Newton's Laws of Motion: Third Law and Applications
Newton's Third Law of Motion
Definition: For every action, there is an equal and opposite reaction. This means the force of object A on object B is equal in magnitude and opposite in direction to the force of object B on object A.
Key Characteristics:
Forces always occur in pairs.
These forces act on different objects. Therefore, they can never produce a net force on a single object (i.e., they cannot cancel each other out when analyzing the motion of one object).
There are absolutely no exceptions to this law.
Examples:
Moth vs. Windshield: When a moth hits a car windshield, the force of the moth on the windshield is equal in magnitude to the force of the windshield on the moth. The car's much larger mass results in a negligible acceleration for the car, while the moth experiences a huge acceleration (and unfortunate outcome).
Horse and Cart (Conceptual): A horse pulling a cart. The force of the horse pulling on the cart and the force of the cart pulling on the horse form an action-reaction pair. The horse can move the cart because when analyzing the horse's motion, other forces (like the road pushing on the horse) are also considered.
Book on a Table: The Earth pulls the book down (gravity), and the book pulls the Earth up. These form an action-reaction pair and are equal in magnitude. Similarly, the table pushes up on the book, and the book pushes down on the table; these also form an action-reaction pair.
Action-Reaction Pairs vs. Balanced Forces (Newton's Second Law)
Newton's Third Law (Action-Reaction Pairs):
Involves two forces acting on two different objects.
Example: Force of horse on cart vs. Force of cart on horse.
Newton's Second Law (Balanced Forces on a Single Object):
Involves multiple forces acting on a single object.
If an object is at rest or moving at a constant velocity (i.e., not accelerating), the net force on that object is zero. This means the forces acting on it are balanced (equal in magnitude and opposite in direction), according to Newton's Second Law: \Sigma \vec{F} = m\vec{a} If \vec{a} = 0, then \Sigma \vec{F} = 0.
Example: A car at rest on the ground. The gravitational pull of the Earth on the car (downward) is equal and opposite to the upward normal force of the ground on the car. These two forces act on the car and are balanced because the car is not accelerating vertically.
Solving Problems with Multiple Objects (Newton's Second Law)
General Approach (Recommended for Beginners): Treat each object separately.
Draw a separate free-body diagram for each object.
Apply Newton's Second Law ( \Sigma \vec{F} = m\vec{a} ) to each object individually.
Use Newton's Third Law to relate forces between interacting objects (e.g., if object A pushes B, B pushes A with equal magnitude).
Solve the resulting system of equations (one for each object).
Internal vs. External Forces:
Internal Forces: Forces that act between objects within a defined system. These forces cancel out within the system and do not affect the system's overall acceleration.
External Forces: Forces exerted on the system by objects outside the system. Only external forces cause a change in the motion of the entire system.
When treating each object separately, the forces between them are considered external to that particular object's free-body diagram.
Example: Blocks Pushed by a Force
Scenario: A 4 \text{ kg} block (M4) and a 2 \text{ kg} block (M2) on a frictionless horizontal surface. A 12 \text{ N} force ( F_P ) pushes M4 against M2. Determine the force M2 exerts on M4.
Given:
m_2 = 2 \text{ kg}
m_4 = 4 \text{ kg}
F_P = 12 \text{ N}
Free-Body Diagram for Block 2 (M2):
Only one horizontal force: The force of block 4 on block 2 ( F_{4 \text{ on } 2} ), acting in the positive x-direction.
Newton's Second Law for M2: \Sigma F{x2} = m2 a2 \implies F{4 \text{ on } 2} = m2 a2 (Equation 1)
Free-Body Diagram for Block 4 (M4):
Force of Push ( F_P ): 12 \text{ N}, in the positive x-direction.
Force of Block 2 on Block 4 ( F{2 \text{ on } 4} ): By Newton's Third Law, F{2 \text{ on } 4} has the same magnitude as F_{4 \text{ on } 2}, but acts in the opposite direction (negative x-direction) on M4.
Newton's Second Law for M4: \Sigma F{x4} = m4 a4 \implies FP - F{2 \text{ on } 4} = m4 a_4 (Equation 2)
Relating Accelerations and Forces:
Since the blocks move together, their accelerations are the same: a2 = a4 = a.
From Newton's Third Law: F{4 \text{ on } 2} = F{2 \text{ on } 4}. Let's call this force F_{contact}.
Substitution and Solution:
From Equation 1: a = \frac{F{contact}}{m2}
Substitute 'a' into Equation 2: FP - F{contact} = m4 \left(\frac{F{contact}}{m_2}\right)
Rearrange to solve for F{contact}: FP = F{contact} + m4 \left(\frac{F{contact}}{m2}\right)
FP = F{contact} \left(1 + \frac{m4}{m2}\right)
F{contact} = \frac{FP}{1 + \frac{m4}{m2}}Plug in values: F_{contact} = \frac{12 \text{ N}}{1 + \frac{4 \text{ kg}}{2 \text{ kg}}} = \frac{12 \text{ N}}{1 + 2} = \frac{12 \text{ N}}{3} = 4 \text{ N}
Result: The force that the 2 \text{ kg} block exerts on the 4 \text{ kg} block ( F_{2 \text{ on } 4} ) is 4 \text{ N}.
Accelerations:
a = \frac{12 \text{ N}}{m2 + m4} = \frac{12 \text{ N}}{6 \text{ kg}} = 2 \text{ m/s}^2
Check for M2: F{4 \text{ on } 2} = m2 a = (2 \text{ kg})(2 \text{ m/s}^2) = 4 \text{ N}.
Example: Astronaut Pushing a Spaceship
Scenario: Astronaut pushes a spaceship with 36 \text{ N}. Astronaut mass mA = 92 \text{ kg}, spaceship mass mS = 11,000 \text{ kg}. (Assume no gravity/external forces in space apart from the interaction).
Newton's Third Law: The force of the astronaut on the spaceship (F{A \text{ on } S}) has a magnitude of 36 \text{ N}. The force of the spaceship on the astronaut (F{S \text{ on } A}) also has a magnitude of 36 \text{ N}.
Free-Body Diagram for Spaceship:
Force: F_{A \text{ on } S} = 36 \text{ N} (let's say in the negative x-direction if the astronaut pushes right and is moving left).
Newton's Second Law: \Sigma FS = mS aS \implies -F{A \text{ on } S} = mS aS
Acceleration of Spaceship: aS = \frac{-F{A \text{ on } S}}{m_S} = \frac{-36 \text{ N}}{11,000 \text{ kg}} \approx -0.0033 \text{ m/s}^2
Free-Body Diagram for Astronaut:
Force: F_{S \text{ on } A} = 36 \text{ N} (in the positive x-direction).
Newton's Second Law: \Sigma FA = mA aA \implies F{S \text{ on } A} = mA aA
Acceleration of Astronaut: aA = \frac{F{S \text{ on } A}}{m_A} = \frac{36 \text{ N}}{92 \text{ kg}} \approx 0.39 \text{ m/s}^2
Conclusion: The forces are equal and opposite, but the accelerations are vastly different due to the different masses. The astronaut accelerates significantly more than the spaceship.
Important Implication: Equal forces do not mean equal accelerations when masses are different (a = F/m). This explains why a moth disintegrates upon hitting a car, while the car's motion is barely affected.
Tension in a String
Tension as a Force: Tension is an internal force within a string or rope when it is pulled taut.
Example: Block on a Pulley System
Case A: A 5 \text{ kg} block hangs from a string over a pulley, with the other end of the string attached to a wall.
Case B: Two 5 \text{ kg} blocks are connected by a string over two pulleys, with one block hanging on each side.
Analysis for Tension (Case A and B are identical in terms of tension):
Draw a free-body diagram for one of the 5 \text{ kg} blocks.
Forces acting on the block:
Gravity (downward): F_g = mg = (5 \text{ kg})(9.8 \text{ m/s}^2) = 49 \text{ N}
Tension (upward): T
Since the block is stationary (a = 0):
\Sigma Fy = may \implies T - Fg = 0 \implies T = Fg = 49 \text{ N}Conclusion: The tension in the string in both cases is the same, equal to the weight of the 5 \text{ kg} block (49 \text{ N}). The wall in Case A provides the same opposing force as the second 5 \text{ kg} block in Case B.
Motion on an Inclined Plane (Frictionless)
Key Strategy: Always choose one axis in the direction of acceleration. For inclined planes, this usually means choosing the x-axis parallel to the incline and the y-axis perpendicular to it.
Free-Body Diagram:
Gravity ( F_g = mg ): Always points straight down.
Normal Force ( F_N ): Always perpendicular to the surface of contact (i.e., perpendicular to the incline).
No Friction: No force opposing motion along the incline due to friction.
Resolving Forces (Tilting Axes):
Gravity ( mg ) needs to be resolved into components along the tilted x and y axes.
If the incline angle is \theta (angle with the horizontal):
Component parallel to incline (x-direction): mg \sin\theta (This component causes the object to slide down the incline).
Component perpendicular to incline (y-direction): mg \cos\theta
F{gx} = mg \sin\theta (positive down the incline) F{gy} = -mg \cos\theta (negative into the incline)
Applying Newton's Second Law:
In the y-direction (perpendicular to incline): There is no acceleration (ay = 0). \Sigma Fy = may \implies FN - mg \cos\theta = 0 \implies F_N = mg \cos\theta (Note: Normal force is not equal to mg on an incline).
In the x-direction (parallel to incline): The object accelerates down the incline.
\Sigma Fx = max \implies mg \sin\theta = max \implies ax = g \sin\theta
Experiment to Determine 'g':
Procedure: Place a cart on a frictionless inclined plane (e.g., air track). Measure the distance it travels ( Xf - Xi ) and the time ( t ) it takes, starting from rest (v_i = 0).
Kinematic Equation: Xf - Xi = v_i t + \frac{1}{2} a t^2
Since vi = 0: Xf - Xi = \frac{1}{2} a t^2 \implies a = \frac{2(Xf - X_i)}{t^2}
Relating 'a' to 'g': We found a = g \sin\theta \implies g = \frac{a}{\sin\theta}
Substituting 'a': g = \frac{2(Xf - Xi)}{t^2 \sin\theta}
Measuring \sin\theta: If the length of the incline is L and the height of the raised end is h, then \sin\theta = \frac{h}{L}.
Example Calculations:
Measured distance ( Xf - Xi ) = 1 \text{ meter}.
Measured height ( h ) = 0.019 \text{ meters}. (Assuming L = 1 \text{ meter} for simplicity, so \sin\theta = 0.019).
Measured time ( t ) = 3.3 \text{ seconds}.
g = \frac{2(1 \text{ m})}{(3.3 \text{ s})^2 (0.019)} \approx 9.7 \text{ m/s}^2
Significance: This experiment demonstrates how Newton's Second Law combined with kinematics can be used to determine fundamental physical constants like the acceleration due to gravity.