AP Physics C Mechanics- 2.3 Newton’s Third Law- Study Notes- New Syllabus
AP Physics C Mechanics- 2.3 Newton’s Third Law – Study Notes
AP Physics C Mechanics- 2.3 Newton’s Third Law – Study Notes – per latest Syllabus.
Key Concepts:
- Newton’s Third Law of Motion
- Internal Forces and Motion of the Center of Mass
- Tension in Strings, Cables, and Pulleys
Newton’s Third Law of Motion
Newton’s Third Law states that when two objects interact, they exert equal and opposite forces on each other. These forces are known as an action–reaction pair.
\( \mathrm{\vec{F}_{A\,on\,B} = -\vec{F}_{B\,on\,A}} \)
- The forces are equal in magnitude and opposite in direction.
- They act on different objects — never cancel each other because they do not act on the same body.
- These paired forces always occur simultaneously as part of the same interaction.
- Examples include gravitational attraction, contact pushes, and tension between connected masses.
Every force arises from a mutual interaction — if object A exerts a force on object B, object B exerts an equal and opposite force on object A.
Example
A horse pulls a cart forward with a horizontal force of \( \mathrm{200\,N} \). What is the force that the cart exerts on the horse?
▶️ Answer / Explanation
Step 1: Identify the action–reaction pair.
- Action: Horse exerts a forward force of \( \mathrm{200\,N} \) on the cart.
- Reaction: Cart exerts an equal backward force of \( \mathrm{200\,N} \) on the horse.
Step 2: Apply Newton’s third law:
\( \mathrm{\vec{F}_{horse\,on\,cart} = -\vec{F}_{cart\,on\,horse}} \)
Result: The two forces are equal in magnitude (200 N) and opposite in direction. They do not cancel because they act on different objects.
Internal Forces and Motion of the Center of Mass
Forces that objects within a system exert on each other are called internal forces. According to Newton’s third law, these forces always occur in equal and opposite pairs and therefore cancel out when analyzing the motion of the system as a whole.
- Only external forces can change the motion of a system’s center of mass.
- Internal forces may change the motion of parts within the system but not its overall momentum or acceleration.
- This principle is key to the conservation of momentum and system-level analysis.
Key Relationship:
\( \mathrm{\sum \vec{F}_{ext} = M \vec{a}_{cm}} \)
Here, \( \mathrm{\vec{F}_{ext}} \) represents the vector sum of all external forces on the system; internal forces do not appear in this equation because they cancel.
Example
Explain when Two ice skaters (Skater A and Skater B) push off each other while standing on frictionless ice.
▶️ Answer / Explanation
Step 1: Identify internal and external forces.
- Each skater exerts a force on the other — these are internal forces within the skater–skater system.
- No external horizontal forces act on the two-skater system (ignoring friction).
Step 2: Apply Newton’s Third Law.
The forces are equal and opposite, so the internal forces cancel when considering the entire system.
Step 3: Analyze system motion.
The skaters move apart in opposite directions, but the center of mass remains stationary because no external force acts on the system.
Result: Internal forces can change individual motions, but the total momentum and motion of the center of mass are unaffected.
Tension in Strings, Cables, and Pulleys
Tension is the macroscopic net effect of the forces that infinitesimal segments of a string, rope, or cable exert on each other when the material is stretched by an external force. At the microscopic level, these forces arise from the electromagnetic interactions between molecules resisting separation.
- Tension acts along the length of the string or rope, always directed away from the object on which it acts.
- It is transmitted through the medium to connect two interacting bodies (e.g., masses on a pulley or an object being pulled).
Ideal String Assumptions
An ideal string has negligible mass and does not stretch under tension. This simplification allows all forces transmitted by the string to be considered instantaneous and uniform throughout its length.
- The string only transmits force — it does not absorb or store significant energy.
- It cannot exert any force perpendicular to its length.
Example
A 2 kg mass is suspended vertically by a light, inextensible string. Find the tension in the string when the system is at rest.
▶️ Answer / Explanation
In equilibrium, the upward tension balances the downward weight:
\( \mathrm{T = mg = (2.0)(9.8) = 19.6\,N.} \)
Since the string is ideal (massless and non-stretching), the tension is the same throughout and acts equally on both ends.
Constant Tension in an Ideal String
In an ideal string, the tension is the same at every point along its length because the string’s mass is negligible and it cannot accelerate differentially along its body.
- If one end of the string pulls with force \( \mathrm{T} \), the other end must pull back with the same \( \mathrm{T} \).
- This allows connected masses in pulley or linear systems to experience identical tension magnitudes.
Example
Two blocks are connected by a light string over a frictionless pulley — Block A (3 kg) hangs vertically, and Block B (2 kg) rests on a frictionless surface. Find the tension in the string.
▶️ Answer / Explanation
Step 1: Forces on each block:
- Block A: \( \mathrm{m_A g – T = m_A a} \)
- Block B: \( \mathrm{T = m_B a} \)
Step 2: Combine equations to find acceleration:
\( \mathrm{a = \dfrac{m_A g}{m_A + m_B} = \dfrac{3(9.8)}{5} = 5.88\,m/s^2.} \)
Step 3: Find tension:
\( \mathrm{T = m_B a = (2)(5.88) = 11.76\,N.} \)
Result: Tension is uniform throughout the ideal string — \( \mathrm{T = 11.8\,N.} \)
Tension in a String with Nonnegligible Mass
If the string has non-negligible mass, its different segments can have different accelerations and thus experience different tensions. Tension increases toward the point where the external force is applied.
- Each small segment must support the weight (or acceleration) of the segments below it.
- This causes tension to vary along the string’s length.
Example
A uniform rope of mass \( \mathrm{0.5\,kg} \) and length \( \mathrm{2\,m} \) hangs vertically from a ceiling. Find the tension at the top and halfway down the rope.
▶️ Answer / Explanation
Step 1: Mass per unit length: \( \mathrm{\lambda = \dfrac{0.5}{2} = 0.25\,kg/m.} \)
Step 2: Tension at a point a distance \( \mathrm{x} \) from the bottom: \( \mathrm{T(x) = \lambda g x.} \)
Step 3: At the top (\( \mathrm{x = 2\,m} \)): \( \mathrm{T = 0.25(9.8)(2) = 4.9\,N.} \)
At halfway (\( \mathrm{x = 1\,m} \)): \( \mathrm{T = 0.25(9.8)(1) = 2.45\,N.} \)
Result: Tension increases linearly from the bottom to the top due to the rope’s own weight.
Ideal Pulley Assumptions
An ideal pulley is a pulley with negligible mass and no friction in its axle. It merely changes the direction of the tension force without altering its magnitude.
- The tension entering and leaving the pulley is equal: \( \mathrm{T_1 = T_2.} \)
- The pulley itself exerts no net torque on the string.
- Forces act tangentially along the string in opposite directions on either side of the pulley.
Example
In a system where a 3 kg mass hangs on one side of an ideal pulley and a 2 kg mass on the other, find the acceleration of the system and the tension in the string.
▶️ Answer / Explanation
Step 1: Write force equations for each side:
\( \mathrm{m_1 g – T = m_1 a} \) and \( \mathrm{T – m_2 g = m_2 a.} \)
Step 2: Combine equations to find acceleration:
\( \mathrm{a = \dfrac{(m_1 – m_2)g}{m_1 + m_2} = \dfrac{(3 – 2)(9.8)}{5} = 1.96\,m/s^2.} \)
Step 3: Find tension:
\( \mathrm{T = m_2(g + a) = 2(9.8 + 1.96) = 23.5\,N.} \)
Result: In an ideal pulley, the tension is the same throughout the string; it only changes direction, not magnitude.