Motion and Force Dynamics Study Guide

Introduction to Force and Dynamics

• Dynamics explores the underlying causes of changes in motion, extending beyond pure kinematics (horizontal straight-line, vertical straight-line, and circular motion) to analyze why objects start moving, stop, accelerate, or maintain constant velocity.

• The study of dynamics integrates key kinematic quantities—displacement, velocity, and acceleration—with the foundational physical properties of force and mass.

• Core objectives of force dynamics:

  • Define force in terms of the acceleration imparted to a standard body.
  • Assign mass to objects to explain why different bodies experience different accelerations when subjected to identical environmental conditions.
  • Calculate net forces acting on a body based on object properties and environmental interactions.

Concepts of Force and Mass

• A force is defined in everyday terms as a push or a pull.

• Force is a vector quantity, possessing both magnitude and direction.

• A spring balance is a standard instrument used for measuring the magnitude of a force.

• Vector resolution of forces allows any angled force to be split into perpendicular horizontal ($x$) and vertical ($y$) components:

• Example of force component calculation:
  • For a block of mass $m = 50.0\,\text{kg}$ pulled by an applied force $F_A = 400\,\text{N}$ at an angle of $57^\circ$ above the horizontal:
  • Horizontal component ($F_{Ax}$):

$F_{Ax} = F_A \cos(57^\circ) = 400\,\text{N} \times \cos(57^\circ) = 218\,\text{N}$

• Vertical component ($F_{Ay}$):

$F_{Ay} = F_A \sin(57^\circ) = 400\,\text{N} \times \sin(57^\circ) = 335\,\text{N}$

Newton's Laws of Motion

• Newton's First Law of Motion (Law of Inertia):
  • A body remains in a state of rest or continues in uniform motion in a straight line at a constant velocity unless acted upon by a net external force.
  • Inertia is the inherent property of a body to resist changes in its state of rest or motion.
  • When the vector sum of all forces acting on a body is zero, the body is in translational equilibrium:

$\sum \mathbf{F} = 0$

$\sum F_x = 0$

$\sum F_y = 0$

• In equilibrium, an object's acceleration is zero ($a = 0\,\text{m/s}^2$), meaning it is either permanently at rest or moving with constant velocity.

  • Newton's Second Law of Motion:

• If a net external force acts on a body, the body accelerates in the direction of that net force. The net force vector equals the product of the mass and the acceleration of the body:

$\sum \mathbf{F} = m \mathbf{a}$

$\sum F_x = m a_x$

$\sum F_y = m a_y$

• The SI unit of force is the Newton ($\text{N}$):

$1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$

• One Newton is defined as the force required to impart an acceleration of $1\,\text{m/s}^2$ to a mass of $1\,\text{kg}$.
• Worked Example (Car Deceleration):
  • Problem: Calculate the net force required to bring a $1500\,\text{kg}$ car to rest from an initial velocity of $30.6\,\text{m/s}$ within a stopping distance of $55\,\text{m}$.
  • Kinematic equation with constant acceleration:

$v_f^2 = v_i^2 + 2 a \Delta x$

- Solving for acceleration ($a$):

$a = \frac{v_f^2 - v_i^2}{2 \Delta x} = \frac{0^2 - (30.6\,\text{m/s})^2}{2(55\,\text{m})} = \frac{-936.36}{110\,\text{m}} = -8.5127\,\text{m/s}^2$

- Calculating required force ($F$):

$F = m a = (1500\,\text{kg}) \times (-8.5127\,\text{m/s}^2) = -12769\,\text{N} \approx -12800\,\text{N}$

- The negative sign indicates that the net force opposes the initial velocity vector.

• Newton's Third Law of Motion:
  • For every action, there is an equal and opposite reaction.
  • If body A exerts a force $\mathbf{F}_{AB}$ on body B, body B simultaneously exerts an equal and opposite force $\mathbf{F}_{BA}$ on body A:

$\mathbf{F}_{AB} = -\mathbf{F}_{BA}$

• Action and reaction forces act on two entirely different objects, meaning they never directly cancel each other out on a single free body diagram.

Fundamental Types of Forces

• Weight ($w$):
  • The gravitational pull exerted on a body by Earth (or another astronomical body).
  • From Newton's second law, for a freely falling body, $\mathbf{a} = \mathbf{g}$:

$w = m g$

• Weight acts vertically downward toward the center of the gravitational source.

  • Normal Force ($N$ or $F_N$):

• A contact force exerted by a surface on an object touching it.

• It always acts perpendicular to the surface of contact.

  • Tension ($T$):

• The pulling force exerted by each end of a stretched rope, cord, string, or cable.

• Tension pulls away from the attached object along the string line. In a continuous string connecting two objects, tension acts towards opposite directions at the respective endpoints.

  • Friction ($f$):

• A contact force exerted parallel to the interface between an object and a surface, opposing sliding or intended motion.

• Static Friction ($f_s$): Prevents an object at rest from starting to move. Its maximum possible value is:

$f_{s,\text{max}} = \mu_s N$

- Motion does not begin until the applied force exceeds $f_{s,\text{max}}$.

• Kinetic Friction ($f_k$): Retards motion for an object actively sliding across a surface:

$f_k = \mu_k N$

- $\mu_s$ is the coefficient of static friction and $\mu_k$ is the coefficient of kinetic friction.

• Worked Example (Box of Bananas on Horizontal Surface):
  • Given: Box weight $w = 40.0\,\text{N}$, $\mu_s = 0.40$, $\mu_k = 0.20$.
  • Normal force: $N = w = 40.0\,\text{N}$.
  • Maximum static friction limit: $f_{s,\text{max}} = \mu_s N = 0.40 \times 40.0\,\text{N} = 16.0\,\text{N}$.
  • Kinetic friction force: $f_k = \mu_k N = 0.20 \times 40.0\,\text{N} = 8.00\,\text{N}$.
  • Scenario Analyses:
  • a) If no horizontal force is applied at rest, the friction force is $16.0\,\text{N}$ static limit threshold.
  • b) If a monkey applies a horizontal force of $6.0\,\text{N}$ to the box while static, the recorded friction threshold is $16.0\,\text{N}$.
  • c) Minimum horizontal force needed to sustain constant velocity once moving equals the kinetic friction force $f_k = 8.00\,\text{N}$.
  • d) If a horizontal force of $18.0\,\text{N}$ is applied:
    • Net horizontal force: $\sum F_x = F_{\text{applied}} - f_k = 18.0\,\text{N} - 8.00\,\text{N} = 10.0\,\text{N}$.
    • Mass of box: $m = \frac{w}{g} = \frac{40.0\,\text{N}}{9.8\,\text{m/s}^2} = 4.0816\,\text{kg}$.
    • Acceleration calculation:

$a = \frac{\sum F_x}{m} = \frac{10.0\,\text{N}}{4.0816\,\text{kg}} = 2.45\,\text{m/s}^2$

Applications of Newton's First Law

• Equilibrium systems require analyzing force components separately along orthogonal axes such that $\sum F_x = 0$ and $\sum F_y = 0$.

• Example 1: Weight Suspended by Three Cords:

• Given: Block weight $w = 600\,\text{N}$. Three cords meet at origin point $O$. Cord 1 makes $60^\circ$ north of east with the ceiling, Cord 2 is horizontal to the left (west), and Cord 3 hangs vertically supporting the block.
• Step 1: Equilibrium of the suspended block:

$\sum F_y = T_3 - w = 0 \implies T_3 = w = 600\,\text{N}$

• Step 2: Equilibrium at junction $O$:
  • Vertical equilibrium:

$\sum F_y = T_{1y} - T_3 = 0 \implies T_1 \sin(60^\circ) - T_3 = 0$

$T_1 = \frac{T_3}{\sin(60^\circ)} = \frac{600\,\text{N}}{\sin(60^\circ)} = 693\,\text{N}$

- Horizontal equilibrium:

$\sum F_x = T_{1x} - T_2 = 0 \implies T_1 \cos(60^\circ) - T_2 = 0$

$T_2 = T_1 \cos(60^\circ) = 693\,\text{N} \times \cos(60^\circ) = 347\,\text{N}$

• Example 2: Asymmetric Hanging Sphere Equilibrium:

• Given: Mass $m = 5.0\,\text{kg}$. Cord 1 makes an angle of $40^\circ$ with the ceiling ($40^\circ$ north of west). Cord 2 makes an angle of $50^\circ$ with the ceiling ($50^\circ$ north of east). Cord 3 hangs vertically attached to the mass.
• Step 1: Force on suspended block:

$T_3 = w = m g = 5.0\,\text{kg} \times 9.8\,\text{m/s}^2 = 49\,\text{N}$

• Step 2: Forces acting on the junction point:
  • Horizontal components:

$\sum F_x = T_{2x} - T_{1x} = 0 \implies T_2 \cos(50^\circ) = T_1 \cos(40^\circ)$

$T_2 = T_1 \frac{\cos(40^\circ)}{\cos(50^\circ)}$

- Vertical components:

$\sum F_y = T_{1y} + T_{2y} - T_3 = 0 \implies T_1 \sin(40^\circ) + T_2 \sin(50^\circ) - T_3 = 0$

- Substituting $T_2$ expression into vertical equation:

$T_1 \sin(40^\circ) + T_1 \left(\frac{\cos(40^\circ)}{\cos(50^\circ)}\right) \sin(50^\circ) = T_3$

$T_1 \left(\frac{\sin(40^\circ)\cos(50^\circ) + \cos(40^\circ)\sin(50^\circ)}{\cos(50^\circ)}\right) = T_3$

- Utilizing trigonometric angle addition identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$:

$\sin(40^\circ + 50^\circ) = \sin(90^\circ) = 1$

$T_1 \left(\frac{\sin(90^\circ)}{\cos(50^\circ)}\right) = T_3 \implies T_1 = T_3 \cos(50^\circ)$

$T_1 = 49\,\text{N} \times \cos(50^\circ) = 31.5\,\text{N}$

- Calculating $T_2$:

$T_2 = 31.5\,\text{N} \times \frac{\cos(40^\circ)}{\cos(50^\circ)} = 37.5\,\text{N}$

Applications of Newton's Second Law

• Example 1: Single Block Pulled at an Angle with Surface Friction:

• Given: Box mass $m = 10.0\,\text{kg}$, cord tension $T = 40.0\,\text{N}$ at $\theta = 30^\circ$ north of east, coefficient of kinetic friction $\mu_k = 0.20$.
• Weight of box: $w = m g = 10.0\,\text{kg} \times 9.8\,\text{m/s}^2 = 98.0\,\text{N}$.
• Tension components:

$T_x = T \cos(30^\circ) = 40.0\,\text{N} \times \cos(30^\circ) = 34.64\,\text{N}$

$T_y = T \sin(30^\circ) = 40.0\,\text{N} \times \sin(30^\circ) = 20.0\,\text{N}$

• (a) Normal Force ($N$):

$\sum F_y = N + T_y - w = m a_y = 0$

$N = w - T_y = 98.0\,\text{N} - 20.0\,\text{N} = 78.0\,\text{N}$

• (b) Acceleration ($a$):
  • Friction force: $f_k = \mu_k N = 0.20 \times 78.0\,\text{N} = 15.6\,\text{N}$.
  • Horizontal dynamics:

$\sum F_x = T_x - f_k = m a_x$

$a = \frac{T_x - f_k}{m} = \frac{34.64\,\text{N} - 15.6\,\text{N}}{10.0\,\text{kg}} = \frac{19.04\,\text{N}}{10.0\,\text{kg}} = 1.90\,\text{m/s}^2$

• Example 2: Two Connected Blocks over a Pulley:

• Given: Sliding block mass $m_1 = 4.0\,\text{kg}$ on table with $\mu_k = 0.15$, connected via string over pulley to hanging block $m_2 = 2.0\,\text{kg}$.
• Forces on sliding block ($m_1$):
  • Vertical forces: $N = m_1 g = 4.0\,\text{kg} \times 9.8\,\text{m/s}^2 = 39.2\,\text{N}$.
  • Kinetic friction: $f_k = \mu_k N = 0.15 \times 39.2\,\text{N} = 5.88\,\text{N}$.
  • Horizontal force equation: $T - f_k = m_1 a \implies T - 5.88 = 4.0 a$
• Forces on hanging block ($m_2$):
  • Weight: $w_2 = m_2 g = 2.0\,\text{kg} \times 9.8\,\text{m/s}^2 = 19.6\,\text{N}$.
  • Vertical force equation: $w_2 - T = m_2 a \implies 19.6 - T = 2.0 a$
• (a) Acceleration of the two blocks:
  • Summing system equations:

$(T - 5.88) + (19.6 - T) = 4.0 a + 2.0 a$

$13.72 = 6.0 a \implies a = \frac{13.72}{6.0} = 2.29\,\text{m/s}^2$

• (b) Tension in the cord ($T$):

$T = 19.6 - 2.0(2.2867) = 15.0\,\text{N}$

Applications of Newton's Third Law

• Example: Three Contact Blocks Pushed by External Force:

• Given: Three blocks each of mass $m_1 = m_2 = m_3 = 12.0\,\text{kg}$ placed side-by-side on a frictionless surface. External horizontal force $F = 96.0\,\text{N}$ is applied directly to block 1.
• (a) Acceleration of the system:
  • Combined mass: $M_{\text{total}} = m_1 + m_2 + m_3 = 12.0\,\text{kg} + 12.0\,\text{kg} + 12.0\,\text{kg} = 36.0\,\text{kg}$.

$a = \frac{F}{M_{\text{total}}} = \frac{96.0\,\text{N}}{36.0\,\text{kg}} = 2.67\,\text{m/s}^2$

• (b) Contact forces on each block:
  • Free body diagram for block 1 ($m_1$):

$\sum F_x = F - F_{12} = m_1 a \implies 96.0\,\text{N} - F_{12} = (12.0\,\text{kg})(2.67\,\text{m/s}^2) = 32.0\,\text{N}$

$F_{12} = 96.0\,\text{N} - 32.0\,\text{N} = 64.0\,\text{N}$

- Free body diagram for block 3 ($m_3$):

$\sum F_x = F_{23} = m_3 a = (12.0\,\text{kg})(2.67\,\text{m/s}^2) = 32.0\,\text{N}$

- Verification on block 2 ($m_2$):

$\sum F_x = F_{12} - F_{23} = 64.0\,\text{N} - 32.0\,\text{N} = 32.0\,\text{N} = m_2 a$

Newton's Law of Universal Gravitation

• Statement: Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance separating their centers.

• Formula:

$F = G \frac{m_1 m_2}{r^2}$

• $m_1, m_2$ are the particle masses.
• $r$ is the separation distance between centers.
• $G$ is the universal gravitational constant:

$G = 6.674 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2$

• Example Calculation:
  • Given: $m_1 = 12.0\,\text{kg}$, $m_2 = 25.0\,\text{kg}$, $r = 1.20\,\text{m}$.
  • Gravitational force calculation:

$F = (6.674 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2) \times \frac{(12.0\,\text{kg})(25.0\,\text{kg})}{(1.20\,\text{m})^2}$

$F = 6.674 \times 10^{-11} \times \frac{300}{1.44} = 6.674 \times 10^{-11} \times 208.333 = 1.39 \times 10^{-8}\,\text{N}$

Centripetal Force

• Definition: The net inward force required to maintain an object of mass $m$ in uniform circular motion at speed $v$ along a circular path of radius $r$

• Formula:

$F_c = \frac{m v^2}{r} = m a_c$

• Where $a_c = \frac{v^2}{r}$ is the centripetal acceleration.

  • Direction: Always points radially inward toward the center of the circular path, continuously changing direction as the body moves.

  • Worked Example (Ball Twirled on String):

• Given: Ball mass $m = 12.0\,\text{g} = 0.0120\,\text{kg}$, radius $r = 10.0\,\text{cm} = 0.100\,\text{m}$, period $T = 0.500\,\text{s}$.

• Linear speed ($v$):

$v = \frac{2 \pi r}{T} = \frac{2 \pi (0.100\,\text{m})}{0.500\,\text{s}} = 0.400 \pi\,\text{m/s} = 1.2566\,\text{m/s}$

• (a) Tension in string ($T_{\text{string}} = F_c$):

$T_{\text{string}} = \frac{m v^2}{r} = \frac{(0.0120\,\text{kg})(1.2566\,\text{m/s})^2}{0.100\,\text{m}} = 0.190\,\text{N}$

• (b) Effect of doubling speed:
  • Because $F_c \propto v^2$, doubling the speed ($v' = 2v$) results in:

$F_c' = \frac{m (2v)^2}{r} = 4 \left(\frac{m v^2}{r}\right) = 4 F_c$

- Thus, tension does not merely double; it increases by a factor of 4.