Newton’s Laws of Motion and Friction are important NEET Physics topics, but numerical questions can be difficult when multiple forces act on a body or several objects move together. Problems involving pulleys, inclined planes, connected blocks and friction require you to identify the forces correctly and account for their directions. Accelerating reference frames, spring systems and changing contact conditions can add further complexity to force and motion calculations.
These notes cover Newton’s three laws, common forces, frictional conditions and their applications through essential formulas, tables and problem-solving methods. Revise the concepts with free-body diagrams and numerical examples to understand how to calculate acceleration, determine tension and normal force, and approach different mechanics questions more accurately.
Newton’s laws explain how forces affect the motion of objects. Before solving numerical questions, identify the forces acting on each body and choose a suitable direction for applying the equations.
|
Law |
Statement |
Mathematical Form |
|
First law |
A body remains at rest or moves with constant velocity unless acted upon by a non-zero net external force. |
(\vec F_{\text{net}}=0 \Rightarrow \vec a=0) |
|
Second law |
The net external force equals the rate of change of momentum. |
(\vec F_{\text{net}}=\frac{d\vec p}{dt}) |
|
Second law for constant mass |
The net force equals mass multiplied by acceleration. |
(\vec F_{\text{net}}=m\vec a) |
|
Third law |
Every action force has an equal and opposite reaction force acting on another body. |
(\vec F_{AB}=-\vec F_{BA}) |
Important: Action and reaction forces act on different bodies, so they do not cancel each other in the free-body diagram of a single object.
|
Force |
Direction or Formula |
Application |
|
Gravitational force |
(mg), vertically downward near Earth's surface |
Weight of a body |
|
Normal force |
Perpendicular to the contact surface |
Contact between blocks or surfaces |
|
Tension |
Along a rope or string, pulling away from the body |
Connected bodies and pulleys |
|
Friction |
Opposes relative motion or its tendency at the contact |
Rough surfaces |
|
Spring force |
(F=-kx) |
Stretched or compressed springs |
|
Buoyant force |
(F_B=\rho Vg) for a fully submerged body in a fluid of density (\rho) |
Objects in liquids |
|
Pseudo force |
(\vec F_p=-m\vec a_{\text{frame}}) |
Problems in accelerating reference frames |
A free-body diagram (FBD) shows all external forces acting on one selected object.
Isolate the body from its surroundings.
Draw its weight vertically downward.
Add normal force perpendicular to the contact surface, if present.
Draw tension along the rope and away from the body.
Add friction opposite to the actual or impending relative motion.
Resolve forces into suitable horizontal and vertical components.
Apply Newton’s second law separately along each axis.
For a block on a horizontal surface with no vertical acceleration, if the only vertical forces are weight and normal force, (N=mg). This relationship changes when additional vertical forces or vertical acceleration are present.
A body is in translational equilibrium when its net external force is zero. It may remain at rest or move with constant velocity. [\sum F_x=0,\qquad \sum F_y=0]
|
Situation |
Important Relationship |
|
Block on a horizontal surface, with no other vertical forces |
(N=mg) |
|
Block held against a vertical wall by a horizontal force (F) |
(N=F) |
|
Block suspended vertically at rest by a rope |
(T=mg) |
|
Rope supporting several stationary masses |
Tension depends on the total weight supported below the point |
|
Rope making angle (\theta) with the vertical and supporting a mass against horizontal force (F) |
(\tan\theta=\frac{F}{mg}), when vertical and horizontal forces balance |
For a massless rope passing over an ideal, frictionless pulley, tension has the same magnitude throughout a continuous rope.
For a system of connected masses, draw an FBD for each mass and apply the relevant force equations. A shortcut using the total system can help calculate acceleration, but separate equations are usually needed to find tension.
When the net force on a body is non-zero, it accelerates according to (F_{\text{net}}=ma). The direction of acceleration should be chosen consistently while writing the equations.
An Atwood machine consists of two masses connected by a light, inextensible string passing over an ideal pulley.
For (m_1>m_2), the acceleration of (m_1) is downward and that of (m_2) is upward.
[m_1g-T=m_1a]
[T-m_2g=m_2a]
Solving these equations gives:
[a=\frac{(m_1-m_2)g}{m_1+m_2}]
[T=\frac{2m_1m_2g}{m_1+m_2}]
Suppose a block of mass (m_1) lies on a smooth horizontal surface and is connected to a block of mass (m_2) on a smooth incline of angle (\theta). If (m_2g\sin\theta) drives the system, then
[a=\frac{m_2g\sin\theta}{m_1+m_2}]
The equation assumes an ideal string and pulley, no friction, and motion in the direction driven by the inclined block's weight component.
For two blocks of masses (m) and (m) on a smooth horizontal surface, pulled together by an external force (F),
[a=\frac{F}{2m}]
If the force is applied to the leading block, the tension between the blocks is (F/2). The tension differs if the force is applied to the other block, so always check which block is being pulled.
Constraint motion occurs when the motion of one body depends on another because of a string, pulley or other mechanical connection.
For an ideal inextensible string, the total length remains constant. This condition relates the displacements, velocities and accelerations of connected bodies.
Define a positive direction for each block.
Write the string-length constraint.
Differentiate the constraint to obtain the velocity relation.
Differentiate again to obtain the acceleration relation.
Draw separate FBDs and apply Newton’s second law.
Solve the simultaneous equations for tension and acceleration.
In a simple fixed-pulley system, the connected masses have equal acceleration magnitudes in opposite directions. In movable-pulley systems, the acceleration ratio depends on the number and arrangement of string segments supporting the moving pulley.
Friction acts between surfaces in contact and opposes their relative motion or the tendency of relative motion.
|
Type |
Formula or Condition |
Meaning |
|
Static friction |
(0\leq f_s\leq\mu_sN) |
Adjusts to prevent slipping |
|
Limiting friction |
(f_{s,\max}=\mu_sN) |
Maximum static friction just before slipping |
|
Kinetic friction |
(f_k=\mu_kN) |
Friction during sliding in the usual dry-friction model |
|
Coefficients of friction |
Usually (\mu_k<\mu_s) |
Kinetic friction is generally lower than maximum static friction |
Here, (N) is the normal reaction, (\mu_s) is the coefficient of static friction, and (\mu_k) is the coefficient of kinetic friction.
Important: Static friction is not always equal to (\mu_sN). It takes the value required for equilibrium, up to its maximum possible value.
When a block rests on another block, friction can accelerate the upper block along with the lower block. The blocks move together only if the required friction does not exceed the maximum static friction.
Case 1: Force applied to the lower block
Let the upper block have mass (m_1) and the lower block have mass (m_2). If the surface beneath the lower block is smooth, the acceleration is
[a=\frac{F}{m_1+m_2}]
The friction needed to accelerate the upper block is (f=m_1a). Therefore, the maximum acceleration without slipping is
[a_{\max}=\mu_sg]
and the maximum applied force for common motion is
[F_{\max}=(m_1+m_2)\mu_sg]
Case 2: Force applied to the upper block
For the same two blocks on a smooth lower surface, the friction required to accelerate the lower block is (m_2a), where (a=F/(m_1+m_2)).
Thus, the no-slip condition is [a_{\max}=\frac{\mu_sm_1g}{m_2}]
[F_{\max}=\frac{(m_1+m_2)\mu_sm_1g}{m_2}]
These expressions assume horizontal motion, a smooth floor and no other horizontal forces.
For a block of mass (m) on a horizontal surface:
Force applied horizontally: (F_{\min}=\mu_smg).
Force applied upward at an angle (\theta) above the horizontal: the normal force is (N=mg-F\sin\theta).
At impending motion,
[F\cos\theta=\mu_s(mg-F\sin\theta)]
The minimum force over all possible upward application angles is [F_{\min}=\frac{\mu_smg}{\sqrt{1+\mu_s^2}}]
This result assumes the force is applied above the horizontal and the block remains in contact with the surface.
For a block on an incline of angle (\theta), resolve the weight into two components:
[mg\sin\theta \quad \text{along the incline}]
[mg\cos\theta \quad \text{perpendicular to the incline}]
If there are no other forces perpendicular to the plane, the normal reaction is (N=mg\cos\theta).
|
Situation |
Acceleration or Required Force |
|
Block sliding down |
(a=g(\sin\theta-\mu_k\cos\theta)) |
|
Block moving up |
Retardation magnitude (=g(\sin\theta+\mu_k\cos\theta)) |
|
Minimum force applied parallel to the plane to move a stationary block upward |
(F_{\min}=mg\sin\theta+\mu_smg\cos\theta) |
|
Minimum additional force parallel to the plane needed to prevent downward slipping |
(F_{\min}=mg\sin\theta-\mu_smg\cos\theta), if positive |
The last expression applies when the required additional force acts up the plane. If the block can remain at rest without an external force, the minimum additional force is zero.
The angle of repose is the greatest inclination at which a block remains at rest just before sliding down.
[\tan\theta_r=\mu_s]
where (\theta_r) is the angle of repose.
This result applies at impending downward motion on an incline without other forces affecting the normal reaction or the force along the plane.
A pseudo force is introduced when Newton’s laws are applied from an accelerating, non-inertial reference frame. It acts opposite to the acceleration of that frame.
[\vec F_{\text{pseudo}}=-m\vec a_{\text{frame}}]
For a lift accelerating vertically, the effective gravity changes according to the direction of its acceleration.
|
Lift Condition |
Effective Gravity |
|
Accelerating upward with acceleration (a) |
(g_{\text{eff}}=g+a) |
|
Accelerating downward with acceleration (a) |
(g_{\text{eff}}=g-a) |
|
Moving with constant velocity |
(g_{\text{eff}}=g) |
|
Freely falling with acceleration (g) |
(g_{\text{eff}}=0) |
For an object at rest relative to the lift, the apparent weight measured by a scale is (N=mg_{\text{eff}}), provided the object remains in contact with the scale.
Consider a block at rest relative to a smooth wedge that accelerates horizontally. In the wedge's frame, the pseudo force acts opposite to the wedge's acceleration.
For a suitable wedge orientation and a block remaining at rest relative to the wedge, the force components can balance when
[a=g\tan\theta]
The direction of wedge acceleration depends on the orientation of the incline. This result assumes a frictionless surface and a block stationary relative to the wedge.
An ideal spring exerts a restoring force that opposes displacement from its natural length.
[F_s=-kx]
Here, (k) is the spring constant and (x) is the displacement from the natural length.
|
Arrangement |
Equivalent Spring Constant |
|
Springs in parallel |
(k_{\text{eq}}=k_1+k_2+\cdots) |
|
Springs in series |
(\frac{1}{k_{\text{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}+\cdots) |
|
Two identical springs in series |
(k_{\text{eq}}=k/2) |
|
Two identical springs in parallel |
(k_{\text{eq}}=2k) |
When a spring is cut, the tension in that spring becomes zero. However, the force exerted by an intact ideal spring does not change instantaneously if its extension or compression has not changed.
To solve such problems:
Draw the FBD of every block before the cut.
Determine the spring force immediately before the cut.
Remove the force transmitted by the cut string or disconnected element.
Keep the intact spring's force at its initial value at the instant after the cut.
Apply (F_{\text{net}}=ma) to each block.
Do not assume that all spring forces disappear when a string is cut. Only the force transmission through the severed connection is removed.
Newton’s second law can also be expressed in terms of momentum. This approach is useful when forces act for a short interval, such as during collisions or reflection from a wall.
[\vec p=m\vec v]
[\vec J=\Delta\vec p=\int \vec F,dt]
For an average force over a time interval (\Delta t),
[\vec F_{\text{avg}}=\frac{\Delta\vec p}{\Delta t}]
|
Situation |
Magnitude of Momentum Change |
|
Body of mass (m) reverses its velocity with the same speed (v) |
(2mv) |
|
Body rebounds with initial speed (u) and final speed (v) in the opposite direction |
(m(u+v)) |
|
Body strikes a smooth wall and reflects with the same speed at angle (\theta) to the normal |
(2mv\cos\theta) |
For the reflection result, (\theta) is measured from the normal to the wall, not from the wall's surface.
The area under a force-time graph gives impulse, which equals the change in momentum. The slope of a momentum-time graph gives the net force.
Variable-mass problems require attention to momentum transfer. The force associated with mass entering or leaving a system depends on the mass flow rate and the relative velocity of the transferred material.
For an ideal rocket, the thrust magnitude is
[F_{\text{thrust}}=u\left|\frac{dm}{dt}\right|]
where (u) is the exhaust speed relative to the rocket and (dm/dt) is the rate of change of rocket mass. The magnitude of the rocket's mass decreases during fuel ejection.
For vertical lift-off, thrust must exceed the rocket's weight to produce upward acceleration:
[F_{\text{thrust}}-mg=ma]
For a conveyor belt that receives material initially at rest and accelerates it to belt speed (v), the ideal horizontal force needed to maintain constant belt speed is
[F=v\frac{dm}{dt}]
and the corresponding power is (P=Fv). This assumes the incoming material has zero horizontal velocity before joining the belt and that other losses are neglected.
Regular problem-solving is important because questions often combine multiple concepts rather than testing a formula in isolation.
Start with free-body diagrams: Identify every force before writing equations.
Revise friction conditions: Distinguish between static friction, limiting friction and kinetic friction.
Practise connected-body questions: Solve problems involving ideal pulleys, multiple blocks and string constraints.
Check directions and signs: Choose axes carefully and interpret negative acceleration in relation to your initial assumption.
Understand accelerating frames: Practise lift, wedge and pseudo-force questions separately.
Use momentum when appropriate: For collisions and short-duration forces, impulse may be more efficient than a direct acceleration calculation.
Attempt timed practice: Solve mixed questions under a time limit, then review incorrect answers to identify gaps in your reasoning.
Newton’s Laws of Motion and Friction become easier to apply when you identify the forces correctly and choose the right equations for each situation. Revise free-body diagrams, friction limits, pulley constraints, inclined-plane motion and pseudo force alongside the important formulas. Practising a variety of numerical questions will help you recognise common patterns, avoid sign errors and develop a more reliable approach to NEET Physics mechanics problems.
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