Torque & Static Equilibrium

A force is not the whole story

Push on a door right next to its hinge and it barely moves. Push with the same hand, the same effort, out at the handle, and it swings. The force is the same. What changed is where you put it.

That is the gap this lesson closes. Newton's second law tells you how a body accelerates as a whole, and it does not care where on the body you push. But a real object is not a point. A wrench, a ladder, a shelf bracket, a plank across two sawhorses: these things can turn, and whether they turn depends on distance as much as on force. The last lesson, angular kinematics, gave you the language of turning: angle $\theta$, angular velocity $\omega$, angular acceleration $\alpha$. Now you need the cause. The cause is called torque.

Torque: force times the distance that counts

Start with the door. You are pushing with force $F$ at a distance $r$ from the hinge. Double $F$ and you clearly double the turning effect. Double $r$ and experience says you also double it: a long wrench beats a short one by exactly the ratio of their lengths. So the turning effect is proportional to the product $rF$.

But only when you push perpendicular to the door. Push along the door, straight towards the hinge, and nothing turns at all, no matter how hard you shove. So only the component of $F$ perpendicular to the line from the pivot does any turning. If the force makes an angle $\theta$ with that line, the perpendicular component is $F\sin\theta$. Putting the two pieces together:

$\tau = rF\sin\theta$

This is the torque, symbol $\tau$ (the Greek letter tau), measured in newton metres (N·m). Here $r$ is the distance from the pivot to the point where the force is applied, $F$ is the size of the force, and $\theta$ is the angle between the direction of $\vec r$ and the direction of $\vec F$.

Check the extremes. At $\theta = 90^\circ$, $\sin\theta = 1$ and you get the maximum, $\tau = rF$. At $\theta = 0$ or $180^\circ$, $\sin\theta = 0$ and the torque vanishes: a force pointing straight at or straight away from the pivot never turns anything.

There is a second way to group the same three numbers, and in practice it is faster. Instead of resolving the force, extend the force's line of action (the infinite straight line the arrow lies along) and measure the perpendicular distance from the pivot to that line. Call it $r_\perp$. A little geometry gives $r_\perp = r\sin\theta$, so

$\tau = r_\perp F$

That perpendicular distance is called the lever arm (or moment arm). Drawing it is often a one-second job on a sketch where computing $\theta$ would take a minute. It also makes the zero cases obvious: if the line of action passes through the pivot, the lever arm is zero, and so is the torque.

Torque needs a sign, because a body can be turned either way. Pick one sense as positive and stick with it for the whole problem. The standard choice is counterclockwise positive, clockwise negative. Nothing physical depends on the choice; only consistency matters.

Two conditions for staying put

An object that is not moving and not starting to move must satisfy two separate requirements, and beginners routinely use only the first.

The first is the one you already know. If the centre of mass is not accelerating, the forces must cancel:

$\sum \vec F = 0$

The second is new. If the body is not starting to spin, the torques must cancel too:

$\sum \tau = 0$

These really are independent. Imagine two equal and opposite forces applied at opposite ends of a rod: one up on the left, one down on the right. The vector sum is exactly zero, so the centre of mass stays put. The rod still spins. That arrangement is called a couple, and it is the proof that force balance alone is not enough.

In two dimensions the two conditions unpack into three scalar equations: $\sum F_x = 0$, $\sum F_y = 0$, and $\sum \tau = 0$. Three equations means you can find at most three unknowns. If a problem hands you four unknown forces with no extra condition, it is not hard, it is statically indeterminate, and no amount of cleverness with a rigid body will fix it.

One more fact does most of the work in this topic. If a body is in equilibrium, the net torque about it is zero about every point, not merely about the real hinge. That is not obvious, but it follows quickly: if you shift the pivot by a vector $\vec d$, every torque changes by an amount involving $\vec d$ crossed with the individual force, and those extra terms add up to $\vec d$ crossed with $\sum \vec F$, which is zero. So the total is unchanged.

You are therefore free to put the pivot anywhere you like. Use that freedom ruthlessly: choose the pivot at the point where an unknown you do not care about is applied. That unknown gets a zero lever arm and drops out of the equation entirely.

Weight acts at the centre of mass

Gravity pulls on every atom of an object, so strictly there are millions of tiny forces. For torque purposes you can replace all of them with a single force $mg$ acting at the centre of mass. For a uniform bar, plank, ladder or rod, that is the geometric middle.

This one line turns an impossible problem into a one-liner, and forgetting it is the single most common mistake in equilibrium. The weight of a beam is not at the hinge and not at the far end. It is at the middle.

Worked example 1: a plank on two supports

A uniform plank of mass $30\ \mathrm{kg}$ and length $5.0\ \mathrm{m}$ rests on supports at each end. A person of mass $70\ \mathrm{kg}$ stands $1.5\ \mathrm{m}$ from the left support. Find the force each support pushes up with. Use $g = 9.8\ \mathrm{m/s^2}$.

First, the sketch. Four forces: the left support pushes up with $N_L$ at $x = 0$; the right support pushes up with $N_R$ at $x = 5.0\ \mathrm{m}$; the plank's weight $30 \times 9.8 = 294\ \mathrm{N}$ pulls down at the middle, $x = 2.5\ \mathrm{m}$; the person's weight $70 \times 9.8 = 686\ \mathrm{N}$ pulls down at $x = 1.5\ \mathrm{m}$.

You want $N_R$, so take torques about the left support, where $N_L$ has zero lever arm. Everything here acts vertically on a horizontal plank, so each lever arm is just the horizontal distance. Counting counterclockwise as positive, the upward $N_R$ at the right turns the plank counterclockwise about the left end, and the two weights turn it clockwise:

$N_R(5.0) - (294)(2.5) - (686)(1.5) = 0$

$5.0\,N_R = 735 + 1029 = 1764$

$N_R = \frac{1764}{5.0} = 352.8\ \mathrm{N}$

Now the vertical force balance gives the other one for free:

$N_L + N_R = 294 + 686 = 980\ \mathrm{N}$

$N_L = 980 - 352.8 = 627.2\ \mathrm{N}$

Sanity check: the person stands nearer the left support, so the left support should carry more. It does. Notice also that you could have found $N_L$ by taking torques about the right support instead, and got the same number: a free way to check your arithmetic.

This also runs in reverse. If a beam of unknown shape sits on two scales $4.0\ \mathrm{m}$ apart reading $180\ \mathrm{N}$ and $300\ \mathrm{N}$, the total weight is $480\ \mathrm{N}$, and torques about the left scale give $480\,x = 300 \times 4.0$, so the centre of mass sits at $x = 1200/480 = 2.5\ \mathrm{m}$ from the left end. Scales locate the centre of mass.

Worked example 2: a boom with a cable, and the hinge force

A uniform horizontal boom weighing $180\ \mathrm{N}$ and $4.0\ \mathrm{m}$ long is hinged to a wall at its left end. A cable runs from the far end up to the wall, making $40^\circ$ with the boom. A $320\ \mathrm{N}$ load hangs from the far end. Find the horizontal component of the force the hinge exerts.

The hinge force is an unknown in both magnitude and direction, so write it as two components, $H_x$ and $H_y$. That is three unknowns in total with the cable tension $T$, and you have exactly three equations.

Take torques about the hinge. Both hinge components vanish there. The cable pulls at $40^\circ$ at a distance $4.0\ \mathrm{m}$, so its torque is $T(4.0)\sin 40^\circ$, counterclockwise. The boom's weight acts down at $2.0\ \mathrm{m}$, and the load acts down at $4.0\ \mathrm{m}$, both clockwise:

$T(4.0)\sin 40^\circ - (180)(2.0) - (320)(4.0) = 0$

$\sin 40^\circ = 0.6428$, so $4.0 \times 0.6428 = 2.571$:

$2.571\,T = 360 + 1280 = 1640$

$T = \frac{1640}{2.571} = 637.9\ \mathrm{N}$

Now horizontal forces. The cable pulls the boom's far end back towards the wall, with horizontal component $T\cos 40^\circ$. The only other horizontal force is the hinge, so it must pull the other way with the same size:

$H_x = T\cos 40^\circ = 637.9 \times 0.766 = 488.6\ \mathrm{N}$

Note how big that is: larger than either the boom's weight or the load. A shallow cable angle forces an enormous tension, and the hinge has to match its horizontal pull. That is why brackets pull out of walls.

Worked example 3: the ladder

This is the standard hard problem, and once you have done one you have done them all.

A uniform ladder of mass $15\ \mathrm{kg}$ and length $5.0\ \mathrm{m}$ leans at $65^\circ$ to the horizontal against a frictionless wall. The coefficient of static friction with the floor is $\mu_s = 0.40$. How far up can a $75\ \mathrm{kg}$ person climb before the ladder slips?

Four forces act. The wall is frictionless, so it can only push horizontally: call that $N_w$, at the top. The floor pushes up with $N_f$ and rubs horizontally with friction $f$, both at the bottom. The ladder's weight $15g = 147\ \mathrm{N}$ acts at the middle, $2.5\ \mathrm{m}$ along. The person's weight $75g = 735\ \mathrm{N}$ acts at distance $d$ along the ladder, which is what you want.

Vertical balance: the floor carries everything, since the wall pushes horizontally only.

$N_f = 147 + 735 = 882\ \mathrm{N}$

Horizontal balance: friction is the only thing opposing the wall's push, so $f = N_w$.

At the point of slipping, friction is at its maximum:

$f_{\max} = \mu_s N_f = 0.40 \times 882 = 352.8\ \mathrm{N}$

so $N_w = 352.8\ \mathrm{N}$ at that instant.

Now take torques about the foot of the ladder, killing both $N_f$ and $f$ at once. For the wall force at the top, the height is $L\sin 65^\circ$, and the force is horizontal, so the lever arm is that height. For the weights, which point down, the lever arm is the horizontal distance out from the foot: $2.5\cos 65^\circ$ for the ladder, $d\cos 65^\circ$ for the person. With $\sin 65^\circ = 0.9063$ and $\cos 65^\circ = 0.4226$:

$N_w (5.0)(0.9063) = (147)(2.5)(0.4226) + (735)\,d\,(0.4226)$

Left side: $352.8 \times 4.532 = 1598.8$. First term on the right: $147 \times 1.0566 = 155.3$. So

$1598.8 - 155.3 = 735 \times 0.4226 \times d = 310.6\,d$

$d = \frac{1443.5}{310.6} = 4.65\ \mathrm{m}$

The person can climb about $4.6\ \mathrm{m}$ up a $5.0\ \mathrm{m}$ ladder before it goes. Climb further and the required friction exceeds what the floor can supply.

The same setup answers the other classic question. Take the empty ladder: mass $m$, length $L$, angle $\phi$. Then $N_f = mg$, $f = N_w$, and torques about the foot give $N_w L\sin\phi = mg(L/2)\cos\phi$, so $N_w = \tfrac{1}{2}mg\cot\phi$. The required friction coefficient is

$\mu_{\min} = \frac{f}{N_f} = \frac{N_w}{mg} = \frac{1}{2}\cot\phi$

Everything cancels: not the mass, not the length. At $\phi = 60^\circ$, $\cot 60^\circ = 0.5774$, so $\mu_{\min} = 0.2887$. Run it the other way and $\tan\phi = 1/(2\mu)$ gives the shallowest angle a given floor will hold. With $\mu = 0.30$: $\tan\phi = 1.667$, $\phi = 59.0^\circ$. With $\mu = 0.50$: $\tan\phi = 1.0$, $\phi = 45^\circ$ exactly.

Tipping, and the edge of the base

A crate does not need to slide to fall over. It can tip. Both failures come out of the same two equilibrium conditions, and which one happens first is a comparison.

An object tips about the edge of its base that it is rotating over. Right at the tipping point, the whole normal force has migrated to that edge and the weight's line of action passes exactly through it. So: an object is on the verge of tipping when its centre of mass is directly above the edge of its base.

That solves the plank question immediately. A uniform $4.0\ \mathrm{m}$ plank on a single support tips when its centre of mass, at the $2.0\ \mathrm{m}$ mark, passes the support, so it can hang over by $2.0\ \mathrm{m}$.

Now the comparison. Take a uniform cube of side $a$ and weight $W$ on a floor with coefficient $\mu_s$, pushed horizontally with force $P$ at the top edge. Sliding starts when $P = \mu_s W$. Tipping starts when the torque about the front bottom edge balances: the push acts at height $a$, the weight at horizontal distance $a/2$ from that edge, so $Pa = W(a/2)$, giving $P = W/2$. The two thresholds are equal when $\mu_s W = W/2$, that is $\mu_s = 0.5$. Below that the crate slides first; above it, it tips.

Stacking uses the same centre-of-mass rule repeatedly. With identical bricks of length $\ell$ at a table edge, the top brick may overhang the one below it by $\ell/2$. That pair, treated as one body, has its centre of mass $\ell/4$ from the edge of the second brick, so it may overhang by $\ell/4$. The three-brick group gives $\ell/6$, and so on. Four bricks of length $0.20\ \mathrm{m}$ give a total overhang of

$\frac{\ell}{2} + \frac{\ell}{4} + \frac{\ell}{6} = 0.10 + 0.05 + 0.0333 = 0.1833\ \mathrm{m}$

beyond the edge of the bottom brick, and the bottom brick itself can be pushed $\ell/8 = 0.025\ \mathrm{m}$ past the table, for $0.2083\ \mathrm{m}$ in total. The series never converges, which is why you can in principle overhang a table by any distance you like with enough bricks.

How to know this is the problem type

Reach for torque and static equilibrium when the problem has all three of these marks:

• Nothing is moving, or something is on the verge of moving. Words like "at rest", "just about to slip", "maximum load before", "minimum angle".
• The object has extent, and forces act at different places on it. A beam, a ladder, a plank, a bracket, a crate, a wrench. If everything acts at a single point, plain force balance is enough and there is no torque to write.
• You are asked for a force or a distance, not a speed or a time. Support reactions, hinge forces, cable tensions, how far along, how far up, what angle.

The procedure is always the same four steps. Draw the body alone and mark every force at the point where it acts, including the weight at the centre of mass. Choose the pivot at whichever unknown you want gone. Write $\sum \tau = 0$ about that pivot and solve. Then, if you need more, use $\sum F_x = 0$ and $\sum F_y = 0$ to mop up the rest. When a problem says "on the verge of slipping", add the single extra fact $f = \mu_s N$; when it says "on the verge of tipping", add instead that the normal force sits at the edge.

Everything so far has been about bodies that refuse to turn. Next you take the brakes off and ask what happens when the torques do not cancel, which means asking how hard a given body is to spin up, and that is moment of inertia.