Newton's Third Law and Free-Body Diagrams Newton's Third Law and Free-Body Diagrams Every push you make is answered by an equal push back. That is Newton’s Third Law in daily life: when your foot pushes the ground backward, the ground pushes you forward; when a table supports a book, the book also pushes the table. The law is precise: if body A exerts a force on body B, then body B exerts an equal in magnitude and opposite in direction force on body A, along the same line of action, at the same instant. Crucially, the two forces act on different bodies, so they never cancel in a single body's free-body diagram. To solve real problems, especially with connected bodies, we must isolate one body at a time and draw all forces on it. That picture is a free-body diagram (FBD). It helps us apply F = m a without mixing up internal and external forces. This is how we compute tension in strings, normal reactions, and accelerations in pulley systems. Once you master FBDs and Third Law pairs, ‘complex’ multi-body questions become a chain of simple F = ma steps linked by constraints. You will also learn when common shortcuts apply: when tension is uniform in a massless, frictionless string and pulley; when the normal equals mg (and when it doesn’t); and how to choose a "system" so internal forces cancel, giving quick results for acceleration. Throughout, always check the boundaries: ideal strings and pulleys are assumptions; friction can break symmetry; and action–reaction pairs live on different bodies, not within one diagram. These habits prevent classic mistakes and align with how NEET frames multi-step dynamics questions. remember Real-world analogy: Walking works because your foot pushes the ground backward and the ground pushes you forward with equal force. Skating, rowing, and rocket thrusts are the same action–reaction story seen in different settings. Two forces of equal magnitude and opposite direction that two bodies exert on each other simultaneously, along the same line of action, acting on different bodies. Action–Reaction Pair Free-Body Diagram (FBD) A sketch of a single isolated body showing all external forces on it, with directions and labels, used to apply F = m a . A chosen collection of one or more bodies considered together; internal forces between parts of the system cancel when you apply F external = M a CM . System External forces come from outside the chosen system and affect its motion; internal forces are between parts of the system and cancel in the net sum when using the whole system. External vs Internal Forces The pulling force along a taut string or rope. In an ideal massless, inextensible string over a frictionless, massless pulley, the tension is uniform. Tension The contact force from a surface on a body, acting perpendicular to the surface at the point of contact. Normal Reaction Weight The gravitational force on a body near Earth’s surface, W = m g , directed vertically downward. A kinematic link between displacements/velocities/accelerations of connected bodies due to string length or geometry (e.g., in pulleys). Constraint Relation Newton’s Third Law is compact but easy to misapply. It does not say forces on the same body cancel out. Instead, it pairs forces on different bodies. For example, the book on a table: the table exerts an upward normal on the book, and the book exerts an equal downward force on the table. In the book’s FBD, you draw weight downward and normal upward; these are not an action–reaction pair. They can balance if the book is at rest, but they are different interactions: gravity and contact. Whenever you identify a force on a body, ask: which other body provides it? Immediately, you have found its partner force acting on that other body. This perspective helps in multi-body problems: contact forces between blocks are equal and opposite on the two blocks, tensions at a rope–block interface are equal and opposite, and interaction forces within a chosen system will cancel if you treat the entire set of bodies as one system. Force on B due to A equals in magnitude and opposite in direction to the force on A due to B. Third Law (vector form) In multi-body systems, forces always appear in pairs that are equal and opposite, regardless of whether the system is moving or at rest. Action–reaction forces act along the same line; they are simultaneous; and they depend on the nature of interaction (contact, gravitational, electric, magnetic). They do not require motion: even in static contact, the forces exist and are equal–opposite. However, equality of forces between two bodies does not imply equal accelerations; accelerations depend on mass via a = F /m . That is why when you kick a football, both your foot and the ball experience equal forces, but the ball accelerates much more due to its smaller mass. Keep this separation clear: Third Law pairs connect forces on different bodies; Second Law connects force to acceleration for a single body or system. Third Law applies to any two bodies interacting: contact or at a distance. It does not require equilibrium, motion, or rest. The pair always lies on different bodies and along the same line. tip The free-body diagram method turns word problems into equations. Steps: pick your body (or system), replace every contact or field interaction by its force vector on that body, choose axes aligned with likely motion, resolve forces, and apply F = ma and F = ma . For multi-body setups, write one equation per body. If you choose the entire connected set as the system, internal forces (tensions or contact forces) cancel in the net, and only external forces remain. This can give acceleration quickly. Then, return to individual bodies to find internal forces like tension or contact force by plugging the found acceleration in a single-body equation. Choose the body or system. Draw a clean outline. Mark all external forces with arrows from the body: weight, normal(s), friction(s), tension(s), applied push/pull, spring force. Pick axes wisely (often along surface or motion). Resolve forces into components along axes. Apply F = m a component-wise. For connected bodies, write equations for each or for the whole system when you want acceleration fast. Use constraint relations to link accelerations (pulley strings, rigid rods). FBD workflow you can trust Choosing the system is powerful. If two blocks touch, and you choose both together as the system, the internal contact forces cancel in the net force. If an external horizontal pull F acts, then the acceleration of the two-block system is a = F/(m 1 + m 2) on a smooth surface. Only after you know a do you go back to a single block to find the contact force. This divide-and-conquer prevents algebraic clutter and exposes key physics at a glance. neet-alert Classic trap: Action–reaction pairs NEVER appear in the same free-body diagram. Forces that balance in one body’s FBD are not Third Law pairs; they are often different interactions (e.g., weight vs normal on a resting block). The normal reaction is not always equal to mg . On a horizontal rough surface with no vertical acceleration, yes, N = mg . But on an incline, N = mg if there is no vertical acceleration across the plane. If the body is in an elevator accelerating vertically, or if someone pushes downward or pulls upward, the normal adjusts: N = mg F vertical - m a vertical depending on directions. Always compute N from F = ma along the direction perpendicular to contact. Special case (horizontal rest) Applies only if the surface is horizontal and the body has zero vertical acceleration with no extra vertical forces. tip Boundary check: If the body is on an incline of angle with no motion perpendicular to the plane, then N = mg . If a vertical force F pushes down on the block on a horizontal surface, N = mg + F . Tension in a light, inextensible string over a smooth, massless pulley is the same on both sides. This follows from two idealizations: the string has negligible mass (so any small element has zero net force unless tensions are equal), and the pulley is frictionless and massless (so it requires no torque difference to rotate). If the pulley has mass or bearing friction, or the string has mass, the tensions can differ. In NEET-style problems unless stated otherwise, assume uniform tension for a single massless string over frictionless, massless pulleys. Valid for a massless, inextensible string over a frictionless, massless pulley. Uniform tension (ideal) Pulley constraints come from fixed string length. If one end moves up by x , another may move down by x (simple pulley), or by x/2 or 2x for multiple pulleys. Accelerations follow similar ratios. These geometric relations tie the FBD equations together, letting you solve for unknown accelerations and tensions with consistency. Atwood machine: acceleration and tension (ideal string and pulley) a = (m 2 - m 1) g m 1 + m 2 , T = 2 m 1 m 2 m 1 + m 2 g Two masses m 1 and m 2 connected by a light inextensible string over a frictionless, massless pulley Uniform tension T on both sides Take m 2 > m 1 so motion is m 2 downward, m 1 upward Downward positive for the heavier mass Upward positive for the lighter mass Add equations to eliminate T Collect terms Acceleration magnitude Back-substitute to get T The Atwood machine derivation shows the FBD logic: write F = ma for each mass, relate accelerations by the string, eliminate internal forces to get acceleration, then back-substitute for tension. This pattern extends to inclines, added pulls, or friction on one side, with only small changes in the forces drawn. In any circular motion analysis, the net radial force on the body must equal m v 2 / r . In FBD terms, the radial components of real forces (tension, normal, gravity) sum to provide this centripetal value. Centripetal force requirement F c = m v 2 r Instantaneous circular motion of radius r and speed v Inertial frame; a c directed toward center Magnitude of centripetal acceleration Newton's Second Law in radial direction Net inward force equals m v 2 / r For frictionless design banking, the horizontal component of normal provides m v 2 / r , giving = v 2/(r g) . In the FBD, only weight and normal appear. = v 2 r g Vehicle treated as point mass on a banked curve of radius r No friction; steady speed v Inertial frame; axes along and perpendicular to plane Banked road (no friction): design angle–speed relation Vertical balance (no vertical acceleration) Horizontal component provides centripetal force Divide second by first Even in circular motion, Third Law thinking helps: the tyre pushes the road sideways, and the road pushes the tyre inward (centripetal). On a frictionless bank, the inward component of the normal is the source of centripetal force. Your FBD clarifies which real forces supply m v 2/r ; the ‘centripetal force’ is not a new force, it is the role played by the net radial component of existing forces. For each, name one correct action–reaction pair and state on which bodies they act. These act on different bodies. Equal in magnitude. Field interaction at a distance. A book rests on a table. A person pushes a wall with hands. A magnet attracts an iron nail. easy Use Newton’s Third Law: if A exerts a force on B, then B exerts equal and opposite on A. Identify action–reaction pairs in everyday setups. Notice the forces that balance the book (weight vs normal) are not an action–reaction pair; they act on the same body. The Third Law pair for the normal is the book’s equal and opposite push on the table. This distinction is crucial when drawing the correct FBD and when deciding which forces can cancel in a system choice. System method (internal forces cancel). Only horizontal external force on m 2 is contact from m 1 . Cross-check with the other block. Acceleration a of the pair and contact force C on m 2 by m 1 . m/ s 2 , N m 1 = 2 , kg m 2 = 3 , kg F = 10 , N Surface smooth (no friction) medium Treat both blocks as a single system to find a , then isolate one block to get the internal contact force. Two blocks m 1 = 2 , kg and m 2 = 3 , kg touch on a smooth horizontal table. A horizontal force F = 10 , N pulls m 1 to the right. Find the acceleration of the system and the contact force between the blocks. This example shows the power of a system FBD: acceleration is quick. Then, a single-body FBD gives the internal force. Both approaches agree, as they must. On an exam, label directions clearly; if you guess the wrong direction for C , your algebra will return a negative sign, which simply means the actual direction is opposite to your guess. m = 5 , kg on table M = 3 , kg hanging k = 0.2 g = 9.8 , m/s 2 hard Draw FBDs. Choose rightward for the table block and downward for the hanging mass as positive. Friction opposes impending/actual motion of the table block. A block of mass m = 5 , kg lies on a rough horizontal table (coefficient of kinetic friction k = 0.2 ). It is connected over a light, frictionless pulley to a hanging mass M = 3 , kg . Find the acceleration of the system and the tension in the string after release. Acceleration a (direction of motion) and tension T . m/ s 2 , N Kinetic friction magnitude. Horizontal forces on the table block. Downward positive. Eliminate T to get a. Back-substitute to find tension. NEET-style connected bodies with friction Before committing to kinetic friction, check if motion truly occurs. For limiting static friction, you would compare the pull required to move with f s = s N . If M g were too small to overcome f s , the system would remain at rest. In this problem, kinetic friction is given, so motion is ensured and f k is used directly. This pre-check is a common NEET step. Constraint relations in pulleys arise from fixed string length. For a simple single pulley with one mass on each side, the accelerations of the two masses are equal in magnitude and opposite in direction. In a movable pulley arrangement where one mass is supported by two string segments, the support point moves differently: if the movable pulley rises by x , each free end shortens by x/2 , so a connected mass elsewhere may move by 2x . Converting displacement relations to accelerations gives you the kinematic links needed to close the equations from each FBD. Equal-magnitude, opposite-direction accelerations for two ends of the same ideal string over a single fixed pulley. Simple fixed pulley constraint Use this value in the free-body diagram when determining if the pulley system is on the verge of sliding or moving. Forces: T (→), f (←), N (↑), mg (↓) FBD of m1 m1 Forces: m2 g (↓), T (↑) FBD of m2 m2 Free-body diagrams for a table–pulley–hanger system. Each FBD shows only forces on that body. Schematic FBDs: m1 on a smooth table pulled by tension T to the right; friction f opposing motion if present; normal N up, weight mg down. Hanging mass m2 with tension T up and weight m2 g down. dependent m1 dependent m2 derived Horizontal axis fbd Vertical axis Common Forces Lexicon Force Symbol Direction Rule Nature (Contact/Field) Agent Remember 'W-N-T-F': Weight is down, Normal is out, Tension pulls, and Friction opposes! Weight ( W ) Vertically downwards towards the center of Earth Field force (Non-contact) Earth/Gravity Normal Force ( N ) Perpendicular to the contact surface, pushing into the body Contact force Surface Tension ( T ) Along the string/rope, always pulling away from the body Contact force String/Rope/Chain Static Friction ( f s ) Parallel to surface, opposite to the direction of impending motion Contact force Rough Surface Kinetic Friction ( f k ) Parallel to surface, opposite to the direction of relative sliding Contact force Rough Surface Spring Force ( F s ) Opposite to the displacement ( x ) from mean position ( F = -kx ) Contact force Deformed Spring Buoyant Force ( F B ) Vertically upwards through the center of buoyancy Contact force (Pressure gradient) Fluid (Liquid/Gas) Pseudo Force ( F p ) Opposite to the direction of acceleration of the non-inertial frame Non-inertial (Fictitious) Accelerating Frame Viscous Force ( F v ) Opposite to the relative velocity between fluid layers Contact force Fluid Layers Electrostatic Force ( F e ) Along the line joining two charges; attractive or repulsive Field force Point Charges Magnetic Force ( F m ) Perpendicular to both velocity ( v ) and magnetic field ( B ) Field force Magnetic Field Thrust ( F thrust ) Opposite to the direction of ejected mass (exhaust) Contact force (Reaction) Ejected Mass/Fuel Air Resistance ( D ) Opposite to the direction of velocity ( v ) Contact force Air Molecules Centripetal Force ( F c ) Towards the center of the circular path Net Resultant (Not a separate force) Tension/Gravity/Friction Centrifugal Force ( F cf ) Radially outwards (only in rotating frame) Non-inertial (Fictitious) Rotating Frame common forces lexicon Normal reaction always equals the weight mg . Only on a horizontal surface with no other vertical forces and zero vertical acceleration. On an incline N = mg ; in an accelerating elevator, N mg ; added pushes/pulls adjust N . Action and reaction cancel each other, so nothing moves. They act on different bodies. Motion depends on the net external force on a chosen body/system, not on forces acting on other bodies. Do not pair weight of a block with normal on the same block as ‘action–reaction’. The Third Law partner of the block’s weight is the upward gravitational pull on Earth by the block. neet-alert Use PAIR to remember Third Law essentials. PAIR: Pairs act Across bodies, Identical in size, Reverse in direction. How to use the pulley visualizer: set m 1 on the table, m 2 as the hanging mass, and choose friction. Predict acceleration using your system FBD before pressing play. Then isolate one mass to get tension and check against the tool. Change friction to see when motion starts or stalls (compare M g vs f s ). Common FBD mistakes to avoid Drawing both forces of an action–reaction pair on the same FBD. Forgetting weight even on a horizontal surface. Placing normal not exactly perpendicular to contact. Assuming N = mg on inclines or in accelerating frames. Forgetting that tension pulls away from the body along the string. Mixing internal and external forces when using the system method. Third Law analysis complements, not replaces, F = ma . Use it to check internal forces: if m 1 pushes m 2 with C , then m 2 pushes m 1 with C opposite. In a rope, the block pulls the rope with T and the rope pulls the block with T opposite. When you select both as a system, these equal–opposite forces cancel, leaving only external agents to determine the acceleration of the system’s center of mass. A quick selection trick: To find acceleration of connected bodies on a smooth surface pulled by a force F , take the whole set as the system. Then a = F/ m . To find the tension between any two bodies, pick one of them and apply F = ma along the line of pull using the same a you just found. This two-step is both fast and robust under exam pressure. Effective mass (system on smooth surface) When a force F pulls a train of blocks on a smooth surface, treat the system to get a , then isolate a portion to get internal tension. This angle defines the limiting equilibrium condition where the applied force overcomes static friction, causing impending motion. Edge cases matter: An exactly massless string connecting a finite force to a finite mass is ideal; however, if you tried to accelerate a massless object by a finite net force, a = F/0 would be undefined—this flags why we never treat an entire massless rope as a separate body without its attachments. Similarly, a massless, frictionless pulley cannot sustain a torque difference; hence equal tensions are a logical consequence, not an extra assumption. remember Big takeaways: Draw one FBD per body; Third Law pairs live on different bodies; choose the whole system to get acceleration fast; return to single bodies for internal forces; check constraints and assumptions before plugging numbers. Worked reasoning practice: Suppose two blocks on an incline are tied together and pulled upward by a rope. If you take both as a system, friction on each block adds to total external resistance, and the rope pull must overcome the sum of components of weight down the plane plus friction to accelerate them. If you need the tension in the tie between them, isolate either block, keep the same acceleration, and solve along the incline. This mirrors the table–pulley examples and builds the same reflexes. Never invent a ‘centrifugal force’ in an inertial frame. In circular motion questions, the inward components of real forces provide m v 2/r . The term ‘centripetal force’ names the role of that net inward force; it is not a separate new force. neet-alert Action–Reaction Pair Equal and opposite forces on two interacting bodies, along the same line, simultaneous. Free-Body Diagram Isolated-body sketch with all external forces, used to apply F = m a . System Chosen set of bodies treated together; internal forces cancel in the net. Pull along a taut string; uniform in an ideal massless string over frictionless, massless pulleys. Tension Normal Reaction Perpendicular contact force from a surface. Constraint Relation Geometric link between motions of connected bodies due to fixed string length or geometry. Centripetal Force Name for the net inward force required for circular motion, m v 2 / r . Frictionless design where = v 2/(r g) ; inward normal component supplies m v 2/r . Banking of Roads Core terms recap