Chapter 4 · Laws of Motion

4.3 Equilibrium of a Particle, Common Forces and Friction

Draw every force on one body, add them as vectors, and set the sum to zero (equilibrium) or to $m\mathbf{a}$. Weight, the normal force, tension, and friction — static up to $\mu_sN$, kinetic at $\mu_kN$.

A painting hangs from two wires; a climber rests on a rope; a parked car stays put on a steep street; a heavy wardrobe refuses to move until you push really hard, then suddenly slides. None of these is accelerating (until the wardrobe moves), yet several forces act on each. Engineers check every bridge, crane and lift cable by the same simple test: the forces on each part must add up to zero. And the force that keeps the car parked and lets you walk at all — friction — has a peculiar rule of its own: it is as large as it needs to be, up to a limit.

This section gives you the working tools of mechanics. First, the free-body diagram, which isolates one body and shows every force on it. Second, the condition for equilibrium of a particle. Third, a catalogue of the forces you will meet in almost every problem: weight, the normal force, tension and friction. With these you can analyse a block on a slope, a crate being dragged, or a lamp hung from two cords — and in Section 4.4, cars on curves and bodies tied together by strings.

Free-body diagrams and equilibrium of a particle

📘 Definition 4.5 — Free-body diagram

A free-body diagram shows one chosen body on its own, drawn as a dot or a simple shape, with an arrow for every force acting on it — and only those forces. Each arrow starts on the body, points in the direction of the force, and is labelled ($m\mathbf{g}$, $\mathbf{N}$, $\mathbf{T}$, $\mathbf{f}$, …). Forces that the body exerts on other things are not drawn.

📐 Result 4.6 — Equilibrium of a particle

A particle is in equilibrium (at rest, or moving with constant velocity) if and only if the net force on it is zero:

$$\mathbf{F}_1+\mathbf{F}_2+\dots+\mathbf{F}_n=\mathbf{0}\iff\sum F_x=0,\quad\sum F_y=0,\quad\sum F_z=0 .$$

For two forces this means they are equal and opposite. For three forces in a plane it means that, drawn head to tail, they form a closed triangle; each force is equal and opposite to the resultant of the other two. (Applies to a particle, or to a body whose size we can ignore; rigid bodies also need a torque condition — Chapter 6.)

🧭 Problem-solving strategy — Free-body diagrams
  1. Isolate the body you are interested in; sketch it alone.
  2. List the forces: its weight $mg$ (always, straight down); then go round the boundary and add one force for each thing that touches it — a surface (normal force, and friction if rough), a string (tension, pulling away from the body), a hand (a push or pull).
  3. Choose axes to make the algebra easy — usually along and perpendicular to the acceleration or the surface.
  4. Resolve every force into components (Section 3.2) and write $\sum F_x=ma_x$, $\sum F_y=ma_y$ (with $a=0$ for equilibrium).
  5. Solve, then check signs, units and a limiting case.
✏️ Example 1 — A lamp hung by two cords
A $10\ \text{kg}$ lamp hangs at rest from two light cords attached to a ceiling. The left cord makes $30^\circ$ with the ceiling and the right cord makes $60^\circ$ with it. Find the tension in each cord.
FBD On the knot: weight $98\ \text{N}$ down; $T_1$ up and to the left at $30^\circ$ above the horizontal; $T_2$ up and to the right at $60^\circ$ above the horizontal.
x $T_2\cos60^\circ=T_1\cos30^\circ$, so $T_2=\sqrt3\,T_1$.
y $T_1\sin30^\circ+T_2\sin60^\circ=98$: $T_1\left(\tfrac12+\tfrac32\right)=98$, so $T_1=49\ \text{N}$ and $T_2=49\sqrt3=84.9\ \text{N}$.
Check The steeper cord carries more of the load. The two tensions add to $134\ \text{N}$, more than the weight — tensions are vectors, and only their vertical components share the $98\ \text{N}$.
✏️ Example 2 — Pulling a hanging mass aside
A $6.0\ \text{kg}$ bob hangs from a rope. A horizontal force $F$ pulls it aside and holds it at rest with the rope at $30^\circ$ to the vertical. Find $F$ and the tension in the rope.
Equations Vertical: $T\cos30^\circ=mg=58.8\ \text{N}$. Horizontal: $T\sin30^\circ=F$.
Tension $T=\dfrac{58.8}{\cos30^\circ}=67.9\ \text{N}$.
Force $F=T\sin30^\circ=mg\tan30^\circ=33.9\ \text{N}$.
Check Triangle of forces: $\sqrt{58.8^2+33.9^2}=67.9\ \text{N}$ ✓. As the angle approaches $90^\circ$, $F$ and $T$ grow without limit — a rope can never be pulled perfectly horizontal while holding a weight.

Weight, normal force and tension

The weight $W=mg$ of a body is the gravitational pull of the Earth on it, directed vertically down. All the other everyday forces are contact forces: electrical forces between the atoms of two surfaces in contact, which we describe by their components.

📘 Definition 4.6 — Normal force and tension

When two surfaces touch, the component of the contact force perpendicular to the surfaces is the normal force $N$ (normal reaction); it pushes the surfaces apart and is as large as needed to stop one body moving into the other. The component parallel to the surfaces is friction.

A taut string, rope or cable pulls on the bodies at its ends with a force called tension $T$, directed along the string and away from the body. For an ideal string (massless and inextensible) the tension is the same all along it, and it passes unchanged over an ideal (massless, frictionless) pulley; only its direction changes.

The normal force is not always equal to $mg$. It equals $mg$ only for a body on a horizontal surface with no other vertical forces and no vertical acceleration. Push down on a box and $N$ grows; pull up on it and $N$ shrinks; tilt the surface and $N$ becomes $mg\cos\theta$.

✏️ Example 3 — A block on a smooth slope
A $2.0\ \text{kg}$ block is on a smooth (frictionless) slope inclined at $30^\circ$ to the horizontal. (a) Find the normal force and the block's acceleration if it is released. (b) What tension in a string parallel to the slope would hold it at rest?
Axes Take $x$ down the slope and $y$ perpendicular to it. Resolve the weight $mg=19.6\ \text{N}$: $mg\sin30^\circ=9.8\ \text{N}$ along $x$; $mg\cos30^\circ=17.0\ \text{N}$ into the slope.
(a) No acceleration perpendicular to the slope: $N=mg\cos30^\circ=17.0\ \text{N}$. Along the slope: $ma=mg\sin30^\circ$, so $a=g\sin30^\circ=4.9\ \text{m/s}^2$ down the slope.
(b) Equilibrium along the slope: $T=mg\sin30^\circ=9.8\ \text{N}$.
Check $\theta=0$: $a=0$, $N=mg$; $\theta=90^\circ$: $a=g$, $N=0$ — free fall beside a wall ✓.

Friction: static, kinetic and rolling

Push gently on a heavy wardrobe and nothing happens. The floor must be pushing back on it, sideways, with a force exactly equal and opposite to your push — otherwise the wardrobe would accelerate. Push harder, and this force grows to match. Only when your push passes a certain limit does the wardrobe start to slide; then you find that a slightly smaller push keeps it moving. This sideways contact force is friction. It comes from the microscopic bumps of the two surfaces catching and from the atoms at the points of real contact sticking to each other.

📘 Definition 4.7 — Static and kinetic friction

Static friction $f_s$ acts between surfaces that are not sliding over each other. It opposes the tendency to slide and takes whatever value (up to a maximum) is needed to prevent sliding. Its maximum value is called limiting friction.

Kinetic (sliding) friction $f_k$ acts between surfaces that are sliding over each other, opposing their relative motion.

The dimensionless constants $\mu_s$ and $\mu_k$ in the laws below are the coefficients of static and kinetic friction; they depend on the pair of surfaces (and their condition), not on the size of the body.

📐 Law 4.7 — Laws of friction

For two dry surfaces pressed together by a normal force $N$:

$$f_s\le\mu_sN\qquad(\text{static: adjusts itself, up to the limit }f_s^{\max}=\mu_sN),$$ $$f_k=\mu_kN\qquad(\text{kinetic: while sliding}),$$

with $\mu_k\lt\mu_s$ for most pairs of surfaces. To a good approximation both are independent of the area of contact, and $f_k$ is nearly independent of the sliding speed. These are empirical laws, valid for dry, unlubricated surfaces and moderate speeds.

The inequality is the part students most often forget. If a $10\ \text{kg}$ box with $\mu_s=0.50$ sits on a floor and you push it sideways with $20\ \text{N}$, the friction is $20\ \text{N}$, not $\mu_sN=49\ \text{N}$ — a $49\ \text{N}$ friction would push the box towards you! Use $f=\mu_sN$ only when you know the body is on the point of sliding.

Angle of repose. Tilt a board with a block on it. The block stays put while static friction balances $mg\sin\theta$, which requires $mg\sin\theta\le\mu_s\,mg\cos\theta$, i.e. $\tan\theta\le\mu_s$. The largest such angle is the angle of repose.

📐 Result 4.8 — Angle of repose

A block on a rough incline stays at rest for all angles up to the angle of repose $\theta_r$, where

$$\tan\theta_r=\mu_s .$$

Beyond $\theta_r$ it slides down with acceleration $a=g(\sin\theta-\mu_k\cos\theta)$. It slides at constant velocity (once started) on a slope with $\tan\theta=\mu_k$. Neither result depends on the mass.

Rolling friction. A wheel or ball that rolls without slipping does not slide at the point of contact, so there is no kinetic friction there. A small resistance still arises because the wheel and the ground deform slightly where they touch; this rolling friction is typically a hundred times smaller than sliding friction. That is why the wheel, ball bearings and roller conveyors are such important inventions. Lubricants work by replacing solid–solid contact with a thin fluid layer.

Friction is not only a nuisance. Static friction from the ground pushes you forward when you walk, drives a car forward through its tyres, and stops it when braking; it lets you grip a pen and keeps a knot tied. On ice, with friction nearly zero, none of these work.

Interactive 1 — When Does the Block Start to Move?

A $5.0\ \text{kg}$ block rests on a rough floor ($\mu_s=0.50$, $\mu_k=0.40$). Drag the slider to set the horizontal pull, or press Play to make the pull grow steadily. While the block is at rest, static friction (blue) exactly matches the pull. At $\mu_sN=24.5\ \text{N}$ it breaks free; friction drops to $\mu_kN=19.6\ \text{N}$ and the block accelerates. The graph traces friction against the pull.

15 N
pull $P$—
friction $f$—
limit $\mu_sN$—
acceleration—
✏️ Example 4 — Moving a crate
A $40\ \text{kg}$ crate rests on a level floor with $\mu_s=0.50$ and $\mu_k=0.35$. (a) What friction acts if you push horizontally with $100\ \text{N}$? (b) What horizontal push will just start it moving? (c) If you keep pushing with that force once it moves, what is its acceleration? (d) What push keeps it moving at constant speed?
Normal force $N=mg=40\times9.8=392\ \text{N}$; $\mu_sN=196\ \text{N}$, $\mu_kN=137\ \text{N}$.
(a) $100\ \text{N}\lt196\ \text{N}$, so the crate stays put and static friction is $100\ \text{N}$ (opposite to the push).
(b), (c) It starts when the push exceeds $196\ \text{N}$. Once sliding: $a=\dfrac{196-137.2}{40}=1.47\ \text{m/s}^2$.
(d) Constant speed needs zero net force: push $=\mu_kN=137\ \text{N}$.
Check The jolt as a heavy object "breaks free" is the drop from $196\ \text{N}$ to $137\ \text{N}$ of friction — the same drop you see in Interactive 1.
✏️ Example 5 — A block on a rough slope
A $2.0\ \text{kg}$ block rests on a plank with $\mu_s=0.70$ and $\mu_k=0.40$. (a) The plank is at $30^\circ$. Does the block slide? What is the friction on it? (b) Find the angle of repose. (c) The plank is raised to $40^\circ$. Find the block's acceleration.
(a) $\tan30^\circ=0.577\lt0.70$: it stays at rest. Friction balances the downhill pull: $f_s=mg\sin30^\circ=19.6\times0.5=9.8\ \text{N}$ up the slope. (The limit is $\mu_smg\cos30^\circ=11.9\ \text{N}$; friction is below it.)
(b) $\theta_r=\tan^{-1}0.70=35.0^\circ$.
(c) $40^\circ\gt35.0^\circ$, so it slides: $a=g(\sin40^\circ-\mu_k\cos40^\circ)=9.8(0.643-0.306)=3.30\ \text{m/s}^2$ down the slope.
Check Part (a) is the default of Interactive 2; set the angle to $40^\circ$ there to see the block slide with $3.30\ \text{m/s}^2$.
✏️ Example 6 — Pulling a sled at an angle
A child pulls a $20\ \text{kg}$ sled across level snow with a rope at $30^\circ$ above the horizontal. The tension is $100\ \text{N}$ and $\mu_k=0.20$. Find the normal force, the friction and the acceleration.
Vertical $N+T\sin30^\circ=mg$: $N=196-50=146\ \text{N}$ — less than $mg$, because the rope lifts a little.
Friction $f_k=\mu_kN=0.20\times146=29.2\ \text{N}$.
Horizontal $T\cos30^\circ-f_k=ma$: $a=\dfrac{86.6-29.2}{20}=2.87\ \text{m/s}^2$.
Check Pushing down on the sled at $30^\circ$ instead would raise $N$ to $246\ \text{N}$ and friction to $49.2\ \text{N}$, giving only $1.87\ \text{m/s}^2$: it is easier to pull a roller or a sled than to push it.
Interactive 2 — Free-Body Diagram on an Incline

A block sits on a rough incline; a string parallel to the slope can pull it uphill. All forces on the block are drawn to the same scale: weight $m\mathbf{g}$ (black), normal force $\mathbf{N}$ (green), tension $\mathbf{T}$ (purple) and friction $\mathbf{f}$ (blue). The dashed red arrow is the net force. Friction is computed exactly as Law 4.7 says: static and adjusting while the required value is below $\mu_sN$, otherwise kinetic and opposing the sliding. Press Play to watch a sliding block move.

30°2 kg0.700.400 N
normal force $N$—
friction—
net force—
state—
⚠️ Common mistakes

1. Always writing $f=\mu_sN$. Static friction is only as large as needed. Use $\mu_sN$ only at the point of slipping (Example 4(a), Example 5(a)).

2. Always writing $N=mg$. On a slope $N=mg\cos\theta$; with a rope pulling up at an angle or a hand pushing down, $N$ changes (Example 6). Find $N$ from the perpendicular equation every time.

3. Friction "always opposes motion". It opposes relative sliding (or its tendency) between the two surfaces. When you walk, static friction on your foot points forward, in the direction you move.

4. Drawing forces that act on other bodies. A free-body diagram contains only forces on the chosen body. "The force of the block on the slope" does not belong in the block's diagram.

5. Adding a "force of motion". A sliding block has no forward force on it unless something pushes it; its velocity is not a force (Aristotle again).

Exercises

Take $g=9.8\ \text{m/s}^2$. Strings are light and inextensible.

I
Equilibrium of a particle.
  • 1 Forces of $3.0\ \text{N}$ along $+x$ and $4.0\ \text{N}$ along $+y$ act on a particle. Find the third force that keeps it in equilibrium.
  • 2 A $4.0\ \text{kg}$ lamp hangs from two cords, each at $45^\circ$ to the vertical. Find the tension in each.
  • 3 A $5.0\ \text{kg}$ mass on a string is pulled aside by a horizontal force of $20\ \text{N}$ and held at rest. Find the angle of the string with the vertical and its tension.
  • 4 A $1.0\ \text{kg}$ wet shirt hangs from the middle of a clothes line, which sags so that each half makes $10^\circ$ with the horizontal. Find the tension in the line. Why can a line holding a weight never be perfectly straight?
  • 5 A particle is in equilibrium under three forces: $6.0\ \text{N}$ east, $8.0\ \text{N}$ north and a third. Find the third.
  • 6 A $3.0\ \text{kg}$ block rests on a smooth $25^\circ$ incline, held by a string parallel to the slope. Find the tension and the normal force.
  • 7 A $60\ \text{kg}$ gymnast hangs at rest from two rings. Find the tension in each rope if the ropes are (a) vertical, (b) each at $15^\circ$ to the vertical.
1 $\mathbf{F}_3=(-3.0\,\hat{\mathbf{i}}-4.0\,\hat{\mathbf{j}})\ \text{N}$: $5.0\ \text{N}$ at $53.1^\circ$ below the $-x$ axis ($233.1^\circ$ from $+x$).
2 $2T\cos45^\circ=39.2$: $T=27.7\ \text{N}$.
3 $\tan\alpha=\dfrac{20}{49}$: $\alpha=22.2^\circ$; $T=\sqrt{49^2+20^2}=52.9\ \text{N}$.
4 $2T\sin10^\circ=9.8$: $T=28.2\ \text{N}$, almost three times the weight. A straight line would need $2T\sin0^\circ=mg$, i.e. infinite tension.
5 $10\ \text{N}$ directed $53.1^\circ$ south of west (opposite to the resultant of the other two).
6 $T=mg\sin25^\circ=29.4\times0.423=12.4\ \text{N}$; $N=mg\cos25^\circ=26.6\ \text{N}$.
7 (a) $T=\dfrac{588}{2}=294\ \text{N}$; (b) $2T\cos15^\circ=588$, $T=304\ \text{N}$.
II
Weight, normal force and tension.
  • 1 Find the weight of a $50\ \text{kg}$ astronaut on the Earth and on the Moon ($g_\text{Moon}=1.62\ \text{m/s}^2$). Does her mass change?
  • 2 A $10\ \text{kg}$ box rests on a level floor. Find the normal force when a hand pushes straight down on it with $30\ \text{N}$.
  • 3 Find the normal force on the same box when a rope pulls straight up on it with $30\ \text{N}$.
  • 4 A $10\ \text{kg}$ block is released on a smooth $37^\circ$ incline. Find the normal force and the acceleration.
  • 5 A block starts from rest at the top of a smooth $30^\circ$ incline $2.0\ \text{m}$ long. Find its speed at the bottom.
  • 6 A $20\ \text{kg}$ bucket is hauled up a well with an upward acceleration of $2.0\ \text{m/s}^2$. Find the tension in the rope.
1 Earth: $490\ \text{N}$; Moon: $50\times1.62=81\ \text{N}$. Her mass stays $50\ \text{kg}$.
2 $N=98+30=128\ \text{N}$.
3 $N=98-30=68\ \text{N}$.
4 $N=98\cos37^\circ=78.3\ \text{N}$; $a=g\sin37^\circ=5.90\ \text{m/s}^2$ down the slope.
5 $a=4.9\ \text{m/s}^2$; $v=\sqrt{2\times4.9\times2.0}=4.43\ \text{m/s}$.
6 $T-mg=ma$: $T=20(9.8+2.0)=236\ \text{N}$.
III
Friction.
  • 1 A $10\ \text{kg}$ box rests on a floor with $\mu_s=0.40$. Find the least horizontal force that will start it moving.
  • 2 With $\mu_k=0.30$, what horizontal force keeps the box moving at constant speed?
  • 3 The box is at rest and you push it horizontally with $20\ \text{N}$. Find the friction force.
  • 4 Once the box is sliding, you push with $50\ \text{N}$. Find its acceleration.
  • 5 A car moving at $15\ \text{m/s}$ brakes hard with its wheels locked; $\mu_k=0.50$. Find its deceleration and stopping distance.
  • 6 Find the angle of repose for a block with $\mu_s=0.36$.
  • 7 A block, once started, slides down a $30^\circ$ incline at constant speed. Find $\mu_k$.
  • 8 A $1.0\ \text{kg}$ book is pressed against a vertical wall by a horizontal push $P$; $\mu_s=0.50$. Find the least $P$ that stops it sliding down.
  • 9 Explain why it is easier to pull a heavy garden roller by its handle than to push it.
1 $\mu_smg=0.40\times98=39.2\ \text{N}$ (just over this).
2 $\mu_kmg=0.30\times98=29.4\ \text{N}$.
3 $20\ \text{N}$ (static, below the $39.2\ \text{N}$ limit), opposite to the push.
4 $a=\dfrac{50-29.4}{10}=2.06\ \text{m/s}^2$.
5 $a=\mu_kg=4.9\ \text{m/s}^2$; $s=\dfrac{v^2}{2a}=\dfrac{225}{9.8}=23\ \text{m}$.
6 $\theta_r=\tan^{-1}0.36=19.8^\circ$.
7 Constant speed: $mg\sin30^\circ=\mu_kmg\cos30^\circ$, so $\mu_k=\tan30^\circ=0.58$.
8 $N=P$ and friction must hold the weight: $\mu_sP\ge mg$, so $P\ge\dfrac{9.8}{0.50}=19.6\ \text{N}$.
9 Pulling along a handle tilted upward has an upward component that reduces $N$ and hence friction; pushing along a handle tilted downward adds a downward component that increases $N$ and friction (Example 6).
IV
Harder problems and applications.
  • 1 A $25\ \text{kg}$ trunk is dragged across a floor by a rope at $37^\circ$ above the horizontal with tension $120\ \text{N}$; $\mu_k=0.25$. Find its acceleration.
  • 2 A block is projected up a $30^\circ$ incline at $10\ \text{m/s}$; $\mu_k=0.25$ and $\mu_s=0.30$. How far up does it go? Does it then slide back, and if so with what acceleration?
  • 3 What force parallel to a $30^\circ$ incline ($\mu_k=0.20$) moves a $5.0\ \text{kg}$ block up it at constant speed? What force parallel to the incline lets it slide down at constant speed?
  • 4 A box sits on the flat bed of a truck; $\mu_s=0.40$ between box and bed. What is the greatest acceleration the truck can have without the box sliding?
  • 5 (Modelling) A car at $25\ \text{m/s}$ brakes to rest. With anti-lock brakes the tyres keep gripping, so static friction with $\mu_s=0.80$ acts; with locked wheels kinetic friction with $\mu_k=0.60$ acts. Compare the stopping distances.
  • 6 Explain why ball bearings reduce friction in a machine, and why tyres need a tread pattern on wet roads but racing cars use smooth tyres on dry tracks.
1 $N=245-120\sin37^\circ=172.8\ \text{N}$; $f_k=43.2\ \text{N}$; $a=\dfrac{120\cos37^\circ-43.2}{25}=\dfrac{95.8-43.2}{25}=2.1\ \text{m/s}^2$.
2 Going up, gravity and friction both act down the slope: $a=g(\sin30^\circ+0.25\cos30^\circ)=7.02\ \text{m/s}^2$; distance $=\dfrac{100}{2\times7.02}=7.1\ \text{m}$. At the top, $\tan30^\circ=0.577\gt\mu_s=0.30$, so it slides back with $a=g(\sin30^\circ-0.25\cos30^\circ)=2.78\ \text{m/s}^2$.
3 Up: $P=mg(\sin30^\circ+\mu_k\cos30^\circ)=49(0.5+0.173)=33.0\ \text{N}$. Down at constant speed (holding back): $P=49(0.5-0.173)=16.0\ \text{N}$ up the slope.
4 Only static friction accelerates the box: $\mu_smg=ma$, so $a_\text{max}=\mu_sg=3.92\ \text{m/s}^2$.
5 ABS: $a=7.84\ \text{m/s}^2$, $s=\dfrac{625}{15.68}=39.9\ \text{m}$. Locked: $a=5.88\ \text{m/s}^2$, $s=53.1\ \text{m}$ — over $13\ \text{m}$ more, and a skidding car cannot be steered.
6 Bearings replace sliding with rolling, whose friction is far smaller. On wet roads the tread channels water away so rubber touches the road; on a dry track a smooth tyre maximises the real contact area and grip.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Distinguish between static and kinetic friction. Define the angle of repose and show that it is related to the coefficient of static friction by $\tan\theta_r=\mu_s$.

Distinction Static friction acts when there is no sliding; it is self-adjusting, $f_s\le\mu_sN$. Kinetic friction acts during sliding; $f_k=\mu_kN$, constant and usually smaller. [1]
Definition The angle of repose is the largest angle of an inclined plane at which a body placed on it just remains at rest. [1]
Proof At $\theta_r$ friction is limiting: along the plane $mg\sin\theta_r=\mu_sN$; perpendicular, $N=mg\cos\theta_r$. Dividing, $\tan\theta_r=\mu_s$. [1]
Exam tip Draw the free-body diagram — it carries marks, and shows clearly that $N=mg\cos\theta$, not $mg$.

E2. NEET A $2.0\ \text{kg}$ block lies on a horizontal floor with $\mu_s=0.50$. A horizontal force of $5.0\ \text{N}$ is applied to it. The friction force on the block is (a) $9.8\ \text{N}$ (b) $5.0\ \text{N}$ (c) zero (d) $14.8\ \text{N}$

Limit $\mu_smg=0.50\times19.6=9.8\ \text{N}\gt5.0\ \text{N}$, so the block does not move.
Equilibrium Static friction balances the applied force: $5.0\ \text{N}$.
Answer (b).
Exam tip (a) is the most common wrong answer: $\mu_sN$ is a maximum, not the actual value.

E3. JEE Main A block rests on an incline of angle $37^\circ$ ($\sin37^\circ=0.6$, $\cos37^\circ=0.8$). The least coefficient of static friction that keeps it from sliding is (a) $0.60$ (b) $0.75$ (c) $0.80$ (d) $1.33$

Condition $mg\sin\theta\le\mu_smg\cos\theta$, so $\mu_s\ge\tan37^\circ=\dfrac{0.6}{0.8}=0.75$.
Answer (b).
Exam tip (d) is $\cot37^\circ$ — the ratio upside down. A quick check: steeper slopes need larger $\mu_s$, and $\tan$ grows with the angle.

E4. JEE Main (Numerical value) A $5.0\ \text{kg}$ block is held against a vertical wall by a horizontal force $F$. The coefficient of static friction between block and wall is $0.50$. Taking $g=10\ \text{m/s}^2$, find the minimum $F$ (in N) that keeps the block from sliding down.

FBD Horizontally $N=F$; vertically friction (up) balances the weight $50\ \text{N}$.
Limit $\mu_sF\ge50$, so $F\ge\dfrac{50}{0.50}=100\ \text{N}$.
Answer $100$.
Exam tip Here the normal force is the push, not the weight. With a larger push the friction stays $50\ \text{N}$ — it does not grow beyond what is needed.

E5. JEE Advanced (One or more correct) A $2.0\ \text{kg}$ block is at rest on a rough horizontal floor with $\mu_s=0.50$ and $\mu_k=0.40$; take $g=10\ \text{m/s}^2$. A horizontal force $F$ is applied to it. Which statements are correct? (A) For $F=8.0\ \text{N}$, the friction on the block is $8.0\ \text{N}$. (B) For $F=12\ \text{N}$, the block accelerates at $2.0\ \text{m/s}^2$. (C) For $F=10\ \text{N}$, the block remains at rest. (D) For $F=12\ \text{N}$, the friction on the block is $10\ \text{N}$.

Numbers $N=20\ \text{N}$: $\mu_sN=10\ \text{N}$, $\mu_kN=8.0\ \text{N}$.
(A) $8.0\lt10$: at rest, static friction $8.0\ \text{N}$ ✓.
(B), (D) $12\gt10$: it slides, friction is kinetic, $8.0\ \text{N}$; $a=\dfrac{12-8.0}{2.0}=2.0\ \text{m/s}^2$. (B) ✓, (D) ✗.
(C) $F=\mu_sN$ exactly: friction can just supply $10\ \text{N}$, so it stays at rest (on the verge of sliding) ✓.
Answer (A), (B), (C).

E6. Board · Case study Anti-lock braking systems (ABS) stop a car's wheels from locking, so that the tyres keep rolling and gripping the road instead of skidding. A $1200\ \text{kg}$ car is moving at $20\ \text{m/s}$ on a dry road where $\mu_s=0.80$ and $\mu_k=0.60$. (i) Find the stopping distance if the wheels lock. (2 marks) (ii) Find the shortest stopping distance with ABS. (1 mark) (iii) Give one other advantage of preventing the wheels from locking. (1 mark)

(i) Skidding: $a=\mu_kg=5.88\ \text{m/s}^2$ [1]; $s=\dfrac{v^2}{2a}=\dfrac{400}{11.76}=34.0\ \text{m}$ [1].
(ii) Rolling tyres use static friction up to $\mu_sg=7.84\ \text{m/s}^2$: $s=\dfrac{400}{15.68}=25.5\ \text{m}$. [1]
(iii) The driver can still steer, because the rolling tyres can also provide sideways friction; a skidding car slides wherever its momentum takes it. [1]
Exam tip The mass does not appear: both friction and $ma$ are proportional to $m$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. How slip-resistant are everyday shoes on everyday floors — and is a wet floor really as dangerous as the warning signs suggest?

Why it matters. Slips and falls are one of the commonest causes of injury at work and in hospitals worldwide. Floor-safety engineers measure a floor's slip resistance as a friction coefficient; a value of about $0.5$ or more is widely treated as safe for walking, while values below about $0.3$ are hazardous. Shoe makers test soles in the same way. You can measure $\mu_s$ with nothing more than a board and a protractor, using the angle of repose.

What to do.

  1. Put a shoe (weighted with a small book inside) on a smooth board covered with a sample of floor material. Raise one end slowly until the shoe just starts to slide; measure the angle $\theta_r$ (or the height and length of the board: $\tan\theta_r=h/\sqrt{L^2-h^2}$). Repeat five times and average.
  2. Find $\mu_s=\tan\theta_r$ for at least three shoes on at least two surfaces (tile, wood, carpet, …), dry and with a little water or soapy water spread on the surface.
  3. Check one result another way: drag the shoe on the level surface with a spring balance and use $\mu_s=F_\text{max}/mg$.
  4. Does the added weight in the shoe change $\theta_r$? What does Law 4.7 predict?

Tools: Plank, protractor or ruler, shoes, floor samples, water, spring balance; a phone inclinometer app helps.

Hand in: A results table of $\mu_s$ (mean and spread) for each shoe–floor pair, dry and wet, and a short safety note for a school canteen: which floor and footwear combinations would you flag?

Going further. Walking needs friction of roughly a fifth to a quarter of body weight at heel strike. Using your values, decide which combinations would make you slip, and explain why people take shorter steps on ice.

Instructor note: typical results are $\theta_r\approx30^\circ$–$40^\circ$ ($\mu_s\approx0.6$–$0.8$) for rubber soles on dry wood or tile, falling to $12^\circ$–$20^\circ$ ($\mu_s\approx0.2$–$0.35$) on wet glazed tile — below the $0.5$ guideline. Adding weight should leave $\theta_r$ unchanged (mass cancels in $\tan\theta_r=\mu_s$); small differences show the limits of the empirical law for soft rubber.

🧪 Self-check quiz

0/5 correct

Q1. A book rests on a horizontal table and nothing pushes it sideways. The friction force on it is

Q2. A $10\ \text{kg}$ box on a floor has $\mu_s=0.60$. A horizontal push of $30\ \text{N}$ is applied. The friction on the box is

Q3. A block just begins to slide when an incline is raised to $30^\circ$. The coefficient of static friction is

Q4. For most pairs of surfaces,

Q5. A block of mass $m$ rests on a rough incline of angle $\theta$. The normal force on it is

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