in various scenarios.
- Applying Newton's laws using free-body diagrams.
Exam Tips:
- Time Management: Quickly draw free-body diagrams; they are the foundation for many problems.
- Common Pitfalls: Misidentifying forces or drawing them incorrectly (e.g., not at the center of mass).
- Strategies: Practice drawing free-body diagrams for various situations (inclines, connected objects, etc.).
#Practice Questions
Practice Question
Multiple Choice Questions
-
A block of mass m is pulled across a rough horizontal surface by a force F at an angle θ above the horizontal. The coefficient of kinetic friction between the block and the surface is μ. What is the magnitude of the frictional force acting on the block?
(A) μmg (B) μ(mg - Fsinθ) (C) μ(mg + Fsinθ) (D) μFcosθ
-
A car is moving at a constant velocity along a straight, level road. Which of the following statements about the forces acting on the car is correct?
(A) There are no forces acting on the car. (B) The net force acting on the car is zero. (C) The force of the engine is greater than the frictional forces. (D) The force of the engine is equal to the normal force.
-
A box is placed on an inclined plane. As the angle of the incline increases, the component of the gravitational force parallel to the plane:
(A) Increases (B) Decreases (C) Remains the same (D) Becomes zero
Free Response Question
A 2.0 kg block is pushed up a 30° incline by a horizontal force of 20 N, as shown in the figure. The coefficient of kinetic friction between the block and the incline is 0.20. 
(a) Draw a free-body diagram for the block, showing all the forces acting on it. (3 points) (b) Calculate the component of the applied force parallel to the incline. (2 points) (c) Calculate the normal force acting on the block. (3 points) (d) Calculate the frictional force acting on the block. (2 points) (e) Calculate the acceleration of the block up the incline. (3 points)
Answer Key and Scoring Rubric
Multiple Choice Answers:
- (B)
- (B)
- (A)
Free Response Question:
(a) Free-body diagram (3 points):
- Correctly drawn and labeled weight (mg) acting downwards.
- Correctly drawn and labeled normal force (N) perpendicular to the incline.
- Correctly drawn and labeled applied force (F) horizontally.
- Correctly drawn and labeled frictional force (f) parallel to the incline, opposing motion.
(b) Component of applied force parallel to the incline (2 points):
- F_parallel = Fcos(30°) = 20N * cos(30°) = 17.3 N
- Correctly calculated value with units.
(c) Normal force (3 points):
- N = mgcos(30°) + Fsin(30°)
- N = (2kg)(9.8m/s^2)cos(30°) + 20N sin(30°)
- N = 16.97N + 10N = 26.97 N
(d) Frictional force (2 points):
- f = μN = 0.20 * 26.97N = 5.39 N
- Correctly calculated value with units.
(e) Acceleration (3 points):
- ΣF_parallel = F_parallel - mgsin(30°) - f = ma
- 17.3N - (2kg)(9.8m/s^2)sin(30°) - 5.39N = 2kg * a
- a = (17.3N - 9.8N - 5.39N) / 2kg = 1.055 m/s^2
Remember, you've got this! Use this guide to refresh your understanding, and go into the exam with confidence. Good luck! 🎉
Explore more resources
How are we doing?
Give us your feedback and let us know how we can improve