Laws of Motion

JEE Main and NEET Physics notes. RelaxPrep is not affiliated with the NTA or the IITs.

Revision sheet

Newton's first law: with zero net force a body keeps constant velocity (rest is the special case), as seen from an inertial frame. Mass measures this inertia.
Newton's second law: $\vec F_{net} = d\vec p/dt$, which is $m\vec a$ for constant mass. Only the net force on the body appears, and it works component by component.
Newton's third law: forces between two bodies are equal and opposite, of the same kind, and act on different bodies, so they never cancel inside one free-body diagram.
Weight and the normal force are not a third-law pair. The pair of the weight of a book is the book's gravitational pull on the Earth.
Static friction adjusts itself from $0$ up to $\mu_s N$; kinetic friction is $\mu_k N$. Friction depends on $N$, not on contact area, and its direction opposes relative sliding (or its tendency).
On an incline: $N = mg\cos\theta$ and the driving component is $mg\sin\theta$. A block slides from rest only if $\tan\theta > \mu_s$.
A light inextensible string over a smooth pulley carries one tension; the connected bodies share the speed and acceleration magnitude. When the other end of the string is fixed, a movable pulley carrying a load moves half as far as the free end of the string.
Impulse equals change in momentum and equals the area under the $F$-$t$ graph. Momentum along a direction is conserved when the external force along it is zero.
Centripetal force is the net radial component of real forces, $mv^2/r$ toward the centre. It is not an extra force to draw in a free-body diagram.
A spring force cannot change instantly; a string tension can. This decides the acceleration just after a string is cut.
In a frame with acceleration $\vec a_0$ every body feels a pseudo force $-m\vec a_0$; this force has no reaction partner and is used only in that frame.
Use $g = 10\,\text{m s}^{-2}$ in the worked numbers of this chapter unless stated.

Second law: $\vec F_{net} = \dfrac{d\vec p}{dt} = m\vec a$ (Constant mass in the last form. $1\,\text{N} = 1\,\text{kg m s}^{-2}$.)
Impulse: $\vec J = \int \vec F\,dt = \Delta \vec p = \vec F_{avg}\,\Delta t$
Limiting and kinetic friction: $f_{s,max} = \mu_s N, \qquad f_k = \mu_k N$ (Angle of friction $\lambda$: $\tan\lambda = \mu_s$.)
Acceleration on a rough incline (down / up the slope): $a_{down} = g(\sin\theta - \mu_k\cos\theta), \qquad a_{up} = g(\sin\theta + \mu_k\cos\theta)$ ($a_{up}$ is the deceleration of a block sliding up.)
Atwood machine: $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}, \qquad T = \dfrac{2m_1m_2\,g}{m_1 + m_2}$
Block on a smooth table pulled by a hanging mass: $a = \dfrac{m_2 g}{m_1 + m_2}, \qquad T = \dfrac{m_1 m_2\,g}{m_1 + m_2}$ ($m_1$ on the table, $m_2$ hanging.)
Lami's theorem: $\dfrac{F_1}{\sin\alpha_1} = \dfrac{F_2}{\sin\alpha_2} = \dfrac{F_3}{\sin\alpha_3}$ (For three concurrent forces in equilibrium, $\alpha_i$ is the angle between the other two forces.)
Centripetal force: $F_c = \dfrac{mv^2}{r} = m\omega^2 r$
Unbanked road, speed limit: $v_{max} = \sqrt{\mu_s\, r g}$
Banked road: $v_{opt}^2 = r g\tan\theta, \qquad v_{max}^2 = rg\,\dfrac{\tan\theta + \mu}{1 - \mu\tan\theta}, \qquad v_{min}^2 = rg\,\dfrac{\tan\theta - \mu}{1 + \mu\tan\theta}$ ($v_{min}$ applies when $\mu < \tan\theta$; for $\mu \ge \tan\theta$ it is $0$. $v_{max}$ is finite only if $\mu\tan\theta < 1$.)
Conical pendulum: $\omega = \sqrt{\dfrac{g}{L\cos\theta}}, \qquad T_{period} = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}$ ($\theta$ from the vertical, $L$ string length.)
Apparent weight in a lift: $N = m(g + a)$ ($a$ is the upward acceleration of the lift (negative when it is downward).)
Pseudo force: $\vec F_{pseudo} = -m\vec a_{frame}$

Topics

Force, inertia and Newton's first law

A force is an interaction that changes, or tries to change, the state of motion of a body. The first law says what happens when no net force acts: the velocity stays constant. A body at rest stays at rest, and a body moving along a straight line keeps its speed and direction. Rest is simply the case of zero constant velocity, so the law does not distinguish between them.

Inertia is the property of resisting a change in velocity, and its measure is mass. A heavier cart is harder to start and harder to stop than a light one under the same push. This is why the first law is also called the law of inertia: it describes a body that has nothing acting on it, and that body just continues.

The law is only valid in some reference frames. A frame in which a force-free body moves with constant velocity is called an inertial frame. Any frame moving with constant velocity relative to an inertial frame is also inertial. The ground is close enough to inertial for most problems; a braking bus or a spinning merry-go-round is not, and in those frames bodies seem to accelerate without any real force. That is the origin of pseudo forces, treated later in this chapter.

A useful working form: if a body has constant velocity in an inertial frame, the vector sum of all real forces on it is zero. This turns many statics problems into the condition $\sum \vec F = 0$.

Key points

Zero net force means zero acceleration (constant velocity), not necessarily zero velocity.
Mass is the measure of inertia; weight is the gravitational force and depends on where the body is.
A frame is inertial if Newton's first law holds in it. A frame with constant velocity relative to an inertial frame is inertial.
A body at rest relative to a braking vehicle seems to slide forward: the vehicle decelerates, the body tends to keep its velocity relative to the ground.
Force is a vector; the SI unit is the newton, $1\,\text{N} = 1\,\text{kg m s}^{-2}$, with dimensions $[MLT^{-2}]$.

Common traps

Constant velocity does not require a force. Friction is the reason everyday objects slow down, not an inherent need for a push.
Zero net force does not mean no forces act. A book on a table has weight and normal force, which cancel.
A body can have zero velocity and non-zero acceleration at an instant (the top of a thrown ball), so zero velocity alone does not imply equilibrium.

Newton's second law and linear momentum

Linear momentum is $\vec p = m\vec v$. The second law states that the net force on a body equals the rate of change of its momentum, $\vec F_{net} = d\vec p/dt$. For a body of constant mass this becomes $\vec F_{net} = m\vec a$, and the acceleration points along the net force.

Only the net force counts. Each force acting on the body is added as a vector, and the result is the single force that the law talks about. The law is a vector equation, so it holds separately along any chosen axis: $\sum F_x = ma_x$ and $\sum F_y = ma_y$. Choosing one axis along the acceleration and the other perpendicular to it removes many unknowns at once.

The form $F = dp/dt$ is the more general one. When the mass changes (sand falling on a moving belt, a rocket expelling gas) one must use $d(mv)/dt$, and the product rule gives an extra term. For JEE Main this appears as the force needed to keep a belt moving at constant speed $v$ while mass is dropped on it at the rate $dm/dt$: $F = v\,dm/dt$.

The law also fixes the meaning of one newton: the force that gives a mass of $1\,\text{kg}$ an acceleration of $1\,\text{m s}^{-2}$. The gravitational unit $1\,\text{kgf} = 9.8\,\text{N}$ is sometimes used and must be converted before use.

Key points

$\vec a$ is parallel to $\vec F_{net}$; for the same force a larger mass gives a smaller acceleration.
Write the law along each axis independently: forces along an axis do not affect the acceleration along a perpendicular axis.
Momentum has units $\text{kg m s}^{-1}$, the same as $\text{N s}$.
If $\vec F_{net}$ is perpendicular to $\vec v$, only the direction of velocity changes and the speed stays constant (uniform circular motion).
The force on a body at an instant depends on what acts on it at that instant, not on its past velocity.

Second law in general and constant-mass forms: $\vec F_{net} = \dfrac{d(m\vec v)}{dt} \;\xrightarrow{\; m \text{ constant}\;}\; \vec F_{net} = m\vec a$
Force to keep a belt at constant speed while mass falls on it: $F = v\,\dfrac{dm}{dt}$ (The added mass is brought from rest to $v$.)

Common traps

In $F = ma$ the $F$ is the vector sum of all forces on that body. A single pull of $10\,\text{N}$ on a body that also has friction of $4\,\text{N}$ gives $F_{net} = 6\,\text{N}$.
$ma$ is not itself a force. It is the result of the net force, and it must not be added to the force list of the same body in an inertial frame.
Weight $mg$ is a force; mass $m$ is not. A body in a lift keeps the same mass while its apparent weight changes.
For variable-mass problems $F = m\,dv/dt$ is incomplete; the full law is $F = dp/dt$.

Newton's third law and action-reaction pairs

Whenever body A exerts a force on body B, B exerts a force on A that has the same magnitude, opposite direction and the same line of action. The two forces are of the same physical kind (both gravitational, both contact, both electrostatic) and they appear at the same instant. Each force is part of one interaction, so there is no cause and effect order.

The decisive property is that the two forces act on different bodies. Because of that they never appear together in the free-body diagram of a single body, and they can never cancel each other out when you apply the second law to that body. When you take two bodies together as one system, the forces between them are internal and their vector sum is zero, which is why internal forces do not change the total momentum.

A horse pulling a cart is the classic puzzle. The horse pulls the cart forward and the cart pulls the horse backward with an equal force; these do not cancel because they act on different bodies. The cart accelerates because the forward pull on it exceeds friction on it. The horse accelerates because the ground pushes it forward with a static friction force larger than the backward pull from the cart. Walking works the same way: the foot pushes backward on the ground, and the ground's static friction pushes the foot forward.

Key points

A third-law pair: equal magnitude, opposite direction, same line, same type, different bodies, simultaneous.
The pair of the force of the Earth on a book is the force of the book on the Earth, not the normal force of the table.
The pair of the normal force of the table on the book is the force of the book on the table (its compression of the table).
Internal forces cancel in pairs for a system, so $\vec F_{net,ext} = d\vec P_{system}/dt$ for the system.
A heavy and a light body in a collision exert equal forces on each other; their accelerations differ because their masses differ.

Common traps

Weight and normal force on a resting book are equal and opposite because of the second law (zero acceleration), not because of the third law. If the table accelerates upward, they stop being equal while the third-law pairs stay equal.
The pair of a force is never on the same body. Two forces that act on the same body and cancel are not an action-reaction pair, whatever their magnitudes.
Equal and opposite forces on two bodies do not imply equal accelerations. Acceleration is force divided by that body's own mass.

Common forces: normal force, tension and spring force

Most mechanics problems use three contact forces besides weight and friction. The normal force $N$ is the part of the contact force perpendicular to the surface. It adjusts to whatever value prevents the bodies from penetrating, so it is found from the equation of motion perpendicular to the surface, not assumed equal to $mg$. On a level floor with other vertical forces absent, $N = mg$; on a slope it is $mg\cos\theta$; in a lift it is $m(g + a)$.

Tension is the pull of a string or rope. A string can only pull, along its own length, and it cannot push. For a string of negligible mass the tension is the same at every point, even when it passes over a smooth massless pulley, because any small piece of the string would otherwise have an unbalanced force on zero mass. If the string has mass, the tension varies along it, being larger where it supports more weight or must accelerate more mass. An inextensible string forces connected points to have the same component of velocity along the string.

A spring exerts a restoring force proportional to its deformation, $F = kx$, directed to restore the natural length. The spring constant $k$ is a property of the spring. The extension of a spring cannot change instantaneously, since the ends would need infinite speed. So the instant after a string is cut, the spring force still has its earlier value while the string tension changes at once. This difference between a spring and a string decides many 'just after cutting' problems.

Springs combine like this. For springs in parallel (the same extension) the constants add: $k = k_1 + k_2$. For springs in series (the same force) the reciprocals add: $1/k = 1/k_1 + 1/k_2$. A spring cut into $n$ equal pieces gives pieces of constant $nk$, since a shorter piece stretches less for the same force.

Key points

A string pulls only; a rigid rod can pull or push; a spring pulls when stretched and pushes when compressed.
Ideal string: tension equal throughout. Ideal smooth pulley: only changes the direction of the string, so the tension is the same on both sides.
Two ends of an inextensible string have equal components of velocity and acceleration along the string.
The force by a massless spring is the same at both its ends, in magnitude.
Normal force and friction are the two components of the same contact force; the total contact force has magnitude $\sqrt{N^2 + f^2}$.

Hooke's law: $F = -kx$ ($x$ is the displacement from the natural length.)
Springs in parallel and series: $k_{par} = k_1 + k_2, \qquad \dfrac{1}{k_{ser}} = \dfrac{1}{k_1} + \dfrac{1}{k_2}$
Spring cut into n equal parts: $k' = nk$

Common traps

The normal force is not always $mg$. It depends on the other forces and on the acceleration perpendicular to the surface.
If a massive rope hangs or is pulled, the tension is not the same at all points; the equal-tension rule is for massless strings.
After a string is cut the spring force is unchanged at that instant; a spring force never jumps when something attached to it changes; it changes only as the length changes.
A spring balance reads the force at its hook (the tension), not the weight of a particular body. In an Atwood machine a balance in the string reads the tension $2m_1m_2g/(m_1 + m_2)$.

Shortcuts

Combining springs in one line. Parallel: $k_1 + k_2$. Series: $k_1k_2/(k_1 + k_2)$. A spring of length $L$ cut to $L/n$ has constant $nk$, so half a spring is twice as stiff.

Free-body diagrams and equilibrium of forces

A free-body diagram (FBD) is a sketch of one body, isolated from everything else, with every force acting on that body drawn as an arrow from it. The method has fixed steps: choose the body, list what touches it (surfaces, strings, springs) and what acts at a distance (gravity), draw each force once, pick axes along the acceleration if there is one, and write $\sum F = ma$ along each axis. Forces that the body exerts on others are not drawn on its diagram.

A body is in translational equilibrium when the net force is zero, so $\sum F_x = 0$ and $\sum F_y = 0$. When exactly three forces keep a point in equilibrium, they must be coplanar and concurrent, and the vectors form a closed triangle. This gives Lami's theorem: each force is proportional to the sine of the angle between the other two. If the angles are not convenient, resolving along perpendicular axes always works.

A heavy load hung from the middle of a stretched string or cable shows how tension depends on geometry. If each half of the string makes an angle $\theta$ with the horizontal, the vertical components must hold the weight: $2T\sin\theta = W$, so $T = W/(2\sin\theta)$. As the string is pulled flatter ($\theta \to 0$) the tension grows without limit, which is why a perfectly horizontal string cannot support a load.

Key points

Draw only forces acting on the chosen body. A force like $ma$ is not drawn in an inertial frame.
Each force in the diagram must be traceable to an object touching the body (or to gravity).
For three concurrent forces in equilibrium: $F_1/\sin\alpha_1 = F_2/\sin\alpha_2 = F_3/\sin\alpha_3$, where $\alpha_i$ is the angle between the other two forces.
The tension in a string at angle $\theta$ to the horizontal supporting weight $W$ at its midpoint is $W/(2\sin\theta)$.
If the system is at rest or at constant velocity, the closed polygon of forces gives the same answer as resolving.

Equilibrium conditions: $\sum F_x = 0, \qquad \sum F_y = 0$
Load at the midpoint of a string: $T = \dfrac{W}{2\sin\theta}$ ($\theta$ is the angle of each half with the horizontal.)

Common traps

In Lami's theorem the angle opposite to a force is the angle between the other two forces, not the angle the force makes with the vertical or with the wall.
In $T = W/(2\sin\theta)$ the angle is from the horizontal. If the angle is measured from the vertical the formula becomes $T = W/(2\cos\alpha)$.
Drawing a force on the wrong body (for example the tension on both the string and the block, or the force the body exerts on the table) gives wrong equations.

Impulse and conservation of linear momentum

A large force that acts for a very short time (a bat striking a ball, a hammer hitting a nail) is described by its impulse, $\vec J = \int \vec F\,dt$. Integrating the second law over the interval gives the impulse-momentum theorem: $\vec J = \Delta\vec p$. On an $F$-$t$ graph the impulse is the area under the curve, and the average force over the contact time $\Delta t$ is $\vec F_{avg} = \vec J/\Delta t$. During an impulsive contact, finite forces such as weight are negligible compared with the contact force.

For a system of bodies the internal forces occur in third-law pairs and cancel, so only external forces change the total momentum: $\vec F_{ext} = d\vec P/dt$. If the net external force is zero along some direction, the total momentum along that direction is constant. Momentum is conserved in every collision and explosion where external impulses are negligible during the event; kinetic energy is conserved only in an elastic collision.

Recoil follows directly. A gun of mass $M$ firing a bullet of mass $m$ at ground speed $v$ recoils at $v_g = mv/M$ in the opposite direction, starting from rest. For two fragments produced from rest, the momenta are equal and opposite, so the kinetic energies are in the inverse ratio of the masses. For a stream of particles striking a surface, force equals momentum delivered per second; a jet that stops on the wall exerts $F = \rho A v^2$, and one that rebounds elastically exerts $F = 2\rho A v^2$.

A rocket works by expelling mass backward at relative speed $u$. The thrust on it is $F = u\,|dm/dt|$, where $dm/dt$ is the rate at which the mass is ejected. Gravity and air drag are then added to find the net force and the acceleration.

Key points

Impulse and momentum have the same unit: $1\,\text{N s} = 1\,\text{kg m s}^{-1}$.
The direction of impulse is the direction of $\Delta\vec p$, not of the initial or final velocity.
For a ball hitting a smooth wall, only the component of momentum perpendicular to the wall changes; the component along the wall is unchanged.
Momentum along a direction is conserved if the external force along it is zero, even if momentum along another direction is not (for example a projectile's horizontal momentum).
Rocket thrust is $u\,|dm/dt|$ with $u$ measured relative to the rocket.

Impulse-momentum theorem: $\vec J = \int_{t_1}^{t_2}\vec F\,dt = \vec p_2 - \vec p_1$
Gun and bullet recoil from rest: $Mv_g = mv \;\Rightarrow\; v_g = \dfrac{m}{M}\,v$
Jet of fluid on a wall (density $\rho$, area $A$, speed $v$): $F_{stop} = \rho A v^2, \qquad F_{rebound} = 2\rho A v^2$
Rocket thrust: $F_{thrust} = u\left|\dfrac{dm}{dt}\right|$

Common traps

A rebounding body has a larger change of momentum than a stopping one: $\Delta p = m(v + v')$ for a head-on rebound, not $m(v - v')$.
Total momentum is conserved with respect to one inertial frame, so all velocities in the equation must be measured in that same frame. A bullet's muzzle speed relative to the gun is not its speed relative to the ground.
Conservation of momentum applies to the system, not to each body. The momentum of one body changes in a collision.
Kinetic energy is not conserved in a perfectly inelastic collision even though momentum is.

Shortcuts

Energy split in an explosion from rest. Two fragments from rest have equal and opposite momentum $p$, so $K = p^2/2m$ and $K_1/K_2 = m_2/m_1$. The lighter piece takes the larger share of the energy.

Friction: static, kinetic and limiting

Friction is the component of the contact force parallel to the surfaces. Two rough surfaces in contact resist sliding with it. When the body is at rest relative to the surface, the static friction adjusts itself to the other forces, taking any value from zero up to a maximum, the limiting friction $f_{s,max} = \mu_s N$. The body stays at rest as long as the force trying to slide it is no more than this. Once sliding starts, the friction is the kinetic friction $f_k = \mu_k N$, which is usually a little smaller, so $\mu_k < \mu_s$.

To a good approximation the friction depends on the normal force and on the materials, but not on the apparent contact area or on the sliding speed. The coefficients $\mu_s$ and $\mu_k$ are dimensionless. Rolling friction is much smaller than sliding friction for the same load, which is why wheels are used.

The direction of friction opposes the relative sliding of the surfaces (or the tendency to slide), not necessarily the motion of the body relative to the ground. A box on the floor of an accelerating truck is pushed forward by static friction; the box is moving forward and the friction is also forward. Walking and driving rely on static friction pointing in the direction of motion.

The contact force $\sqrt{N^2 + f^2}$ makes an angle with the normal that cannot exceed the angle of friction $\lambda$, where $\tan\lambda = f_{s,max}/N = \mu_s$. The angle of repose of a slope, at which a block just begins to slide, equals the angle of friction.

Pulling a block at an angle $\phi$ above the horizontal reduces the normal force to $N = mg - F\sin\phi$, so the force needed to move it is $F = \mu mg/(\cos\phi + \mu\sin\phi)$. This is smallest when $\tan\phi = \mu$, with the value $F_{min} = \mu mg/\sqrt{1 + \mu^2}$. Pushing downward at the same angle instead increases the normal force, $N = mg + F\sin\phi$, giving $F = \mu mg/(\cos\phi - \mu\sin\phi)$; if $\tan\phi \ge 1/\mu$ no force, however large, can start the motion.

Key points

Static friction is self-adjusting: it equals the applied tangential force until the limit $\mu_s N$ is reached.
Friction is proportional to $N$, which may differ from $mg$ (inclines, extra vertical forces, lifts).
Maximum acceleration of a vehicle (or of a block carried by friction) on a surface is $\mu_s g$; the shortest braking distance from speed $v$ is $v^2/(2\mu g)$.
A block on another block: the upper block accelerates only because of friction from the lower one, so its largest possible acceleration without slipping is $\mu_s g$.
Angle of friction $\lambda = \tan^{-1}\mu_s$; angle of repose $= \lambda$.

Limiting and kinetic friction: $0 \le f_s \le \mu_s N, \qquad f_k = \mu_k N$
Minimum force to move a block by pulling: $F_{min} = \dfrac{\mu mg}{\sqrt{1 + \mu^2}} \quad \text{at} \quad \tan\phi = \mu$
Braking distance with locked wheels: $s = \dfrac{v^2}{2\mu_k g}$

Common traps

Friction on a resting body is not $\mu N$ in general. It is whatever the situation needs, as long as it does not exceed $\mu_s N$. A block with a $5\,\text{N}$ push and $\mu_sN = 8\,\text{N}$ has $f = 5\,\text{N}$.
Friction does not always oppose the motion of the body. On the box in an accelerating truck, or the foot of a walker, it acts forward.
Friction does not depend on the contact area: placing a brick on a larger or smaller face leaves $f_k$ unchanged (for the same $N$).
On an incline $N = mg\cos\theta$, so limiting friction is $\mu mg\cos\theta$, not $\mu mg$.

Shortcuts

Smallest pulling force. Pull at the angle $\tan\phi = \mu$ above the horizontal and the minimum force is $\mu mg/\sqrt{1 + \mu^2}$. For $\mu = 0.5$ and $m = 10\,\text{kg}$ it is $44.7\,\text{N}$.

Block carried by another block. The upper block stays put while the common acceleration is at most $\mu_s g$, so with the force on the lower block and a smooth floor the largest force without slipping is $F = (m_1 + m_2)\mu_s g$.

Motion on an inclined plane

For a block on a plane inclined at angle $\theta$, choose axes along and perpendicular to the plane. The weight $mg$ resolves into $mg\sin\theta$ along the plane (down the slope) and $mg\cos\theta$ perpendicular to it (into the plane). There is no acceleration perpendicular to the plane, so the normal force is $N = mg\cos\theta$. The friction limit is therefore $\mu N = \mu mg\cos\theta$.

On a smooth incline the acceleration down the slope is $g\sin\theta$ whatever the mass. On a rough incline a block sliding down has $a = g(\sin\theta - \mu_k\cos\theta)$, while a block sliding up is slowed by both the weight component and friction, $a_{up} = g(\sin\theta + \mu_k\cos\theta)$ (as a deceleration). The block released from rest slides only if $mg\sin\theta > \mu_s mg\cos\theta$, that is $\tan\theta > \mu_s$. If it does not slide, friction equals $mg\sin\theta$, which is less than $\mu_s N$.

A block thrown up a rough slope takes longer to come down than it did to go up, because friction acts down the slope on the way up and up the slope on the way down. The descent time exceeds the ascent time by the factor $\sqrt{(\sin\theta + \mu\cos\theta)/(\sin\theta - \mu\cos\theta)}$ for the same distance. On a smooth slope the two times are equal.

Energy and kinematics complete the picture. A block released from height $h$ on a smooth incline reaches the bottom with speed $\sqrt{2gh}$ for any angle, but the time taken is $t = \dfrac{1}{\sin\theta}\sqrt{2h/g}$, which is longer for a gentler slope.

Key points

$N = mg\cos\theta$ when no other force has a component perpendicular to the plane.
A block just begins to slide when $\tan\theta = \mu_s$; it moves at constant speed down the slope when $\tan\theta = \mu_k$.
Down the slope: $a = g(\sin\theta - \mu_k\cos\theta)$. Up the slope (deceleration): $a = g(\sin\theta + \mu_k\cos\theta)$.
A horizontal push or a pull at an angle changes $N$ as well as the driving force, so recompute $N$ before using $\mu N$.
On a smooth plane $N$ does not enter the equation along the slope, so the acceleration needs no normal force.

Components of weight: $W_\parallel = mg\sin\theta, \qquad W_\perp = mg\cos\theta$
Smooth incline: speed at the foot and time of descent: $v = \sqrt{2gh}, \qquad t = \dfrac{1}{\sin\theta}\sqrt{\dfrac{2h}{g}}$
Force parallel to the slope to move a block up at constant speed: $F = mg(\sin\theta + \mu_k\cos\theta)$

Common traps

The component along the slope is $mg\sin\theta$ and the one perpendicular to it is $mg\cos\theta$. When the angle is measured with the vertical (as in some problems) the two are interchanged.
If a force acts parallel to the slope, $N$ is still $mg\cos\theta$; if the force is horizontal or at an angle with the slope, $N$ changes.
A block that is at rest on a rough slope experiences friction $mg\sin\theta$ (static), not $\mu_s mg\cos\theta$, unless the block is on the verge of slipping.
Friction reverses its direction when the block's motion reverses. A block thrown up the plane feels a downhill friction on the way up and an uphill one on the way down.

Shortcuts

Constant-speed slide gives the coefficient. If a block slides down at constant speed, $\tan\theta = \mu_k$, so $\mu_k$ is read straight from the angle. For $\theta = 37^\circ$, taking $\tan 37^\circ \approx 0.75$, this gives $\mu_k \approx 0.75$.

Smooth incline, speed and time. On a smooth slope the speed at the bottom is $\sqrt{2gh}$ for any angle, and $a = g\sin\theta$ gives $t = \sqrt{2L/(g\sin\theta)}$ for slope length $L$.

Connected bodies, pulleys and string constraints

When bodies are linked by strings, the accelerations are related by the constraint of the string. An inextensible string has constant length, so the speeds of its ends are linked: if the string runs straight, both ends have the same acceleration along it. For a string over a fixed pulley, one body going down by $x$ means the other goes up by $x$. For a movable pulley, the load hung on it moves by half as much as the free end of the string moves. In general, write the total length of string as a sum of segment lengths, hold it constant, and differentiate twice.

The method is the same each time. Draw an FBD for each body, write $\sum F = ma$ along its direction of motion using the constraint to relate the accelerations, and solve. If the question asks only for the acceleration, treat the connected bodies as a single system along the string: the internal tensions cancel and $a = F_{net,driving}/m_{total}$. The tension needs one body's equation afterwards.

The Atwood machine (masses $m_1 > m_2$ over a smooth pulley) has $a = (m_1 - m_2)g/(m_1 + m_2)$ and $T = 2m_1m_2g/(m_1 + m_2)$. The pulley itself feels a downward force $2T$ from the string, which is why a support reading is $4m_1m_2g/(m_1 + m_2)$, less than $(m_1 + m_2)g$ when the masses accelerate. For a mass $m_1$ on a smooth table pulled by a hanging mass $m_2$, $a = m_2g/(m_1 + m_2)$ and $T = m_1m_2g/(m_1 + m_2)$; the string around the pulley corner pulls on the pulley with magnitude $T\sqrt{2}$.

When a force $F$ pushes or pulls a row of blocks on a smooth surface, all share one acceleration $a = F/\sum m$. The force between neighbours is the force needed to accelerate everything beyond the point of contact. For a uniform rope of mass $M$ and length $L$ pulled at one end by $F$ on a smooth surface, the tension at distance $x$ from the pulled end is $F(L - x)/L$, since the part beyond $x$ is the mass that the tension must accelerate.

Key points

For a light string, tension is the same along the string; for a massive rope it changes with position.
A string over a fixed pulley reverses the direction of the force, not its magnitude.
Movable pulley: the load's displacement, velocity and acceleration are half those of the free end of the string. The string tension acts twice on a massless movable pulley, giving $2T$ on the load side.
For two blocks pushed by $F$ on a smooth surface (force on $m_1$, which touches $m_2$) the contact force is $Fm_2/(m_1 + m_2)$.
System shortcut: acceleration $= $ net external driving force along the path divided by the total mass.

Atwood machine, force on the pulley support: $F_{support} = 2T = \dfrac{4m_1m_2\,g}{m_1 + m_2}$
Force transmitted through a string between two blocks: $T = \dfrac{F\,m_1}{m_1 + m_2}$ ($F$ pulls the front block $m_2$ on a smooth surface and the string pulls $m_1$ behind it.)
Tension at distance $x$ from the pulled end of a uniform rope: $T(x) = F\,\dfrac{L - x}{L}$ (Smooth horizontal surface, total length $L$.)

Common traps

In an Atwood machine the tension is not $m_1g$ or $m_2g$ unless the system is at rest. It lies between them.
With a movable pulley the two bodies do not have equal accelerations, and the tension on the load side is $2T$ on the pulley. Using the same acceleration for both is the usual mistake in this type of problem.
For a string around a corner the directions of the accelerations differ but their magnitudes are equal. Use signs that follow the string direction, not the axes.
For a heavy rope on a table with part hanging over the edge, the hanging fraction drives the motion and the acceleration is not constant.

Shortcuts

Atwood and table-pulley accelerations in one step. Take the pulling weight difference over the total mass: $a = (m_1 - m_2)g/(m_1 + m_2)$ for the Atwood machine, and $a = m_2 g/(m_1 + m_2)$ for a smooth table with $m_2$ hanging.

Tension in a string between two blocks on a smooth surface. With $F$ pulling the front block and mass $m_1$ behind the string, $T = Fm_1/(m_1 + m_2)$, since $a = F/(m_1 + m_2)$ and the string only accelerates $m_1$.

Dynamics of circular motion

A body moving on a circle of radius $r$ with speed $v$ has an acceleration $v^2/r$ directed toward the centre. By the second law, the net force along the radius, toward the centre, must be $mv^2/r$. This net inward force is called the centripetal force; it is the vector sum of real forces (tension, normal force, friction, gravity components) and it is not drawn as an additional arrow in a free-body diagram. Write the radial equation as (inward forces) $-$ (outward forces) $= mv^2/r$.

If the speed also changes, the force has a tangential component $m\,dv/dt$ along the motion, and the total acceleration is $\sqrt{(v^2/r)^2 + (dv/dt)^2}$. In uniform circular motion the tangential force is zero.

On a flat road friction provides the centripetal force, so the highest safe speed is $v_{max} = \sqrt{\mu_s rg}$. Banking the road at angle $\theta$ lets the normal force contribute: with no friction, $N\sin\theta = mv^2/r$ and $N\cos\theta = mg$, giving $v^2 = rg\tan\theta$. With friction the range of safe speeds is between $v_{min}$ and $v_{max}$ given in the formula list; a cyclist leans at $\tan\theta = v^2/(rg)$ so that the contact force passes through the centre of mass.

A conical pendulum (a bob whirled in a horizontal circle with the string making angle $\theta$ with the vertical) obeys $T\cos\theta = mg$ and $T\sin\theta = m\omega^2 L\sin\theta$, so $\omega = \sqrt{g/(L\cos\theta)}$ and $T = m\omega^2 L$. The string cannot be horizontal since the vertical component of the tension must carry the weight.

When a body moves in a vertical circle its speed changes from point to point, so the speed at each point has to come from energy conservation. That case (minimum speeds $\sqrt{gR}$ at the top and $\sqrt{5gR}$ at the bottom for a string, the $6mg$ tension difference, the rod case and where a string goes slack) is covered in the Work, Energy and Power hub, topic 'Motion in a vertical circle'. The dynamics step is the same as for any circle: at each point the net force toward the centre equals $mv^2/R$.

Other standard cases: for a car on a convex bridge of radius $R$, $N = m(g - v^2/R)$, so it loses contact at $v = \sqrt{gR}$; in a dip, $N = m(g + v^2/R)$. In a rotor (wall of death), the wall's normal force $m\omega^2R$ supplies the centripetal force and friction supports the weight, so $\mu m\omega^2R \ge mg$ gives $\omega \ge \sqrt{g/(\mu R)}$.

Key points

Centripetal force is the name of the net radial force, not a new kind of force. Do not add it to an FBD that already contains the real forces.
Tangential force changes the speed; radial force changes the direction.
A banked road without friction has exactly one speed $\sqrt{rg\tan\theta}$ at which no friction is needed.
Conical pendulum: the bob's circle has radius $L\sin\theta$ and the period depends on the vertical depth $h = L\cos\theta$ of the plane of the circle below the point of suspension.

Radial equation: $\sum F_{inward} - \sum F_{outward} = \dfrac{mv^2}{r}$
Convex bridge, normal force: $N = m\left(g - \dfrac{v^2}{R}\right)$
Rotor minimum angular speed: $\omega_{min} = \sqrt{\dfrac{g}{\mu R}}$

Common traps

Centrifugal force does not act on a body in an inertial frame. It is a pseudo force that appears only in the rotating frame, and it should not be added to the centripetal force in the same equation.
On a banked road the normal force is not $mg$; it is $mg/\cos\theta$ when no friction acts, and the horizontal component of it is the centripetal force.
At the top of a vertical circle the centripetal force is $T + mg$ (both point down, toward the centre). At the bottom it is $T - mg$.

Shortcuts

Banking angle for no friction. Without friction $\tan\theta = v^2/(rg)$, which gives the angle directly. For $v = 20\,\text{m s}^{-1}$ and $r = 40\,\text{m}$ with $g = 10$, $\tan\theta = 1$ and $\theta = 45^\circ$.

Conical pendulum period. The period is $2\pi\sqrt{h/g}$ where $h = L\cos\theta$ is the vertical drop from the support to the plane of the circle. It does not depend on the mass.

Non-inertial frames and pseudo forces

Newton's laws hold in inertial frames. In a frame that accelerates with $\vec a_0$ relative to an inertial frame, a free body appears to accelerate with $-\vec a_0$. To keep $\vec F = m\vec a$ working in that frame, one adds a pseudo force $\vec F_{pseudo} = -m\vec a_0$ to the real forces on every body. It acts at the centre of mass, points opposite to the frame's acceleration, and has no reaction partner because no other body causes it.

The pseudo force is a bookkeeping device: the physics is the same as in the ground frame, and you may solve either way. In the ground frame, no pseudo force appears and the accelerations are the true ones. Mixing the two (adding a pseudo force and also using the ground-frame acceleration) double counts the effect.

The lift is the standard example. For a body of mass $m$ on a weighing scale in a lift accelerating upward at $a$, the scale reads the normal force $N = m(g + a)$, since in the lift frame the pseudo force $ma$ points down and adds to the weight. If the lift accelerates downward at $a$ the reading is $m(g - a)$, and at free fall ($a = g$) it is zero, which is apparent weightlessness. At constant velocity the reading is $mg$.

A simple pendulum in a car that accelerates at $a$ hangs at an angle with the vertical such that $\tan\theta = a/g$, leaning backward. In the car frame the bob feels gravity $mg$ downward and pseudo force $ma$ backward, so the effective gravity is $g_{eff} = \sqrt{g^2 + a^2}$ in that direction. The tension is $m g_{eff}$ and small oscillations have the period $2\pi\sqrt{L/g_{eff}}$. The same idea explains why a helium balloon in an accelerating car leans forward: the effective gravity points backward and downward, and the balloon rises opposite to it.

In a frame rotating with angular speed $\omega$, a body at distance $r$ from the axis feels a centrifugal pseudo force $m\omega^2r$ directed outward. A body that moves relative to the rotating frame also feels the Coriolis force, which JEE Main questions of this chapter do not need.

Key points

Pseudo force $= -m\vec a_{frame}$, applied on each body in the frame, at its centre of mass.
A frame is non-inertial if it accelerates (including rotating); a frame moving at constant velocity is inertial.
Effective gravity in an accelerating frame: $\vec g_{eff} = \vec g - \vec a_{frame}$. A plumb line points along it.
Lift reading $N = m(g + a)$ with $a$ taken upward; $N = 0$ in free fall.
A block on a smooth wedge that is pushed horizontally stays at rest relative to the wedge when the wedge acceleration is $a = g\tan\theta$.

Pseudo force: $\vec F_{pseudo} = -m\vec a_{frame}$
Pendulum in a horizontally accelerating frame: $\tan\theta = \dfrac{a}{g}, \qquad T = m\sqrt{g^2 + a^2}$
Wedge: block at rest relative to a smooth wedge: $a = g\tan\theta, \qquad F = (M + m)\,g\tan\theta$ (Smooth wedge and smooth ground; $F$ is the horizontal force on the wedge of mass $M$ carrying the block of mass $m$.)

Common traps

A pseudo force has no third-law reaction; there is no body that experiences an equal and opposite force.
Do not include a pseudo force when applying Newton's laws in an inertial frame (the ground).
A weighing scale reads the normal force, not the true weight. In free fall the reading is zero even though gravity still acts.
Centrifugal force acts outward in the rotating frame only. In the inertial frame the body's motion needs a net inward force, with no outward force at all.

Shortcuts

Replace g by an effective g. In a frame with acceleration $a$ along a direction, use $g_{eff} = \sqrt{g^2 + a^2}$ for horizontal $a$, or $g \pm a$ in a lift. Then the pendulum period is $2\pi\sqrt{L/g_{eff}}$.

Angle of a plumb line in an accelerating vehicle. The line deflects by $\tan\theta = a/g$, backward from the acceleration. For $a = 7.5\,\text{m s}^{-2}$ and $g = 10$ this is $\tan\theta = 0.75$, so $\theta \approx 37^\circ$.

Worked examples

A weight hung from two strings at 37 and 53 degrees

A body of weight $W = 100\,\text{N}$ hangs from the junction of two strings. The left string makes $37^\circ$ with the horizontal ceiling and the right string makes $53^\circ$ with it. Find the tension in each string (take $\sin 37^\circ = 0.6$, $\sin 53^\circ = 0.8$).

Draw the free-body diagram of the junction. The junction is in equilibrium under three forces: $W$ downward, $T_1$ up and to the left, $T_2$ up and to the right.
Horizontal balance. $T_1\cos 37^\circ = T_2\cos 53^\circ$, so $0.8\,T_1 = 0.6\,T_2$ and $T_2 = \tfrac43 T_1$.
Vertical balance. $T_1\sin 37^\circ + T_2\sin 53^\circ = W$. Substituting, $0.6\,T_1 + 0.8\cdot\tfrac43 T_1 = 100$, so $\tfrac{5}{3}T_1 = 100$ and $T_1 = 60\,\text{N}$.
The other tension. $T_2 = \tfrac43 \times 60 = 80\,\text{N}$.
Cross-check with Lami's theorem. The strings are at $37^\circ + 53^\circ = 90^\circ$ to each other. The angle between $T_2$ and $W$ is $143^\circ$ (and $\sin 143^\circ \approx 0.6$) and between $T_1$ and $W$ is $127^\circ$ ($\sin 127^\circ \approx 0.8$). Lami gives $T_1 = 100\times 0.6/1 = 60\,\text{N}$ and $T_2 = 100\times 0.8/1 = 80\,\text{N}$.

Answer: $T_1 = 60\,\text{N}$ (the string at $37^\circ$ to the horizontal) and $T_2 = 80\,\text{N}$ (the string at $53^\circ$).

A block released on a rough incline

A $5\,\text{kg}$ block is placed on a plane inclined at $37^\circ$ (take $\sin 37^\circ = 0.6$, $\cos 37^\circ = 0.8$, $g = 10\,\text{m s}^{-2}$). The coefficients of friction are $\mu_s = 0.5$ and $\mu_k = 0.25$. (a) Does it slide? (b) Find its acceleration and the time to slide $2\,\text{m}$ from rest. (c) What force parallel to the incline, directed up the slope, moves it up at constant speed?

Normal force and driving component. $N = mg\cos\theta = 5\times10\times0.8 = 40\,\text{N}$ and $mg\sin\theta = 5\times10\times0.6 = 30\,\text{N}$.
(a) Will it slide?. Limiting friction is $\mu_sN = 0.5\times40 = 20\,\text{N}$, which is less than $30\,\text{N}$. Equivalently $\tan 37^\circ = 0.75 > \mu_s = 0.5$. So it slides.
(b) Acceleration. Kinetic friction is $\mu_kN = 0.25\times40 = 10\,\text{N}$ up the slope. $ma = 30 - 10 = 20\,\text{N}$, so $a = 4\,\text{m s}^{-2}$ (the formula $g(\sin\theta - \mu_k\cos\theta) = 10(0.6 - 0.2) = 4$ agrees).
(b) Time for 2 m. $s = \tfrac12 at^2$ gives $2 = \tfrac12\times4\times t^2$, so $t = 1\,\text{s}$. The speed then is $at = 4\,\text{m s}^{-1}$.
(c) Pulling it up at constant speed. The friction now acts down the slope, and the net force is zero: $F = mg\sin\theta + \mu_kN = 30 + 10 = 40\,\text{N}$.

Answer: (a) Yes, it slides. (b) $a = 4\,\text{m s}^{-2}$ down the slope and $t = 1\,\text{s}$. (c) $F = 40\,\text{N}$ up the slope.

A block on a rough table pulled by a hanging mass

A $4\,\text{kg}$ block rests on a horizontal table with kinetic friction coefficient $0.1$. A light string passes over a smooth pulley at the edge and carries a $1\,\text{kg}$ mass hanging freely ($g = 10\,\text{m s}^{-2}$). Find the acceleration of the system and the tension in the string.

Check that it moves. The driving weight is $m_2g = 10\,\text{N}$. Taking the limiting friction as $\mu mg = 0.1\times4\times10 = 4\,\text{N}$, the $10\,\text{N}$ pull exceeds it, so the system accelerates.
Equation for the hanging mass. Taking downward as positive: $m_2g - T = m_2a$, that is $10 - T = a$.
Equation for the block. Along the table: $T - \mu m_1g = m_1a$, that is $T - 4 = 4a$.
Add the two equations. The tension cancels: $10 - 4 = 5a$, so $a = 1.2\,\text{m s}^{-2}$. This is the system shortcut, $a = (m_2 - \mu m_1)g/(m_1 + m_2)$.
Tension. $T = 10 - a = 8.8\,\text{N}$. Check with the block: $T - 4 = 4.8 = 4\times1.2$.

Answer: $a = 1.2\,\text{m s}^{-2}$ and $T = 8.8\,\text{N}$.

A movable pulley and the string constraint

A string passes over a fixed pulley. On one side it carries a mass $m_1 = 3\,\text{kg}$. The other end goes down around a movable massless pulley that carries a mass $m_2 = 4\,\text{kg}$, and then up to be fixed to the ceiling, as in the figure. Find the acceleration of each mass and the tension ($g = 10\,\text{m s}^{-2}$).

String constraint. The part of the string around the movable pulley has two vertical strands. If $m_2$ rises by $y$, both strands shorten by $y$, so $2y$ of string is released to the other side. That equals the distance $m_1$ falls, $x$. So $x = 2y$ and $a_2 = a_1/2$.
Forces on $m_1$. Taking downward as positive for $m_1$: $m_1g - T = m_1a_1$.
Forces on the movable pulley and $m_2$. The massless pulley with its load has two strands pulling up, $2T$, and the weight $m_2g$ down. Taking upward as positive: $2T - m_2g = m_2a_2 = m_2a_1/2$.
Solve. From the first, $T = m_1(g - a_1) = 3(10 - a_1)$. Substituting: $6(10 - a_1) - 40 = 2a_1$, so $20 = 8a_1$ and $a_1 = 2.5\,\text{m s}^{-2}$. The general form is $a_1 = 2g(2m_1 - m_2)/(4m_1 + m_2)$.
Other results. $a_2 = a_1/2 = 1.25\,\text{m s}^{-2}$ upward and $T = 3\times(10 - 2.5) = 22.5\,\text{N}$. Check: $2T - m_2g = 45 - 40 = 5\,\text{N} = 4\times1.25$.

Answer: $m_1$ moves down with $2.5\,\text{m s}^{-2}$, $m_2$ moves up with $1.25\,\text{m s}^{-2}$, and $T = 22.5\,\text{N}$.

Two blocks one above the other: when does the top one slip?

A $2\,\text{kg}$ block lies on an $8\,\text{kg}$ block, which rests on a smooth floor. The coefficient of friction between the blocks is $0.4$ (static and kinetic). A horizontal force $F$ acts on the lower block. (a) Find the largest $F$ for which the blocks move together. (b) For $F = 30\,\text{N}$ find the friction on the upper block. (c) For $F = 60\,\text{N}$ find both accelerations ($g = 10\,\text{m s}^{-2}$).

What accelerates the upper block. Only friction from the lower block acts horizontally on it, with a maximum of $\mu m g = 0.4\times2\times10 = 8\,\text{N}$. So its largest acceleration is $\mu g = 4\,\text{m s}^{-2}$.
(a) Condition to move together. Both must have $a \le 4$. The whole system of $10\,\text{kg}$ moves with $a = F/10$, so $F_{max} = 10\times4 = 40\,\text{N}$.
(b) F = 30 N. The common acceleration is $30/10 = 3\,\text{m s}^{-2}$. The upper block needs a force $2\times3 = 6\,\text{N}$, which is below the limit of $8\,\text{N}$. The friction on it is $6\,\text{N}$ (forward).
(c) F = 60 N, slipping. Since $60 > 40$ the upper block slips behind. Its acceleration is $a_{up} = \mu g = 4\,\text{m s}^{-2}$ (kinetic friction $8\,\text{N}$ forward).
(c) Lower block. The lower block feels $F$ forward and the friction $8\,\text{N}$ backward (third-law pair): $a_{low} = (60 - 8)/8 = 6.5\,\text{m s}^{-2}$.

Answer: (a) $F_{max} = 40\,\text{N}$. (b) $6\,\text{N}$ forward on the upper block. (c) Upper block $4\,\text{m s}^{-2}$, lower block $6.5\,\text{m s}^{-2}$.

Cutting the string above a spring

A block of mass $m_1 = 1\,\text{kg}$ hangs from the ceiling by a string. A spring connects it to a block of mass $m_2 = 3\,\text{kg}$ hanging below it. The system is at rest, and then the string is cut. Find the acceleration of each block just after the cut ($g = 10\,\text{m s}^{-2}$).

Before the cut. The spring supports $m_2$, so the spring force is $m_2g = 30\,\text{N}$ (stretched). The string supports both blocks: $T = (m_1 + m_2)g = 40\,\text{N}$.
At the instant of the cut. The tension vanishes immediately. The spring length has not changed yet, so the spring force remains $30\,\text{N}$.
Lower block. Forces on $m_2$: spring force $30\,\text{N}$ up and weight $30\,\text{N}$ down. The net force is zero, so $a_2 = 0$.
Upper block. The stretched spring pulls $m_1$ downward with $30\,\text{N}$, and its weight is $10\,\text{N}$ downward. The net force is $40\,\text{N}$ down, so $a_1 = 40/1 = 40\,\text{m s}^{-2}$ downward, which equals $(m_1 + m_2)g/m_1$.

Answer: Upper block $40\,\text{m s}^{-2}$ downward, lower block $0$.

Impulse of a ball striking a wall at an angle

A ball of mass $0.2\,\text{kg}$ moving at $10\,\text{m s}^{-1}$ strikes a smooth vertical wall, with its velocity at $60^\circ$ to the normal to the wall. It rebounds elastically with the same speed, and the contact lasts $0.02\,\text{s}$. Find the impulse and the average force on the ball.

Resolve the velocity. Perpendicular to the wall: $v_\perp = 10\cos60^\circ = 5\,\text{m s}^{-1}$. Along the wall: $v_\parallel = 10\sin60^\circ$.
Component along the wall. The wall is smooth, so no force acts along it, and $v_\parallel$ is unchanged. Its momentum change is zero.
Component along the normal. The ball reverses the normal component: from $+5$ to $-5\,\text{m s}^{-1}$ (elastic). $\Delta p = m(-5 - 5) = 0.2\times(-10) = -2\,\text{kg m s}^{-1}$.
Impulse and force. The magnitude of the impulse is $J = 2\,\text{N s}$, directed along the normal, away from the wall. $F_{avg} = J/\Delta t = 2/0.02 = 100\,\text{N}$.

Answer: Impulse $2\,\text{N s}$ perpendicular to the wall (away from it); average force $100\,\text{N}$.

Safe speed range on a banked curve with friction

A road is banked with $\tan\theta = 3/4$ and has radius $40\,\text{m}$. The coefficient of friction between the tyres and the road is $0.5$. Find the speed at which no friction is needed, and the largest and smallest speeds at which a car can take the curve ($g = 10\,\text{m s}^{-2}$).

No friction needed. $v^2 = rg\tan\theta = 40\times10\times0.75 = 300$, so $v \approx 17.3\,\text{m s}^{-1}$ (about $62\,\text{km h}^{-1}$).
Maximum speed: friction acts down the slope. With $\sin\theta = 0.6$ and $\cos\theta = 0.8$, vertical balance: $N\cos\theta - \mu N\sin\theta = mg$, so $N(0.8 - 0.3) = mg$ and $N = 2mg$.
Maximum speed: radial equation. $N\sin\theta + \mu N\cos\theta = mv^2/r$, so $N(0.6 + 0.4) = 2mg = mv^2/r$. Then $v_{max}^2 = 2gr = 800$ and $v_{max} = 20\sqrt{2} \approx 28.3\,\text{m s}^{-1}$.
Minimum speed: friction acts up the slope. Vertical balance: $N\cos\theta + \mu N\sin\theta = mg$, so $N(0.8 + 0.3) = mg$ and $N = mg/1.1$.
Minimum speed: radial equation. $N\sin\theta - \mu N\cos\theta = mv^2/r$, so $0.2N = mv^2/r$. $v_{min}^2 = 0.2\times rg/1.1 = 80/1.1 \approx 72.7$ and $v_{min} \approx 8.5\,\text{m s}^{-1}$.
Check with the formulas. $v_{max}^2 = rg(\tan\theta + \mu)/(1 - \mu\tan\theta) = 400\times1.25/0.625 = 800$ and $v_{min}^2 = 400\times0.25/1.375 \approx 72.7$, in agreement.

Answer: $v_{opt} \approx 17.3\,\text{m s}^{-1}$, $v_{max} \approx 28.3\,\text{m s}^{-1}$, $v_{min} \approx 8.5\,\text{m s}^{-1}$.

A pendulum in an accelerating car and in a lift

A bob of mass $0.4\,\text{kg}$ hangs from the roof of a car that accelerates forward at $7.5\,\text{m s}^{-2}$ ($g = 10\,\text{m s}^{-2}$). (a) Find the angle the string makes with the vertical and the tension, when the bob is at rest relative to the car. (b) A person of $60\,\text{kg}$ stands on a scale in a lift that accelerates upward at $2\,\text{m s}^{-2}$. What does the scale read?

Car frame: forces on the bob. Real forces: tension $T$ along the string and weight $mg$. Pseudo force: $ma$ backward (opposite to the car's acceleration). The bob is at rest in this frame, so these three balance.
Angle. Horizontal: $T\sin\theta = ma$. Vertical: $T\cos\theta = mg$. Dividing, $\tan\theta = a/g = 0.75$, so $\theta \approx 37^\circ$, with the bob displaced backward.
Tension. $T = m\sqrt{g^2 + a^2} = 0.4\sqrt{100 + 56.25} = 0.4\times12.5 = 5\,\text{N}$.
Check in the ground frame. The bob accelerates forward at $7.5\,\text{m s}^{-2}$, so the horizontal component of $T$ is $ma = 3\,\text{N}$ and the vertical component is $mg = 4\,\text{N}$. The resultant is $\sqrt{9 + 16} = 5\,\text{N}$, in agreement.
(b) Lift. In the lift frame the pseudo force is $ma$ downward. The scale reading is $N = m(g + a) = 60\times12 = 720\,\text{N}$. If the lift accelerated downward at the same rate, the reading would be $60\times8 = 480\,\text{N}$.

Answer: (a) $\theta \approx 37^\circ$ from the vertical (backward) and $T = 5\,\text{N}$. (b) $720\,\text{N}$.

Figure: A block on an incline of angle theta with arrows for weight mg, normal force N, friction f up the slope, and dashed components mg sin theta and mg cos theta.

Figure: Atwood machine: two masses m1 and m2 on a string over a pulley, with tension arrows on each side and acceleration arrows (m1 down, m2 up).

Figure: Block m1 on a table joined by a string over a pulley at the table edge to a hanging mass m2, with tension and acceleration arrows.

Figure: A string over a fixed pulley with mass m1 on one end; the other end goes round a movable pulley carrying m2 and is fixed to the ceiling.

Figure: A weight W hung from the junction of two strings at 37 and 53 degrees to the horizontal ceiling, with tension arrows T1 and T2.

Figure: A car on a banked road of angle theta, with weight mg downward, normal force N perpendicular to the road, and an arrow toward the centre of the circle.

Figure: A pendulum bob in an accelerating car leaning backward at angle theta, with tension T, weight mg and a dashed pseudo force ma pointing backward.

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