2D Kinematics and Dynamics Study Guide

Fundamentals of 2D Kinematics and Dynamics

• Course Context and Overview:

  • Topics cover 2D Kinematics and Dynamics under Mullin, with specific application noted for assisting with the rocket lab.
  • Section focus: 1.1 - Review and Expansion of Dynamics.
  • Vector Dynamics is fundamentally rooted in Newton's Three Laws of Motion.

• Core Equations and Definitions:

  • Newton's Second Law of Motion in vector form:     $\mathbf{F}_{\text{net}} = m \mathbf{a}$
  • Gravitational Force and Weight relationship:     $F_g = m g$$\text{weight} = F_g = m g$

• The Winning Force Principle:

  • The winning force in any dynamic scenario is explicitly defined as the force acting in the direction of the object's motion.
  • Example Application: When a rocket accelerates upwards, the winning force is the thrust force ($F_{TH}$), which opposes the force of gravity ($F_g$).

Applied Dynamics Problems and Worked Examples

• Example 1: Rocket Dynamics and Mass Consumption Rate

  • Problem Description:
  • A $12.0\,\text{kg}$ rocket experiences a $215\,\text{N}$ thrust force ($F_{TH}$) at launch.
  • Part A: Calculate the initial acceleration ($a$) of the rocket.
  • Equation set up using winning force:       $F_{TH} - F_g = m a$
  • Substitution of values ($m = 12.0\,\text{kg}$, $F_{TH} = 215\,\text{N}$, $g = 9.8\,\text{m/s}^2$):       $215 - 12(9.8) = 12 a$$215 - 117.6 = 12 a$$97.4 = 12 a$$a = \frac{97.4}{12} = 8.12\,\text{m/s}^2\text{, up}$
  • Part B: If $25\,\text{s}$ later its acceleration increases to $28\,\text{m/s}^2$, determine the rate of fuel consumption.
  • Set up the dynamic equation to find the new rocket mass ($m$):       $F_{TH} - F_g = m a$$215 - m(9.8) = m(28)$$215 = m(28) + m(9.8)$$215 = m(37.8)$$m = \frac{215}{37.8} = 5.69\,\text{kg}$
  • Calculate mass change (fuel burned):       $\Delta m = m_{\text{final}} - m_{\text{initial}} = 5.69\,\text{kg} - 12.0\,\text{kg} = -6.31\,\text{kg}$
  • Calculate the fuel consumption rate over the time interval ($\Delta t = 25\,\text{s}$):       $\text{rate} = \frac{\Delta m}{\Delta t}$$\text{rate} = \frac{5.69 - 12}{25} = \frac{-6.31}{25} = -0.25\,\text{kg/s}$

• Example 2: Connected Dragging Systems with Friction

  • Problem Description:
  • A $2.0\,\text{kg}$ toy tractor exerts a propulsive force ($F_P$) of $18\,\text{N}$ across a bench while dragging a $3.0\,\text{kg}$ block.
  • The coefficient of friction ($\mu$) between the block and the bench is $0.20$.
  • Part A: Calculate the system acceleration ($a$).
  • System component equations:
    • Tractor ($2.0\,\text{kg}$):         $F_P - F_T = 2 a$
    • Block ($3.0\,\text{kg}$):         $F_T - F_f = 3 a$
  • Sum of system components ($m_{\text{total}} = 5.0\,\text{kg}$):       $F_P - F_f = 5 a$
  • Evaluate friction force ($F_f = \mu m_{\text{block}} g$):       $18 - (0.20)(3.0)(9.8) = 5 a$$18 - 5.88 = 5 a$$12.12 = 5 a$$a = 2.42\,\text{m/s}^2\text{, right}$
  • Part B: Calculate the tension force ($F_T$) in the connecting cord.
  • Substitute acceleration into the block's motion equation:       $F_T - F_f = 3 a$$F_T - (0.20)(3.0)(9.8) = 3(2.42)$$F_T - 5.88 = 7.26$$F_T = 13.14\,\text{N}$

• Example 3: Direct Contact Push Dynamics with Single Friction Source

  • Problem Description:
  • Joe pushes a $55\,\text{kg}$ stove with a force of $325\,\text{N}$.
  • The stove is directly in contact with and pushes a $75\,\text{kg}$ fridge.
  • Coefficient of friction ($\mu$) for the stove is $0.40$.
  • The fridge experiences no friction.
  • Part A: Calculate system acceleration ($a$).
  • Component dynamic equations:
    • Stove ($55\,\text{kg}$):         $F_{\text{push}} - F_C - F_{f,\text{stove}} = 55 a$
    • Fridge ($75\,\text{kg}$):         $F_C = 75 a$
  • Combine equations for system total mass ($130\,\text{kg}$):       $325 - (0.40)(55)(9.8) = 130 a$$325 - 215.6 = 130 a$$109.4 = 130 a$$a = 0.84\,\text{m/s}^2\text{, right}$
  • Part B: Calculate the contact force ($F_C$) between stove and fridge.
  • Substitute acceleration into fridge equation:       $F_C = 75(0.84)$$F_C = 63\,\text{N}$

• Example 4: Direct Contact Push Dynamics with Friction on Both Objects

  • Problem Description:
  • Repeat Example 3 with an added frictional force of $40\,\text{N}$ acting on the fridge.
  • Part A: Calculate system acceleration ($a$).
  • Dynamic equations:
    • Stove ($55\,\text{kg}$):         $F_{\text{push}} - F_C - F_{f,\text{stove}} = 55 a$
    • Fridge ($75\,\text{kg}$):         $F_C - F_{f,\text{fridge}} = 75 a$
  • Combine equations:       $325 - 215.6 - 40 = 130 a$$69.4 = 130 a$$a = 0.53\,\text{m/s}^2\text{, right}$
  • Part B: Calculate the contact force ($F_C$).
  • Substitute into fridge dynamic equation:       $F_C - F_{f,\text{fridge}} = 75 a$$F_C - 40 = 75(0.53)$$F_C - 40 = 39.75$$F_C = 79.75\,\text{N}$
  • Rounded result:       $F_C = 80\,\text{N}$

• Example 5: Single Pulley and Horizontal Table System

  • Problem Description:
  • A $5\,\text{kg}$ object (Object A) sits on a frictionless table connected via a string over a pulley to a $2\,\text{kg}$ object (Object B) hanging vertically.
  • Acceleration and Tension Calculations:
  • Motion equation for Object A:       $F_T = 5 a$
  • Motion equation for Object B:       $F_{gB} - F_T = 2 a$
  • System net force equation:       $F_{gB} = (5 + 2) a$$2(9.8) = 7 a$$19.6 = 7 a$$a = 2.8\,\text{m/s}^2$
  • Solve for tension force ($F_T$):       $F_T = 5(2.8) = 14\,\text{N}$

• Example 6: Double Pulley Pulled Table System

  • Problem Description:
  • A $5.0\,\text{kg}$ block rests on a frictionless table.
  • Attached via a left string over a pulley to a hanging $2.0\,\text{kg}$ block.
  • Attached via a right string over a pulley to a hanging $3.0\,\text{kg}$ block.
  • Part A: Calculate system acceleration ($a$).
  • Dynamic equations:
    • Right hanging block ($3.0\,\text{kg}$):         $F_{g3} - F_{TR} = 3 a$
    • Middle block ($5.0\,\text{kg}$):         $F_{TR} - F_{TL} = 5 a$
    • Left hanging block ($2.0\,\text{kg}$):         $F_{TL} - F_{g2} = 2 a$
  • System combined equation ($m_{\text{total}} = 10.0\,\text{kg}$):       $F_{g3} - F_{g2} = 10 a$$3(9.8) - 2(9.8) = 10 a$$29.4 - 19.6 = 10 a$$9.8 = 10 a$$a = 0.98\,\text{m/s}^2\text{, right}$
  • Part B: Calculate Left String Tension ($F_{TL}$).
  • Set up using left block dynamic equation:       $F_{TL} - F_{g2} = 2 a$$F_{TL} - 2(9.8) = 2(0.98)$$F_{TL} - 19.6 = 1.96$$F_{TL} = 21.56\,\text{N}$
  • Rounded value:       $F_{TL} = 22\,\text{N}$
  • Part C: Calculate Right String Tension ($F_{TR}$).
  • Set up using middle block dynamic equation:       $F_{TR} - F_{TL} = 5 a$$F_{TR} - 21.56 = 5(0.98)$$F_{TR} - 21.56 = 4.9$$F_{TR} = 26.46\,\text{N}$
  • Rounded value:       $F_{TR} = 26\,\text{N}$