Comprehensive Study Guide: Newtonian Mechanics, Dynamics, Free-Body Diagrams, and Vector Analysis

Fundamental Laws and Principles of Forces

• Newton's First Law of Motion (Law of Inertia):

  • An object remains at rest or continues to move at a constant velocity unless acted upon by a net external force.

  • When forces acting on an object are balanced, the net force is zero ($F_{\text{net}} = 0$), resulting in zero acceleration ($a = 0\,\text{m/s}^2$).

• Newton's Second Law of Motion (Force and Acceleration):

  • The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

  • Mathematical Formulation:         $F_{\text{net}} = m \times a$         $a = \frac{F_{\text{net}}}{m}$         $m = \frac{F_{\text{net}}}{a}$

  • Gravitational force (weight) acting on a mass $m$ is calculated using acceleration due to gravity ($g$):         $F_w = m \times g$         (Note: In standard approximations, $g = 10\,\text{m/s}^2$ or $g = 9.8\,\text{m/s}^2$).

• Equilibrium and Equibrant Forces:

  • An object is in static or dynamic equilibrium when the vector sum of all forces acting on it equals zero ($F_{\text{net}} = 0$).

  • Resultant Force ($F_R$): The single combined vector resulting from the addition of two or more individual force vectors.

  • Equibrant / Balancing Force ($F_e$): The force that must be added to a system to produce balanced forces ($F_{\text{net}} = 0$). It is exactly equal in magnitude to the resultant force but acts in the opposite direction ($180^\circ$ away from the resultant angle).

Linear Force Calculations and One-Dimensional Motion

• Balancing a Single Force:

  • Problem: A net force of $35.12\,\text{N}$ acts due west on an object. Determine the added single force required to balance the system.

  • Calculation:         $F_{\text{net}} = F_1 + F_2 = 0$         $35.12\,\text{N} + F_2 = 0 \implies F_2 = -35.12\,\text{N}$

  • Result: A force of $35.12\,\text{N}$ directed due East.

• Friction on a Moving Object at Constant Velocity:

  • Problem: A hockey puck slides across ice at constant velocity driven by an applied force of $18.5\,\text{N}$.

  • Calculation:         $a = 0\,\text{m/s}^2 \implies F_{\text{net}} = F_{\text{applied}} + F_{\text{friction}} = 0$         $18.5\,\text{N} + F_{\text{friction}} = 0 \implies F_{\text{friction}} = -18.5\,\text{N}$

  • Result: The force of friction is $18.5\,\text{N}$ in the direction opposite to motion.

• Acceleration of a Pushed Box:

  • Problem: An $8.75\,\text{kg}$ box is pushed horizontally with a net force of $46.15\,\text{N}$.

  • Calculation:         $a = \frac{F_{\text{net}}}{m} = \frac{46.15\,\text{N}}{8.75\,\text{kg}} = 5.27\,\text{m/s}^2$

• Acceleration with Variable Mass Terms:

  • Problem: A model airplane has a mass of $(1.5 m)\,\text{kg}$ and experiences a forward net propeller force of $(3.75 m)\,\text{N}$.

  • Calculation:         $a = \frac{F_{\text{net}}}{m} = \frac{(3.75 m)\,\text{N}}{(1.5 m)\,\text{kg}} = 2.5\,\text{m/s}^2$

  • Result: Mass term $m$ cancels out completely, yielding $2.5\,\text{m/s}^2$

• Deceleration Force on a Braking Automobile:

  • Problem: A car of mass $1.25 \times 10^3\,\text{kg}$ comes to a stop with a deceleration rate of $6.73\,\text{m/s}^2$.

  • Calculation:         $F_{\text{net}} = m \times a = (1.25 \times 10^3\,\text{kg}) \times (6.73\,\text{m/s}^2) = 8.41 \times 10^3\,\text{N}$

  • Result: $8.41 \times 10^3\,\text{N}$ acting opposite to the direction of motion.

• Water Resistance on a Motorboat:

  • Problem: A boat travels down a river at a constant velocity of $10\,\text{m/s}$ driven by an engine thrust of $25\,\text{N}$ forward.

  • Calculation:         $v = 10\,\text{m/s}\;\text{(constant)} \implies a = 0\,\text{m/s}^2$         $F_{\text{net}} = F_{\text{thrust}} - F_{\text{resistive}} = 0$         $25\,\text{N} - F_{\text{resistive}} = 0 \implies F_{\text{resistive}} = 25\,\text{N}$

  • Result: Acceleration is $0\,\text{m/s}^2$; resistive water force is $25\,\text{N}$ backward.

• Normal Force on a Pushed Stationary Book:

  • Problem: A book weighing $8.25\,\text{N}$ rests motionless on a table while an additional downward force of $3.2\,\text{N}$ is applied to it.

  • Calculation:         $F_{\text{net}} = 0$         $F_{\text{normal}} - F_{\text{weight}} - F_{\text{push}} = 0$         $F_{\text{normal}} - 8.25\,\text{N} - 3.2\,\text{N} = 0 \implies F_{\text{normal}} = 11.45\,\text{N}$

  • Result: Total upward normal force from table is $11.45\,\text{N}$.

• Mass of an Accelerating Bowling Ball:

  • Problem: A bowling ball thrown with a force of $35\,\text{N}$ accelerates at $2.4\,\text{m/s}^2$ to the right.

  • Calculation:         $m = \frac{F}{a} = \frac{35\,\text{N}}{2.4\,\text{m/s}^2} = 14.58\,\text{kg}$

Free-Body Diagrams (FBD): Scenarios and Force Structures

• Rocket Scenarios (Vertical Motion):

  • Rocket on its way up, right before leaving the ground:

    • Forces: Upward Thrust force ($F_{\text{thrust}}$), Downward Weight force ($F_{\text{weight}}$).

    • State: Static equilibrium ($F_{\text{weight}}$ equals thrust, balanced prior to lift-off).

  • Rocket accelerating upward (including air resistance):

    • Forces: Upward Thrust ($F_{\text{thrust}}$, large), Downward Air Resistance ($F_{\text{air}}$), Downward Weight ($F_{\text{weight}}$).

    • State: Unbalanced forces ($F_{\text{thrust}} > F_{\text{air}} + F_{\text{weight}}$).

  • Rocket decelerating while moving upward (including air resistance):

    • Forces: Upward Thrust ($F_{\text{thrust}}$, small), Downward Air Resistance ($F_{\text{air}}$), Downward Weight ($F_{\text{weight}}$).

    • State: Unbalanced forces ($F_{\text{thrust}} < F_{\text{air}} + F_{\text{weight}}$).

  • Rocket at its highest point (peak of flight):

    • Forces: Instantaneous velocity is zero ($v = 0\,\text{m/s} \implies F_{\text{air}} = 0$). Only Downward Weight ($F_{\text{weight}}$) acts.

    • State: Unbalanced forces.

  • Rocket falling downward (including air resistance):

    • Forces: Upward Air Resistance ($F_{\text{air}}$), Downward Weight ($F_{\text{weight}}$, larger). Engine thrust is zero ($F_{\text{thrust}} = 0$).

    • State: Unbalanced forces.

• Aircraft Scenarios:

  • Plane angled upward, accelerating both up and forward right at take-off:

    • Forces: Upward Lift ($F_{\text{lift}}$, large), Forward Thrust ($F_{\text{thrust}}$, large), Backward Air Resistance ($F_{\text{air}}$), Downward Weight ($F_{\text{weight}}$).

    • State: Unbalanced forces.

  • Plane flying at constant altitude and velocity:

    • Forces: Upward Lift ($F_{\text{lift}}$) equals Downward Weight ($F_{\text{weight}}$); Forward Thrust ($F_{\text{thrust}}$) equals Backward Air Resistance ($F_{\text{air}}$).

    • State: Balanced forces.

• Automobile Scenarios:

  • Car towed at constant velocity:

    • Forces: Upward Normal Force ($F_{\text{normal}}$) equals Downward Weight ($F_{\text{weight}}$); Forward Applied/Tension Force ($F_{\text{applied}}$) equals combined Backward Friction and Air Resistance ($F_f + F_{\text{air}}$).

    • State: Balanced forces.

  • Car slowing down (decelerating):

    • Forces: Upward Normal Force ($F_{\text{normal}}$) equals Downward Weight ($F_{\text{weight}}$); Backward resistive forces ($F_f + F_{\text{air}}$) exceed forward applied force.

    • State: Unbalanced forces.

  • Car speeding up (accelerating forward):

    • Forces: Upward Normal Force ($F_{\text{normal}}$) equals Downward Weight ($F_{\text{weight}}$); Forward Applied Force ($F_{\text{applied}}$) is larger than backward friction ($F_f$).

    • State: Unbalanced forces.

• Suspended and Falling Object Scenarios:

  • Object attached by string to ceiling (static):

    • Forces: Upward Tension ($F_{\text{tension}}$) equals Downward Weight ($F_{\text{weight}}$).

    • State: Balanced forces.

  • Object attached by string right after string is cut:

    • Forces: Tension instantly drops to zero ($F_{\text{tension}} = 0$). Only Downward Weight ($F_{\text{weight}}$) acts.

    • State: Unbalanced forces.

• Comprehensive Rock Diagram Analysis (Vertical Plane, Negligible Friction unless specified):

  • Case 1: Rock resting statically on horizontal surface:

    • Normal force ($F_N$) directed upward equals Weight ($F_w$) directed downward.

  • Case 2: Rock hanging statically from a single rope:

    • Tension ($F_T$) directed upward equals Weight ($F_w$) directed downward.

  • Case 3: Rock falling freely with no air friction:

    • Only Weight ($F_w$) directed downward acts on the rock.

  • Case 4: Rock resting statically on an incline:

    • Normal force ($F_N$) perpendicular to incline, Static Friction ($F_f$) pointing up incline, Weight ($F_w$) straight down.

  • Case 5: Rock resting statically suspended by two equal angled ropes:

    • Two symmetric tension forces ($F_{T1}$, $F_{T2}$) directed diagonally upward, balancing downward Weight ($F_w$).

  • Case 6: Rock resting statically supported by two unequal surfaces ($M_1, M_2$):

    • Normal forces ($F_{M1}$, $F_{M2}$) perpendicular to contact surfaces, balancing downward Weight ($F_w$).

  • Case 7: Rock resting statically suspended by two ropes at non-identical angles:

    • Tension forces ($F_{T1}$, $F_{T2}$) at differing angles whose vector sum equals downward Weight ($F_w$).

  • Case 8: Rock pressed statically against a vertical wall:

    • Horizontal pushing force ($F_{\text{push}}$) equals horizontal Normal force ($F_N$); upward Static Friction ($F_f$) equals downward Weight ($F_w$).

  • Case 9: Rock sliding down an incline without friction:

    • Normal force ($F_N$) perpendicular to incline; Weight ($F_w$) straight down.

  • Case 10: Static friction prevents rock from sliding down incline:

    • Normal force ($F_N$) perpendicular to incline, Static friction ($F_f$) directed up the slope, Weight ($F_w$) straight down. Forces are balanced.

  • Case 11: Rock sliding at constant speed across horizontal surface without friction:

    • Normal force ($F_N$) directed upward equals Weight ($F_w$) directed downward. No horizontal forces act.

  • Case 12: Rock falling at constant (terminal) velocity:

    • Upward Air Resistance ($F_{\text{air}}$) equals downward Weight ($F_w$).

  • Case 13: Rock decelerating due to kinetic friction on horizontal ground:

    • Normal force ($F_N$) equals Weight ($F_w$) vertically; Kinetic friction ($F_f$) points opposite to motion horizontally.

  • Case 14: Rock rising along a parabolic trajectory (no air resistance):

    • Only downward Weight ($F_w$) acts on the rock.

  • Case 15: Rock at the top peak of a parabolic trajectory (no air resistance):

    • Only downward Weight ($F_w$) acts on the rock.

  • Case 16: Rock tied to a rope pulled straight upward accelerating at $9.8\,\text{m/s}^2$:

    • Calculation:             $F_{\text{net}} = F_T - F_w = m \times a$             $a = g \implies m \times a = F_w$             $F_T - F_w = F_w \implies F_T = 2 F_w$

    • Result: Upward Tension ($F_T$) is exactly double the downward Weight ($F_w$).

Newton's Second Law: Mass, Force, and Cart Systems

• Cart Acceleration Comparisons ($a = \frac{F}{m}$, Cart A mass $m_A = 1\,\text{kg}$):

  • Scenario 1: Cart A ($m = 1\,\text{kg}, F = 1\,\text{N} \implies a = 1\,\text{m/s}^2$) vs Cart B ($m = 1\,\text{kg}, F = 2\,\text{N} \implies a = 2\,\text{m/s}^2$).

    • Greater Acceleration: Cart B.

  • Scenario 2: Cart A ($m = 1\,\text{kg}, F = 1\,\text{N} \implies a = 1\,\text{m/s}^2$) vs Cart B ($m = 2\,\text{kg}, F = 1\,\text{N} \implies a = 0.5\,\text{m/s}^2$).

    • Greater Acceleration: Cart A.

  • Scenario 3: Cart A ($m = 1\,\text{kg}, F = 1\,\text{N} \implies a = 1\,\text{m/s}^2$) vs Cart B ($m = 2\,\text{kg}, F = 2\,\text{N} \implies a = 1\,\text{m/s}^2$).

    • Greater Acceleration: Same for both.

  • Scenario 4: Cart A ($m = 1\,\text{kg}, F = 1\,\text{N} \implies a = 1\,\text{m/s}^2$) vs Cart B ($m = 3\,\text{kg}, F = 3\,\text{N} \implies a = 1\,\text{m/s}^2$).

    • Greater Acceleration: Same for both.

  • Scenario 5: Cart A ($m = 1\,\text{kg}, F = 4\,\text{N} \implies a = 4\,\text{m/s}^2$) vs Cart B ($m = 4\,\text{kg}, F = 4\,\text{N} \implies a = 1\,\text{m/s}^2$).

    • Greater Acceleration: Cart A.

  • Scenario 6: Cart A ($m = 1\,\text{kg}, F = 2\,\text{N} \implies a = 2\,\text{m/s}^2$) vs Cart B ($m = 4\,\text{kg}, F = 3\,\text{N} \implies a = 0.75\,\text{m/s}^2$).

    • Greater Acceleration: Cart A.

• Dynamics of Pulley Systems ("Drop and Pull"):

  • System Principle: When a hanging weight pulls a horizontal cart over a pulley, the accelerated mass ($m_{\text{total}}$) is the sum of the cart's mass and the hanging mass.

  • Gravitational Equivalence: A $10\,\text{N}$ force corresponds to a $1\,\text{kg}$ mass under $g = 10\,\text{m/s}^2$:         $F_w = m \times g = (1\,\text{kg})(10\,\text{m/s}^2) = 10\,\text{N}$

  • Baseline Comparison:

    • $1\,\text{kg}$ cart pulled directly by $10\,\text{N}$ horizontal force:             $a = \frac{10\,\text{N}}{1\,\text{kg}} = 10\,\text{m/s}^2 = g$

    • $1\,\text{kg}$ cart pulled by $10\,\text{N}$ hanging iron weight ($1\,\text{kg}$):             $m_{\text{total}} = 1\,\text{kg} + 1\,\text{kg} = 2\,\text{kg}$             $a = \frac{10\,\text{N}}{2\,\text{kg}} = 5\,\text{m/s}^2 = \frac{g}{2}$

  • Pulley Calculations with Varying Hanging Masses ($1\,\text{kg}$ cart):

    • Two $10\,\text{N}$ hanging weights ($2\,\text{kg}$ total hanging mass):             $a = \frac{F_{\text{applied}}}{m_{\text{total}}} = \frac{20\,\text{N}}{1\,\text{kg} + 2\,\text{kg}} = \frac{20}{3} = 6.67\,\text{m/s}^2$

    • Three $10\,\text{N}$ hanging weights ($3\,\text{kg}$ total hanging mass):             $a = \frac{F_{\text{applied}}}{m_{\text{total}}} = \frac{30\,\text{N}}{1\,\text{kg} + 3\,\text{kg}} = \frac{30}{4} = 7.5\,\text{m/s}^2$

    • Four $10\,\text{N}$ hanging weights ($4\,\text{kg}$ total hanging mass):             $a = \frac{F_{\text{applied}}}{m_{\text{total}}} = \frac{40\,\text{N}}{1\,\text{kg} + 4\,\text{kg}} = \frac{40}{5} = 8\,\text{m/s}^2$

  • Pulley Calculations with Added Cart Mass and Varying Hanging Mass:

    • $1\,\text{kg}$ iron added to cart ($2\,\text{kg}$ total cart) + One $10\,\text{N}$ hanging weight ($1\,\text{kg}$):             $a = \frac{10\,\text{N}}{2\,\text{kg} + 1\,\text{kg}} = \frac{10}{3} = 3.33\,\text{m/s}^2$

    • $2\,\text{kg}$ iron added to cart ($3\,\text{kg}$ total cart) + One $10\,\text{N}$ hanging weight ($1\,\text{kg}$):             $a = \frac{10\,\text{N}}{3\,\text{kg} + 1\,\text{kg}} = \frac{10}{4} = 2.5\,\text{m/s}^2$

    • $3\,\text{kg}$ iron added to cart ($4\,\text{kg}$ total cart) + One $10\,\text{N}$ hanging weight ($1\,\text{kg}$):             $a = \frac{10\,\text{N}}{4\,\text{kg} + 1\,\text{kg}} = \frac{10}{5} = 2\,\text{m/s}^2$

    • $3\,\text{kg}$ iron added to cart ($4\,\text{kg}$ total cart) + Four $10\,\text{N}$ hanging weights ($4\,\text{kg}$):             $a = \frac{40\,\text{N}}{4\,\text{kg} + 4\,\text{kg}} = \frac{40}{8} = 5\,\text{m/s}^2$

Vector Addition, Resultants, and Equilibrants

• General Protocol for Analytical Vector Resolution:

  1. Resolve all force vectors into orthogonal components:        $F_x = F \times \cos(\theta)$        $F_y = F \times \sin(\theta)$

  2. Calculate net components along each axis:        $R_x = \sum F_x$        $R_y = \sum F_y$

  3. Calculate resultant vector magnitude using the Pythagorean theorem:        $F_R = \sqrt{R_x^2 + R_y^2}$

  4. Calculate reference angle $\theta$:        $\theta = \tan^{-1}\left(\frac{|R_y|}{|R_x|}\right)$

  5. Adjust $\theta$ to standard polar angle (0° to 360°) based on component signs:

     • Quadrant I ($+R_x, +R_y$): Angle $= \theta$

     • Quadrant II ($-R_x, +R_y$): Angle $= 180^\circ - \theta$

     • Quadrant III ($-R_x, -R_y$): Angle $= 180^\circ + \theta$

     • Quadrant IV ($+R_x, -R_y$): Angle $= 360^\circ - \theta$

  6. For equilibrant/balancing forces, set $F_e = F_R$ and direction $= (\text{Angle} + 180^\circ) \pmod{360^\circ}$.

• Detailed Vector Problem Solutions:

  • Problem 1: Single force $F = 5\,\text{N}$ at $0^\circ$.

    • Balancing Force: $5\,\text{N}$ at $180^\circ$

  • Problem 2: Single force $F = 10\,\text{N}$ at $90^\circ$.

    • Balancing Force: $10\,\text{N}$ at $270^\circ$

  • Problem 3: Forces $F_1 = 5\,\text{N}$ at $0^\circ$ and $F_2 = 10\,\text{N}$ at $270^\circ$.

    • $R_x = 5\,\text{N}$, $R_y = -10\,\text{N}$

    • $F_R = \sqrt{5^2 + (-10)^2} = 11.18\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{10}{5}\right) = 63.43^\circ$

    • Direction: $360^\circ - 63.43^\circ = 296.56^\circ$

    • Resultant: $11.18\,\text{N}$ at $296.56^\circ$

  • Problem 4: Wind pushing horizontally at $200\,\text{N}$ against a screen with weight $50\,\text{N}$.

    • $R_x = -200\,\text{N}$, $R_y = -50\,\text{N}$

    • $F_R = \sqrt{(-200)^2 + (-50)^2} = 206.16\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{50}{200}\right) = 14.04^\circ$

    • Direction: $180^\circ + 14.04^\circ = 194.04^\circ$

    • Resultant: $206.16\,\text{N}$ at $194.04^\circ$

  • Problem 5: Force system with components $R_x = -30\,\text{N}$ and $R_y = 50\,\text{N} - 70\,\text{N} = -20\,\text{N}$.

    • $F_R = \sqrt{(-30)^2 + (-20)^2} = 36.06\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{20}{30}\right) = 33.69^\circ$

    • Direction: $180^\circ + 33.69^\circ = 213.69^\circ$

    • Resultant: $36.06\,\text{N}$ at $213.69^\circ$

  • Problem 6: Ring acted on by $10\,\text{N}$ at $0^\circ$ and $8\,\text{N}$ at $135^\circ$.

    • $R_x = 10 + 8 \times \cos(135^\circ) = 10 - 5.66 = 4.34\,\text{N}$

    • $R_y = 0 + 8 \times \sin(135^\circ) = 5.66\,\text{N}$

    • $F_R = \sqrt{4.34^2 + 5.66^2} = 7.06\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{5.66}{4.34}\right) = 52.48^\circ$

    • Added Balancing Force Direction: $180^\circ + 52.48^\circ = 232.48^\circ$

    • Balancing Force: $7.06\,\text{N}$ at $232.48^\circ$

  • Problem 7: Box acted on by forces $(40\,\text{N}, 30\,\text{N})$ right, $10\,\text{N}$ left, $30\,\text{N}$ left, $30\,\text{N}$ up, $30\,\text{N}$ down.

    • $R_x = 40 + 30 - 10 - 30 = 30\,\text{N}$

    • $R_y = 30 - 30 = 0\,\text{N}$

    • Motion: Moves right with a net force of $30\,\text{N}$ at $0^\circ$

  • Problem 8: Box acted on by $3\,\text{N}$ right, $3\,\text{N}$ right, $5\,\text{N}$ left, $3\,\text{N}$ down.

    • $R_x = 3 + 3 - 5 = 1\,\text{N}$

    • $R_y = -3\,\text{N}$

    • $F_R = \sqrt{1^2 + (-3)^2} = 3.16\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{3}{1}\right) = 71.57^\circ$

    • Direction: $360^\circ - 71.57^\circ = 288.43^\circ$

    • Resultant: $3.16\,\text{N}$ at $288.43^\circ$

  • Problem 9: Forces $10\,\text{N}$ right, $7\,\text{N}$ right, $3\,\text{N}$ down, $5\,\text{N}$ down.

    • $R_x = 10 + 7 = 17\,\text{N} \implies F_{ex} = -17\,\text{N}$

    • $R_y = -3 - 5 = -8\,\text{N} \implies F_{ey} = 8\,\text{N}$

    • $F_e = \sqrt{(-17)^2 + 8^2} = 18.79\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{8}{17}\right) = 25.20^\circ$

    • Direction: $180^\circ - 25.20^\circ = 154.80^\circ$

    • Balancing Force: $18.79\,\text{N}$ at $154.80^\circ$

  • Problem 10: Net displacement of a whale swimming $13\,\text{km}$ NE ($45^\circ$), $5\,\text{km}$ at $30^\circ$ W of S ($240^\circ$), and $10\,\text{km}$ S ($270^\circ$).

    • $d_x = 13 \times \cos(45^\circ) + 5 \times \cos(240^\circ) + 10 \times \cos(270^\circ) = 9.192 - 2.5 + 0 = 6.69\,\text{km}$

    • $d_y = 13 \times \sin(45^\circ) + 5 \times \sin(240^\circ) - 10 = 9.192 - 4.330 - 10 = -5.14\,\text{km}$

    • $\text{Displacement} = \sqrt{6.69^2 + (-5.14)^2} = 8.43\,\text{km}$

    • $\theta = \tan^{-1}\left(\frac{5.14}{6.69}\right) = 37.54^\circ$

    • Direction: $360^\circ - 37.54^\circ = 322.46^\circ$

    • Net Displacement: $8.43\,\text{km}$ at $322.46^\circ$

  • Problem 11: Cruise ship moving $3\,\text{m/s}$ N ($90^\circ$), person walking $1\,\text{m/s}$ toward port/west ($180^\circ$), kite flying in wind of $5\,\text{m/s}$ S ($270^\circ$).

    • $v_x = -1\,\text{m/s}$

    • $v_y = 3 - 5 = -2\,\text{m/s}$

    • $v_R = \sqrt{(-1)^2 + (-2)^2} = 2.23\,\text{m/s}$

    • $\theta = \tan^{-1}\left(\frac{2}{1}\right) = 63.43^\circ$

    • Direction: $180^\circ + 63.43^\circ = 243.43^\circ$

    • Resulting Velocity: $2.23\,\text{m/s}$ at $243.43^\circ$

  • Problem 12: Net force on a suitcase pulled with $175\,\text{N}$ upward, $450\,\text{N}$ at $30^\circ$, and $400\,\text{N}$ downward.

    • $F_x = 450 \times \cos(30^\circ) = 389.71\,\text{N}$

    • $F_y = 175 + 450 \times \sin(30^\circ) - 400 = 175 + 225 - 400 = 0\,\text{N}$

    • Resultant Net Force: $389.71\,\text{N}$ at $0^\circ$

  • Problem 13: Motor boat pointing N ($90^\circ$) with engine force $1500\,\text{N}$, water current exerting $2300\,\text{N}$ W ($180^\circ$), wind exerting $700\,\text{N}$ at $40^\circ$ N of W ($140^\circ$).

    • $F_x = -2300 + 700 \times \cos(140^\circ) = -2300 - 536.23 = -2836.23\,\text{N}$

    • $F_y = 1500 + 700 \times \sin(140^\circ) = 1500 + 449.95 = 1949.95\,\text{N}$

    • $F_R = \sqrt{(-2836.23)^2 + (1949.95)^2} = 3441.88\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{1949.95}{2836.23}\right) = 34.51^\circ$

    • Direction: $180^\circ - 34.51^\circ = 145.49^\circ$

    • Net Force: $3441.88\,\text{N}$ at $145.49^\circ$

  • Problem 14: Vector Addition of Vector $A = 500\,\text{N}$ at $0^\circ$ and Vector $B = 400\,\text{N}$ at $60^\circ$.

    • $R_x = 500 + 400 \times \cos(60^\circ) = 500 + 200 = 700\,\text{N}$

    • $R_y = 400 \times \sin(60^\circ) = 346.41\,\text{N}$

    • $R = \sqrt{700^2 + 346.41^2} = 781\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{346.41}{700}\right) = 26^\circ$

    • Resultant Vector: $781\,\text{N}$ at $26^\circ$

  • Problem 15: Vector Addition of Vector $C = 350\,\text{N}$ at $15^\circ$ and Vector $D = 300\,\text{N}$ at $60^\circ$.

    • $R_x = 350 \times \cos(15^\circ) + 300 \times \cos(60^\circ) = 338.08 + 150 = 488\,\text{N}$

    • $R_y = 350 \times \sin(15^\circ) + 300 \times \sin(60^\circ) = 90.58 + 259.81 = 351\,\text{N}$

    • $R = \sqrt{488^2 + 351^2} = 601\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{351}{488}\right) = 36^\circ$

    • Resultant Vector: $601\,\text{N}$ at $36^\circ$

  • Problem 16: Vector Addition of Vector $E = 625\,\text{N}$ at $95^\circ$ and Vector $F = 285\,\text{N}$ at $210^\circ$.

    • $R_x = 625 \times \cos(95^\circ) + 285 \times \cos(210^\circ) = -54.46 - 246.82 = -301\,\text{N}$

    • $R_y = 625 \times \sin(95^\circ) + 285 \times \sin(210^\circ) = 622.62 - 142.50 = 480\,\text{N}$

    • $R = \sqrt{(-301)^2 + 480^2} = 567\,\text{N}$

    • $\theta = \tan^{-1}\left(\frac{480}{301}\right) = 58^\circ$

    • Direction: $180^\circ - 58^\circ = 122^\circ$

    • Resultant Vector: $567\,\text{N}$ at $122^\circ$