Free-Body Diagram (FBD) Practice and Solutions

Fundamentals of Free-Body Diagrams (FBDs)

• Definition of a Free-Body Diagram (FBD): A graphical illustration used to visualize all applied forces, constraint forces, and moments acting on an isolated physical body or system in static or dynamic equilibrium.

• Primary Mechanical Forces:

  • Gravitational Force ($m\,g$): Acts vertically downward toward the center of the Earth, where $m$ represents mass in kilograms and $g$ represents gravitational acceleration ($9.8\,m/s^2$).
  • Normal Force ($N$): The perpendicular contact force exerted by a surface on an object resting or sliding upon it.
  • Tension Force ($T$): The pulling force exerted by a string, rope, or cable, directed along the length of the connector away from the object.
  • Frictional Forces ($f$): Parallel contact forces opposing relative motion or impending motion across a surface.
  • Static Friction ($f_s$): Opposes impending motion of a stationary object up to a maximum value of $f_{s,\max} = \mu_s\,N$.
  • Kinetic Friction ($f_k$): Opposes relative sliding motion of a moving object with magnitude f_k = \mu_k\,N$.\n * **Applied Forces (F_{\text{push}}$,$F_{\text{pull}}):** Contact or non-contact forces applied to the system by external agents.\n\n# Basic Motion and Single-Mass Systems\n\n* **Scenario 1: Stationary Object on a Horizontal Surface**\n * **Description:** A ball or block of mass m resting motionless on a flat horizontal surface.\n * **Forces Acting:**\n * Normal force N directed vertically upward.\n * Gravitational force m\,g directed vertically downward.\n * **Equilibrium Equation:**\n    \sum F_y = N - m\,g = 0 \implies N = m\,g\n    \sum F_x = 0\n\n* **Scenario 2: Object Rolling horizontally to the Right**\n * **Description:** A ball of mass m moving horizontally to the right across a flat surface.\n * **Forces Acting:**\n * Normal force N directed vertically upward.\n * Gravitational force m\,g directed vertically downward.\n * **Equilibrium Equation:** In the absence of horizontal resistive forces, horizontal acceleration is zero (a_x = 0), maintaining constant velocity.\n\n* **Scenario 3: Mass Hanging Stationary by a Cable**\n * **Description:** A mass m suspended at rest from an overhead support by a light rope.\n * **Forces Acting:**\n * Tension force T directed vertically upward along the rope.\n * Gravitational force m\,g directed vertically downward.\n * **Equilibrium Equation:**\n    \sum F_y = T - m\,g = 0 \implies T = m\,g\n\n* **Scenario 4: Mass Moving Downward at Constant Velocity**\n * **Description:** A hanging mass m$lowered vertically at a constant speed ($v = \text{constant}$,$a = 0).\n * **Forces Acting:**\n * Upward tension T.\n * Downward gravitational force m\,g.\n * **Equilibrium Equation:**\n    \sum F_y = T - m\,g = 0 \implies T = m\,g\n\n* **Scenario 5: Mass Accelerating Downward**\n * **Description:** A hanging mass m$accelerating downward ($a > 0).\n * **Forces Acting:**\n * Upward tension T.\n * Downward gravitational force m\,g$.
  • Equation of Motion:$\sum F_y = m\,g - T = m\,a \implies T = m(g - a)$

• Scenario 6: Mass Pulled to the Right across a Frictionless Surface

  • Description: Mass $m$ pulled horizontally to the right by an external horizontal force $F_{\text{pull}}$.
  • Forces Acting:
  • Normal force $N$ directed vertically upward.
  • Gravitational force $m\,g$ directed vertically downward.
  • Pulling force $F_{\text{pull}}$ directed horizontally to the right.
  • Equations of Motion:$\sum F_y = N - m\,g = 0 \implies N = m\,g$$\sum F_x = F_{\text{pull}} = m\,a_x$

Inclined Plane Mechanics

• Problem 1: Mass on Inclined Plane Subjected to Horizontal Push
  • Physical Setup: A block of mass $m$ sits on an inclined plane angled at $\theta$ above the horizontal, subjected to an external horizontal pushing force directed toward the incline.

• Coordinate Transformation: Define the positive $x$-axis parallel to the incline pointing up-slope and the positive $y$-axis perpendicular to the incline pointing away from the ramp surface.
• Force Component Decompositions:
  • Gravitational Force ($m\,g$): Points straight down vertically.
    • Parallel component (down-plane): $-m\,g\,\sin(\theta)$
    • Perpendicular component (into plane): $-m\,g\,\cos(\theta)$
  • Horizontal Push Force ($F_{\text{push}}$): Points straight horizontally to the right.
    • Parallel component (up-plane): $F_{\text{push}}\,\cos(\theta)$
    • Perpendicular component (into plane): $-F_{\text{push}}\,\sin(\theta)$
  • Normal Force ($N$): Directed along positive $y$-axis.

• Governing Equilibrium Equations:$\sum F_y = N - m\,g\,\cos(\theta) - F_{\text{push}}\,\sin(\theta) = 0$$N = m\,g\,\cos(\theta) + F_{\text{push}}\,\sin(\theta)$$\sum F_x = F_{\text{push}}\,\cos(\theta) - m\,g\,\sin(\theta) - f$

  • Problem 2: Stationary Mass on a Rough Inclined Plane

• Physical Setup: A block of mass $m$ remains statically at rest on an inclined plane oriented at angle $\theta$ relative to the horizontal.

• Force Analysis:
  • Upward normal force $N$ perpendicular to incline.
  • Downward gravitational force $m\,g$ pointing vertically downward, subtending angle $\theta$ with the rotated $-y$-axis.
  • Static friction force $f_s$ directed up the incline parallel to the ramp surface to prevent downward sliding.

• Static Equilibrium Equations:$\sum F_y = N - m\,g\,\cos(\theta) = 0 \implies N = m\,g\,\cos(\theta)$$\sum F_x = f_s - m\,g\,\sin(\theta) = 0 \implies f_s = m\,g\,\sin(\theta)$

  • Problem 3: Mass Sliding Down an Incline at Constant Velocity

• Physical Setup: A mass $m$ slides down a ramp angled at $\theta$ with constant velocity ($v = \text{constant}$, acceleration $a = 0$).

• Force Balance and Kinetic Friction Analysis:
  • Normal force $N$ acts perpendicular to the surface.
  • Downward pull of gravity decomposes into $m\,g\,\sin(\theta)$ down the plane and $m\,g\,\cos(\theta)$ into the plane.
  • Kinetic friction $f_k$ acts up the incline opposing downward sliding motion.

• Mathematical Derivations:$\sum F_y = N - m\,g\,\cos(\theta) = 0 \implies N = m\,g\,\cos(\theta)$$\sum F_x = m\,g\,\sin(\theta) - f_k = 0 \implies f_k = m\,g\,\sin(\theta)$$f_k = \mu_k\,N \implies \mu_k\,m\,g\,\cos(\theta) = m\,g\,\sin(\theta)$$\mu_k = \frac{m\,g\,\sin(\theta)}{m\,g\,\cos(\theta)} = \tan(\theta)$

Connected Pulley Systems

• Problem 4: Horizontal Table and Hanging Mass System (Modified Atwood Machine)
  • Physical Setup: Block $m_1$ rests on a horizontal table connected by an ideal string passing over a massless, frictionless pulley to block $m_2$ hanging vertically.

• Free-Body Diagram & Analysis for Mass $m_1$ (on table):

  • Vertical forces: Upward normal force $N$ and downward gravitational force $m_1\,g$.
  • Horizontal forces: Rightward tension $T$ and leftward friction force $f$
  • Axis configuration: $x$ pointing right, $y$ pointing up.     $\sum F_y = N - m_1\,g = 0 \implies N = m_1\,g$$\sum F_x = T - f = m_1\,a$

• Free-Body Diagram & Analysis for Mass $m_2$ (hanging vertically):

  • Vertical forces: Upward tension $T$ and downward gravitational force $m_2\,g$
  • Axis configuration: $x$ pointing downward along direction of motion, $y$ pointing rightward.

* Equations of Motion:

      $\sum F_x = m_2\,g - T = m_2\,a$

• Problem 5: Classic Atwood Machine
  • Physical Setup: Two masses $m_1$ and $m_2$ are suspended vertically on opposite sides of a light string passing over a frictionless, massless pulley.

• Ideal System Assumptions:

  • Massless, inextensible string ensures identical tension $T$ and acceleration magnitude $a$ across both masses.
  • Massless, frictionless pulley requires zero net torque to accelerate.

• Free-Body Diagram for Mass $m_1$:

  • Upward force: Tension $T$
  • Downward force: Gravitational force $m_1\,g$
  • Equation (assuming upward acceleration): $T - m_1\,g = m_1\,a$

• Free-Body Diagram for Mass $m_2$:
  • Upward force: Tension $T$
  • Downward force: Gravitational force $m_2\,g$
  • Equation (assuming downward acceleration for $m_2 > m_1$): $m_2\,g - T = m_2\,a$

• Algebraic Derivations for System Dynamics:
  • Adding equations for $m_1$ and $m_2$ eliminates tension $T$:       $m_2\,g - m_1\,g = (m_1 + m_2)\,a$$a = g\,\frac{m_2 - m_1}{m_1 + m_2}$
  • Substituting acceleration $a$ back into either mass equation yields tension $T$:       $T = \frac{2\,m_1\,m_2\,g}{m_1 + m_2}$