Comprehensive Physics Study Guide: Classical Mechanics, Kinematics, and Dynamics

Static Equilibrium in Multi-Rope Systems

• Static equilibrium requires that both the net horizontal force and net vertical force acting on a system equal zero:   $\sum F_x = 0$$\sum F_y = 0$

System Setup and Parameters

• Mass of performer: $m = 76.0\,\text{kg}$
• Gravitational acceleration: $g = 9.80\,\text{m/s}^2$
• Downward weight force: $w = m g = 76.0\,\text{kg} \times 9.80\,\text{m/s}^2 = 744.8\,\text{N}$
• Rope 1 tension ($T_1$): Oriented at an angle of $15^\circ$ relative to the vertical line.
• Rope 2 tension ($T_2$): Oriented at an angle of $10^\circ$ above the horizontal line.

Force Component Equations

• Horizontal force equilibrium:   $T_2 \cos(10^\circ) - T_1 \sin(15^\circ) = 0$$T_1 \sin(15^\circ) = T_2 \cos(10^\circ)$
• Vertical force equilibrium:   $T_1 \cos(15^\circ) + T_2 \sin(10^\circ) - w = 0$$T_1 \cos(15^\circ) + T_2 \sin(10^\circ) = 744.8\,\text{N}$

Step-by-Step Derivation and Calculation

• Express $T_1$ in terms of $T_2$ using horizontal force balance:   $T_1 = T_2 \frac{\cos(10^\circ)}{\sin(15^\circ)}$$T_1 = T_2 \times \frac{0.98481}{0.25882} \approx 3.8050 T_2$
• Substitute $T_1$ into the vertical force equilibrium equation:   $(3.8050 T_2) \cos(15^\circ) + T_2 \sin(10^\circ) = 744.8\,\text{N}$$3.8050 T_2 (0.96593) + 0.17365 T_2 = 744.8\,\text{N}$$3.6754 T_2 + 0.17365 T_2 = 744.8\,\text{N}$$3.8490 T_2 = 744.8\,\text{N}$$T_2 = \frac{744.8\,\text{N}}{3.8490} \approx 193.5\,\text{N}$
• Calculate $T_1$:   $T_1 = 3.8050 \times 193.5\,\text{N} \approx 736.3\,\text{N}$

One-Dimensional Kinematics and Elevator Dynamics

• The vertical motion of an elevator supported by a cable is governed by Newton's second law along the vertical axis:   $\sum F_y = T - m g = m a$$T = m(g + a)$

System Parameters

• Mass of elevator and passengers: $m = 1.7 \times 10^3\,\text{kg} = 1700\,\text{kg}$
• Acceleration due to gravity: $g = 9.80\,\text{m/s}^2$

Phase (a): Upward Acceleration from Rest

• Acceleration: $a_1 = 1.20\,\text{m/s}^2$ upward for duration $t_1 = 1.50\,\text{s}$.
• Cable tension calculation:   $T_a = m(g + a_1)$$T_a = (1.7 \times 10^3\,\text{kg}) \times (9.80\,\text{m/s}^2 + 1.20\,\text{m/s}^2) = 1700\,\text{kg} \times 11.00\,\text{m/s}^2 = 18700\,\text{N} = 1.87 \times 10^4\,\text{N}$

Phase (b): Constant Velocity Motion

• Acceleration: $a_2 = 0\,\text{m/s}^2$ for duration $t_2 = 8.50\,\text{s}$.
• Cable tension calculation:   $T_b = m g$$T_b = (1.7 \times 10^3\,\text{kg}) \times 9.80\,\text{m/s}^2 = 16660\,\text{N} = 1.67 \times 10^4\,\text{N}$

Phase (c): Deceleration (Acceleration Opposite to Motion)

• Acceleration: $a_3 = -0.600\,\text{m/s}^2$ for duration $t_3 = 3.00\,\text{s}$.
• Cable tension calculation:   $T_c = m(g + a_3)$$T_c = (1.7 \times 10^3\,\text{kg}) \times (9.80\,\text{m/s}^2 - 0.600\,\text{m/s}^2) = 1700\,\text{kg} \times 9.20\,\text{m/s}^2 = 15640\,\text{N} = 1.56 \times 10^4\,\text{N}$

Phase (d): Displacement and Velocity Analysis

• Velocity after Phase 1:   $v_1 = v_0 + a_1 t_1 = 0\,\text{m/s} + (1.20\,\text{m/s}^2)(1.50\,\text{s}) = 1.80\,\text{m/s}$
• Displacement during Phase 1:   $\Delta y_1 = v_0 t_1 + \frac{1}{2} a_1 t_1^2 = 0 + \frac{1}{2}(1.20\,\text{m/s}^2)(1.50\,\text{s})^2 = 1.35\,\text{m}$
• Velocity after Phase 2:   $v_2 = v_1 = 1.80\,\text{m/s}$
• Displacement during Phase 2:   $\Delta y_2 = v_1 t_2 = (1.80\,\text{m/s})(8.50\,\text{s}) = 15.3\,\text{m}$
• Final velocity after Phase 3:   $v_f = v_2 + a_3 t_3 = 1.80\,\text{m/s} + (-0.600\,\text{m/s}^2)(3.00\,\text{s}) = 1.80\,\text{m/s} - 1.80\,\text{m/s} = 0\,\text{m/s}$
• Displacement during Phase 3:   $\Delta y_3 = v_2 t_3 + \frac{1}{2} a_3 t_3^2 = (1.80\,\text{m/s})(3.00\,\text{s}) + \frac{1}{2}(-0.600\,\text{m/s}^2)(3.00\,\text{s})^2 = 5.40\,\text{m} - 2.70\,\text{m} = 2.70\,\text{m}$
• Total height moved above starting point:   $\Delta y_{\text{total}} = \Delta y_1 + \Delta y_2 + \Delta y_3 = 1.35\,\text{m} + 15.3\,\text{m} + 2.70\,\text{m} = 19.35\,\text{m}$
• Final velocity: $v_f = 0\,\text{m/s}$

Frictionless Inclined Plane Motion

• On a frictionless incline inclined at angle $\theta$ to the horizontal, the force component parallel to the incline is $m g \sin(\theta)$.

System Parameters

• Length of track: $d = 30.0\,\text{m}$
• Incline angle: $\theta = 20.0^\circ$
• Initial speed: $v_0 = 0\,\text{m/s}$
• Acceleration of gravity: $g = 9.80\,\text{m/s}^2$

(a) Acceleration Along the Incline

• Formula:   $a = g \sin(\theta)$
• Calculation:   $a = 9.80\,\text{m/s}^2 \times \sin(20.0^\circ) = 9.80\,\text{m/s}^2 \times 0.34202 \approx 3.35\,\text{m/s}^2$

(b) Elapsed Time Calculation

• Kinematic equation:   $d = v_0 t + \frac{1}{2} a t^2$$30.0\,\text{m} = 0 + \frac{1}{2} (3.3518\,\text{m/s}^2) t^2$$t^2 = \frac{2 \times 30.0\,\text{m}}{3.3518\,\text{m/s}^2} \approx 17.8998$$t = \sqrt{17.8998} \approx 4.23\,\text{s}$

Force Analysis and Static Equilibrium in Rock Climbing

System Parameters

• Mass of climber: $m = 52.0\,\text{kg}$
• Climber weight: $w = m g = 52.0\,\text{kg} \times 9.80\,\text{m/s}^2 = 509.6\,\text{N}$
• Rope angle with vertical cliff face: $\theta_{\text{rope}} = 31^\circ$
• Leg force angle relative to horizontal: $\theta_{\text{legs}} = 15^\circ$

(a) Tension and Leg Force Calculations

• Horizontal force equilibrium (taking horizontal axis perpendicular to the cliff):   $F_{\text{legs}} \cos(15^\circ) - T \sin(31^\circ) = 0$$F_{\text{legs}} = T \frac{\sin(31^\circ)}{\cos(15^\circ)}$
• Vertical force equilibrium:   $T \cos(31^\circ) + F_{\text{legs}} \sin(15^\circ) - w = 0$
• Substitute leg force expression into vertical equation:   $T \cos(31^\circ) + T \frac{\sin(31^\circ) \sin(15^\circ)}{\cos(15^\circ)} = w$$T \left( \cos(31^\circ) + \sin(31^\circ) \tan(15^\circ) \right) = 509.6\,\text{N}$$T (0.85717 + 0.51504 \times 0.26795) = 509.6\,\text{N}$$T (0.85717 + 0.13800) = 509.6\,\text{N}$$T (0.99517) = 509.6\,\text{N} \implies T \approx 512\,\text{N}$
• Compute force exerted parallel to legs ($F_{\text{legs}}$):   $F_{\text{legs}} = 512.07\,\text{N} \times \frac{\sin(31^\circ)}{\cos(15^\circ)} = 512.07\,\text{N} \times \frac{0.51504}{0.96593} \approx 273\,\text{N}$

(b) Minimum Coefficient of Static Friction

• The leg force exerted against the cliff face has a normal component $N$ perpendicular to the cliff and a tangential friction force $f_s$ parallel to the cliff face:   $N = F_{\text{legs}} \cos(15^\circ)$$f_s = F_{\text{legs}} \sin(15^\circ)$
• Minimum static coefficient of friction condition:   $\mu_{s,\text{min}} = \frac{f_s}{N} = \frac{F_{\text{legs}} \sin(15^\circ)}{F_{\text{legs}} \cos(15^\circ)} = \tan(15^\circ) \approx 0.268$

Vehicle Acceleration Dynamics on Inclined Slopes with Friction Limits

• When a vehicle accelerates up a slope, the driving force is provided by static friction at the drive wheels.

General Governing Equation

• Slope angle: $\theta = 4.00^\circ$
• Normal force on drive wheels (supporting half the total weight):   $N_{\text{drive}} = \frac{1}{2} m g \cos(4.00^\circ)$
• Maximum available static friction force:   $f_{s,\text{max}} = \mu_s N_{\text{drive}} = \frac{1}{2} \mu_s m g \cos(4.00^\circ)$
• Acceleration balance along incline:   $f_{s,\text{max}} - m g \sin(4.00^\circ) = m a_{\text{max}}$$a_{\text{max}} = g \left( \frac{1}{2} \mu_s \cos(4.00^\circ) - \sin(4.00^\circ) \right)$
• Numerical constant values:   $\cos(4.00^\circ) \approx 0.99756$$\sin(4.00^\circ) \approx 0.06976$$g = 9.80\,\text{m/s}^2$

(a) Acceleration on Dry Concrete ($\mu_s = 1.0$)

• Calculation:   $a_{\text{max}} = 9.80\,\text{m/s}^2 \times \left( 0.5 \times 1.0 \times 0.99756 - 0.06976 \right)$$a_{\text{max}} = 9.80 \times (0.49878 - 0.06976) = 9.80 \times 0.42902 \approx 4.20\,\text{m/s}^2$

(b) Acceleration on Wet Concrete ($\mu_s = 0.7$)

• Calculation:   $a_{\text{max}} = 9.80\,\text{m/s}^2 \times \left( 0.5 \times 0.7 \times 0.99756 - 0.06976 \right)$$a_{\text{max}} = 9.80 \times (0.27932 - 0.06976) = 9.80 \times 0.20956 \approx 2.05\,\text{m/s}^2$

(c) Acceleration on Ice ($\mu_s = 0.100$)

• Calculation:   $a_{\text{max}} = 9.80\,\text{m/s}^2 \times \left( 0.5 \times 0.100 \times 0.99756 - 0.06976 \right)$$a_{\text{max}} = 9.80 \times (0.04988 - 0.06976) = 9.80 \times (-0.01988) \approx -0.195\,\text{m/s}^2$
• Physical Meaning: The negative value demonstrates that static friction is insufficient to counteract gravity down the incline; the vehicle cannot accelerate Uphill and will slide backwards.

Centripetal Force and Seat Normal Force in Vertical Circular Loops

System Parameters

• Mass of child: $m = 40.0\,\text{kg}$
• Radius of circular loop: $r = 7.00\,\text{m}$
• Speed at Point A (top of loop): $v_A = 10.0\,\text{m/s}$
• Speed at Point B ($30^\circ$ past top of loop): $v_B = 10.5\,\text{m/s}$

(a) Force of Car Seat on Child at Point A

• At the top of the loop, both normal force $N_A$ and gravity $m g$ point downward toward the loop center:   $N_A + m g = \frac{m v_A^2}{r}$$N_A = m \left( \frac{v_A^2}{r} - g \right)$
• Computation:   $N_A = 40.0\,\text{kg} \times \left( \frac{(10.0\,\text{m/s})^2}{7.00\,\text{m}} - 9.80\,\text{m/s}^2 \right)$$N_A = 40.0 \times (14.286 - 9.80) = 40.0 \times 4.486 \approx 179\,\text{N}$

(b) Force of Car Seat on Child at Point B

• Point B is positioned at an angle $\theta = 30.0^\circ$ from vertical. Radial inward gravitational component is $m g \cos(30^\circ)$:   $N_B + m g \cos(30^\circ) = \frac{m v_B^2}{r}$$N_B = m \left( \frac{v_B^2}{r} - g \cos(30^\circ) \right)$
• Computation:   $\frac{v_B^2}{r} = \frac{(10.5\,\text{m/s})^2}{7.00\,\text{m}} = \frac{110.25}{7.00} = 15.75\,\text{m/s}^2$$g \cos(30^\circ) = 9.80\,\text{m/s}^2 \times 0.86603 = 8.487\,\text{m/s}^2$$N_B = 40.0\,\text{kg} \times (15.75\,\text{m/s}^2 - 8.487\,\text{m/s}^2) = 40.0 \times 7.263 \approx 291\,\text{N}$

(c) Minimum Speed Required at Point A to Remain in Seat

• The child remains in contact with the seat as long as $N_A \ge 0$. Threshold velocity occurs when $N_A = 0$:   $\frac{m v_{\text{min}}^2}{r} = m g$$v_{\text{min}} = \sqrt{g r}$
• Computation:   $v_{\text{min}} = \sqrt{9.80\,\text{m/s}^2 \times 7.00\,\text{m}} = \sqrt{68.6} \approx 8.28\,\text{m/s}$

Maximum Traversing Speed on Unbanked Curves

• On an unbanked flat curve, centripetal force is supplied entirely by static friction between tires and road.

System Parameters

• Radius of unbanked curve: $r = 65\,\text{m}$
• Coefficient of static friction: $\mu_s = 0.70$
• Gravitational acceleration: $g = 9.80\,\text{m/s}^2$

Derivation and Speed Calculation

• Friction force balance:   $f_{s,\text{max}} = \mu_s N = \mu_s m g$$\frac{m v_{\text{max}}^2}{r} = \mu_s m g$$v_{\text{max}} = \sqrt{\mu_s g r}$
• Computation:   $v_{\text{max}} = \sqrt{0.70 \times 9.80\,\text{m/s}^2 \times 65\,\text{m}} = \sqrt{445.9} \approx 21.1\,\text{m/s}$

Dynamics of Ideally Banked Circular Turns

• An ideally banked curve allows a vehicle to make a turn at a specific speed without relying on friction forces.

System Parameters

• Bank angle: $\theta = 75.0^\circ$
• Bobsled speed: $v = 30.0\,\text{m/s}$
• Acceleration of gravity: $g = 9.80\,\text{m/s}^2$

(a) Radius of the Turn

• Relationship for ideal banking angle:   $\tan(\theta) = \frac{v^2}{g r}$$r = \frac{v^2}{g \tan(\theta)}$
• Computation:   $\tan(75.0^\circ) \approx 3.73205$$r = \frac{(30.0\,\text{m/s})^2}{9.80\,\text{m/s}^2 \times 3.73205} = \frac{900}{36.574} \approx 24.6\,\text{m}$

(b) Centripetal Acceleration

• Formula:   $a_c = \frac{v^2}{r} = g \tan(\theta)$
• Computation:   $a_c = 9.80\,\text{m/s}^2 \times \tan(75.0^\circ) = 9.80 \times 3.73205 \approx 36.6\,\text{m/s}^2$

(c) Acceleration Expressed in g's

• Conversion:   $\text{Acceleration in g's} = \frac{a_c}{g} = \tan(75.0^\circ) \approx 3.73\,\text{g}$

Fluid Drag and Terminal Velocity Analysis

• Falling objects reach terminal velocity when resistive air drag force equals object weight.

Drag Force Governing Model

• Drag force formula:   $F_d = \frac{1}{2} \rho C A v^2$   where:
  • $\rho$ is fluid density.
  • $C$ is dimensionless drag coefficient.
  • $A$ is cross-sectional frontal area.
  • $v$ is velocity relative to fluid.

System Parameters

• Skydiver mass: $m = 80.0\,\text{kg}$
• Cross-sectional area in pike position: $A = 0.14\,\text{m}^2$
• Drag coefficient: $C = 0.70$
• Air density: $\rho = 1.21\,\text{kg/m}^3$
• Gravitational acceleration: $g = 9.80\,\text{m/s}^2$

Terminal Velocity Derivation and Calculation

• Force balance at terminal speed ($v_t$):   $F_d = m g$$\frac{1}{2} \rho C A v_t^2 = m g$$v_t = \sqrt{\frac{2 m g}{\rho C A}}$
• Numerical substitution:   $2 m g = 2 \times 80.0\,\text{kg} \times 9.80\,\text{m/s}^2 = 1568\,\text{N}$$\rho C A = 1.21\,\text{kg/m}^3 \times 0.70 \times 0.14\,\text{m}^2 = 0.11858\,\text{kg/m}$$v_t = \sqrt{\frac{1568\,\text{N}}{0.11858\,\text{kg/m}}} = \sqrt{13223.14} \approx 115\,\text{m/s}$