10.1 Torque as the Rotational Effect of Force

Same Force, Different Effect

Try something right now. Stand up, walk to a door, and push it closed by pressing near the hinge. Then open it again by pushing near the handle.

You used the same hand. You pushed with roughly the same force. But one push swung the door easily, and the other barely moved it.

This is not a subtle effect. Pushing near the hinge feels almost useless. Pushing near the handle feels effortless. The difference is dramatic --- and it has nothing to do with how hard you push.

Something about where and how you push determines how effective the force is at producing rotation.

Before you read on: You push a door with 10 N of force. First, you push at the handle, which is 0.8 m from the hinge. Then you push at a point halfway between the hinge and the handle, 0.4 m from the hinge. Both pushes are perpendicular to the door.

Is the rotational effect of the second push half as much, twice as much, or the same as the first?

Commit to an answer before continuing.

What Makes a Force Good at Turning?

You already have strong physical intuition about this, even if you have never formalized it. Think about the following everyday situations:

Opening a jar lid. You grip at the outer edge of the lid, not near the center. Why? Because pushing at the edge is more effective at turning the lid.

Using a wrench. You grab the end of the wrench handle, as far from the bolt as possible. A longer wrench makes the job easier --- not because you push harder, but because you push farther from the axis.

A seesaw. A small child can balance a heavier adult by sitting farther from the pivot. The child's weight has a greater turning effect because of the greater distance.

In every case, the same pattern appears: distance from the rotation axis matters. More distance means more rotational effect.

But distance is not the whole story. There is a second factor, and it is just as important.

Imagine pushing on that door again, this time at the handle. Push perpendicular to the door --- it swings easily. Now push at a steep angle, almost along the length of the door, toward the hinge. The door barely moves. Push exactly along the door, directly toward the hinge, and the door does not rotate at all.

Same force magnitude. Same distance from the hinge. But the angle of the push changed everything.

Pause and think: Based on the door and wrench examples, what two geometric factors seem to determine how effective a force is at producing rotation?

Exploring Torque

[Interactive: Door Torque Explorer. A top-down view of a door attached to a hinge on the left side. The student can click anywhere on the door to set the point where the force is applied, and drag an arrow to set the force direction and magnitude. A torque meter on the side displays the resulting torque in real time, both as a number and as a bar graph.

The interactive has two guided modes:

Mode 1: Vary the angle. The force application point is fixed at the handle (0.8 m from the hinge), and the force magnitude is fixed at 10 N. The student can only change the angle of the force. The torque meter updates continuously. Prompts appear: - "Push perpendicular to the door. What does the torque meter read?" - "Now angle your push to 45 degrees from the door. What happened to the torque?" - "Push directly along the door, toward the hinge. What is the torque now?" - "Push directly along the door, away from the hinge. What is the torque?" - "At what angle is the torque maximum?"

Mode 2: Vary the distance. The force is fixed at 10 N, perpendicular to the door. The student drags the application point along the door from hinge to handle. The torque meter updates. Prompts appear: - "Apply the force at the handle (0.8 m). Record the torque." - "Move to the midpoint (0.4 m). What is the torque now?" - "Move to one-quarter of the way (0.2 m). What is the torque?" - "Push at the hinge itself (0 m). What happens?" - "Is the relationship between distance and torque linear?"

Mode 3: Free exploration. All parameters are adjustable. The student can test any combination of distance, angle, and force magnitude.]

If you worked through those guided explorations carefully, you discovered two clean relationships:

1. When you vary the angle (keeping distance and force magnitude fixed), the torque is largest when the force is perpendicular to the door and zero when the force points directly along the door, toward or away from the hinge. The torque varies smoothly between these extremes.

2. When you vary the distance (keeping force magnitude and angle fixed), the torque is directly proportional to the distance from the hinge. Half the distance gives half the torque. Zero distance gives zero torque.

These are not approximate trends. They are exact mathematical relationships.

The Torque Equation

The quantity that measures the rotational effect of a force is called torque. It depends on three things:

• The magnitude of the force, $F$
• The distance from the rotation axis to the point where the force is applied, $r$
• The angle $\theta$ between the position vector $\vec{r}$ (pointing from the axis to the application point) and the force vector $\vec{F}$

The torque is:

$$\tau = rF\sin\theta$$

Let's verify this against what you discovered in the exploration.

Angle dependence. When the force is perpendicular to the door, $\theta = 90°$ and $\sin 90° = 1$, so $\tau = rF$. Maximum torque. When the force is along the door, $\theta = 0°$ (or $180°$) and $\sin 0° = 0$, so $\tau = 0$. No torque. This matches exactly.

Distance dependence. When you double $r$, you double $\tau$. When $r = 0$ (pushing at the hinge), $\tau = 0$ regardless of the force. This also matches exactly.

Now go back to the opening prediction. You pushed with $F = 10$ N, perpendicular to the door ($\theta = 90°$).

At the handle: $\tau = (0.8)(10)\sin 90° = 8 \text{ N}!\cdot!\text{m}$.

At the midpoint: $\tau = (0.4)(10)\sin 90° = 4 \text{ N}!\cdot!\text{m}$.

The rotational effect is exactly half as much. If you predicted "half," your intuition is well-calibrated.

The unit of torque is the newton-meter ($\text{N}!\cdot!\text{m}$). Note that this is the same combination of units as energy (joules), but torque and energy are different physical quantities. We write $\text{N}!\cdot!\text{m}$ rather than $\text{J}$ to keep them distinct.

Why Only the Perpendicular Component Matters

The $\sin\theta$ factor in $\tau = rF\sin\theta$ has a clean geometric meaning. You can decompose any force into two components relative to the position vector $\vec{r}$:

• A parallel component, $F_\parallel = F\cos\theta$, which points along the line from the axis to the application point
• A perpendicular component, $F_\perp = F\sin\theta$, which points at right angles to that line

The torque formula then says:

$$\tau = rF\sin\theta = r \cdot F_\perp$$

Only the perpendicular component produces torque. The parallel component pushes directly toward or away from the axis, and pushing toward or away from the axis cannot make something rotate.

Think about it physically. If you push a door directly toward the hinge, you are compressing the door into the hinge --- not rotating it. If you push directly away from the hinge, you are pulling the door out of the hinge --- still not rotating it. Only the sideways component, the part that acts perpendicular to the line from the axis, produces a turning effect.

This is a fundamental geometric fact about rotation: forces along the radial direction do not contribute to rotation. Only the tangential component does.

Connection to What You Already Know

In Chapter 5, you learned that force causes translational acceleration: $F = ma$. The equation is clean and direct. If you push harder, the object accelerates more. The geometry of where you push does not enter the equation --- any force on a point particle accelerates the whole thing.

But a door is not a point particle. It is an extended object that can rotate around a fixed axis. For extended objects, the story is richer. The same force can have very different rotational effects depending on two geometric factors: how far from the axis you apply it, and at what angle.

Torque is the quantity that captures this richer story. It is the rotational analog of force --- the thing that causes changes in rotational motion. And the new ingredient, the one that did not appear in $F = ma$, is geometry.

Translational motion	Rotational motion
Force $F$	Torque $\tau$
Depends on: force magnitude	Depends on: force magnitude, distance from axis, angle
$F = ma$	$\sum \tau = I\alpha$ (coming in Section 10.5)
Push harder $\to$ more acceleration	Push harder, farther out, more perpendicular $\to$ more angular acceleration

This analogy will deepen throughout the chapter. By Section 10.5, you will see the full rotational version of Newton's second law: $\sum \tau = I\alpha$. But that equation needs torque as its input --- and now you know what torque is.

Torque in Action: Three Examples

Before we move to practice problems, let's ground the formula in a few concrete situations.

Example 1: Tightening a bolt with a wrench.

You apply a 50 N force at the end of a 0.3 m wrench, perpendicular to the wrench handle. The torque about the bolt is:

$$\tau = rF\sin\theta = (0.3)(50)\sin 90° = 15 \text{ N}!\cdot!\text{m}$$

Now suppose the wrench is in a tight space, and you can only push at $60°$ to the handle. The torque drops to:

$$\tau = (0.3)(50)\sin 60° = (0.3)(50)(0.866) = 13.0 \text{ N}!\cdot!\text{m}$$

The force is the same, the distance is the same, but the less favorable angle reduces the torque by about 13%.

Example 2: A lever.

You press down on the end of a 1.5 m crowbar with 200 N of force, perpendicular to the bar. The pivot is at the far end.

$$\tau = (1.5)(200)\sin 90° = 300 \text{ N}!\cdot!\text{m}$$

If the crowbar were only 0.5 m long, the torque would be 100 N$\cdot$m --- one-third as much. This is why longer levers are more effective. The ancient Egyptians and Greeks understood this principle: a longer lever arm amplifies the rotational effect of a force.

Example 3: A force applied at the axis.

You try to rotate a wheel by pushing at its center axle with 100 N. Since $r = 0$:

$$\tau = (0)(100)\sin\theta = 0$$

No matter how hard you push, no matter at what angle, you cannot produce torque by pushing at the rotation axis. The distance $r$ must be nonzero for a force to have a rotational effect.

Practice

Layer 1: Concrete

A mechanic applies a 40 N force to the end of a 0.25 m wrench. Compute the torque about the bolt for each of the following angles between the force and the wrench handle:

(a) $\theta = 90°$ (perpendicular)

(b) $\theta = 60°$

(c) $\theta = 30°$

(d) $\theta = 0°$ (along the handle)

Check your answer

Using $\tau = rF\sin\theta$ with $r = 0.25$ m and $F = 40$ N: (a) $\tau = (0.25)(40)\sin 90° = (0.25)(40)(1) = 10 \text{ N}\!\cdot\!\text{m}$ (b) $\tau = (0.25)(40)\sin 60° = (0.25)(40)(0.866) = 8.66 \text{ N}\!\cdot\!\text{m}$ (c) $\tau = (0.25)(40)\sin 30° = (0.25)(40)(0.5) = 5.0 \text{ N}\!\cdot\!\text{m}$ (d) $\tau = (0.25)(40)\sin 0° = (0.25)(40)(0) = 0$ Notice the pattern: as the angle decreases from 90° toward 0°, the torque decreases. At 90°, you get the full effect of the force. At 0°, the force is entirely along the wrench and produces no rotation at all.

Layer 2: Pattern

Five forces are applied to the same door, which is 1.0 m wide (measured from the hinge). Rank the torques from greatest to least.

Case	Force (N)	Distance from hinge (m)	Angle $\theta$
A	20	1.0	90°
B	40	0.5	90°
C	10	1.0	90°
D	20	1.0	30°
E	30	0.8	45°

Check your answer

Computing $\tau = rF\sin\theta$ for each case: | Case | $\tau = rF\sin\theta$ | Torque (N$\cdot$m) | |:---|:---|:---| | A | $(1.0)(20)(1)$ | 20 | | B | $(0.5)(40)(1)$ | 20 | | C | $(1.0)(10)(1)$ | 10 | | D | $(1.0)(20)(0.5)$ | 10 | | E | $(0.8)(30)(0.707)$ | 17.0 | Ranking from greatest to least: **A = B > E > C = D**. Notice that A and B produce the same torque through different combinations of force and distance. And C and D produce the same torque --- C with a small force at a favorable angle, D with a larger force at an unfavorable angle. Torque depends on all three factors together, not any one in isolation.

Layer 3: Structure

Why does only the perpendicular component of force produce torque? Give both a mathematical and a physical explanation.

Check your answer

**Mathematical explanation:** The torque formula is $\tau = rF\sin\theta$. Since $F_\perp = F\sin\theta$, this is equivalent to $\tau = rF_\perp$. The parallel component $F_\parallel = F\cos\theta$ does not appear in the torque. The $\sin\theta$ factor mathematically filters out the parallel part and keeps only the perpendicular part. **Physical explanation:** The parallel component of force points along the line from the axis to the application point --- either directly toward the axis or directly away from it. A force pointing toward the axis pushes the object into the axis. A force pointing away from the axis pulls the object out from the axis. Neither of these directions tends to rotate the object around the axis. Only a force that acts sideways --- perpendicular to the radial line --- pushes the object in a circular arc around the axis, which is what rotation means. You can feel this with a door. Pushing the door toward the hinge compresses it; pushing it away from the hinge pulls it outward. Neither rotates it. Only pushing sideways (perpendicular to the door) swings it open or closed.

Layer 4: Debug

A student is asked to compute the torque produced by a 25 N force applied at 0.6 m from the axis, at an angle of $\theta = 60°$ to the position vector. The student writes:

$$\tau = rF = (0.6)(25) = 15 \text{ N}!\cdot!\text{m}$$

The student used the full force magnitude without the $\sin\theta$ factor.

(a) What is the correct torque?

(b) When does this error produce the right answer?

Check your answer

(a) The correct torque is: $$\tau = rF\sin\theta = (0.6)(25)\sin 60° = (0.6)(25)(0.866) = 13.0 \text{ N}\!\cdot\!\text{m}$$ The student's answer of 15 N$\cdot$m is too large by about 15%. (b) The error $\tau = rF$ gives the correct answer **only when $\theta = 90°$**, because $\sin 90° = 1$, so $rF\sin 90° = rF$. This is the special case where the force is perpendicular to the position vector. Many textbook problems are set up this way (forces perpendicular to lever arms) precisely because it simplifies the arithmetic. The danger is that students practice this special case so often that they forget the general formula includes $\sin\theta$. In any real-world situation where the force is not perfectly perpendicular --- which is most of the time --- omitting $\sin\theta$ gives the wrong answer. The error always *overestimates* the torque (since $\sin\theta \leq 1$ for all angles), which means the student would predict a larger rotational effect than actually occurs.

Reflection

Think about your daily life. What experiences have already given you intuition about torque --- even if you did not use that word?

You might think about: opening doors, turning steering wheels, using wrenches or screwdrivers, pedaling a bicycle, turning a faucet handle, or even the way you instinctively grab a heavy suitcase by the handle rather than near its body.

In each case, ask yourself: what is the axis, where is the force applied, and why does the geometry matter?

Looking Ahead

You now know the torque formula: $\tau = rF\sin\theta$. It tells you the magnitude of the rotational effect of a force, and it depends on three factors --- force, distance, and angle.

But the formula $rF\sin\theta$ can be interpreted in more than one way. You can think of it as $r$ times the perpendicular component of the force ($F_\perp = F\sin\theta$), which is what we emphasized here. Or you can think of it as $F$ times a quantity called the lever arm ($r_\perp = r\sin\theta$), which is the perpendicular distance from the axis to the line of action of the force. These two viewpoints give the same number, but each is more natural in different geometric situations.

And there is a third perspective: the cross product $\vec{\tau} = \vec{r} \times \vec{F}$, which packages both the magnitude and the direction of torque into a single vector operation.

In the next section, you will see all three interpretations side by side, learn when each one is most useful, and understand why they all agree. The cross product will also reveal something we have not yet discussed: torque has a direction, and that direction tells you which way the object tends to rotate.