Mistake Master
Mistake Master
A force pushes; a torque turns. How much a force rotates an object depends on three things: the force itself ($F$), the distance from the axis to where it's applied ($r$), and the angle between them ($\theta$). Combined, $\tau = rF\sin\theta$. Drop any one factor to zero and the torque is zero, no matter how hard you push. The trap: a bigger force doesn't always mean a bigger torque.
Not every force makes an object rotate. Pull on a doorknob, the door swings open. Pull with the same force on the hinge, nothing happens. Same force, different result. The difference is where the force is applied and how it is aimed.
Torque is the rotational version of force. A net force changes how an object moves through space ($\vec F_\text{net} = m\vec a$). A net torque changes how an object rotates. The two are not the same quantity, and treating them as one is the most common mistake in this unit.
For a single force, the magnitude of the torque about a chosen axis is
$$\tau = rF\sin\theta,$$Here $r$ is the distance from the axis to where the force is applied, $F$ is the magnitude of the force, and $\theta$ is the angle between $\vec r$ (the line from axis to application point) and $\vec F$.
Three factors. Drop any one to zero and $\tau$ goes to zero. A 1000 N push right at the hinge ($r = 0$) gives zero torque. A force aimed straight at the axis ($\theta = 0$) gives zero torque too: it pushes the object, but it doesn't twist it. The question is never "how hard?" alone, but "how hard, where, and at what angle?"
Two more rules to flag up front. First, $\tau$ is always taken about a chosen axis. "The torque of this force" is not a number until you name the axis. Second, in this course $\tau$ is a signed scalar: a single number whose sign tells you CW versus CCW. The sign convention is yours to pick; we'll use CCW positive throughout.
The torque from a single force can be written three ways. All three give the same number; they just group the factors differently. Real problems hand you whichever pieces are easiest to read off, so all three are worth recognizing:
Only the component of $\vec F$ perpendicular to $\vec r$ produces torque. The component parallel to $\vec r$ pushes the object toward the axis or away from it, but it doesn't rotate it.
Worked example. A rigid bar is pivoted at its left end. A force of magnitude $F = 50$ N is applied at a point $r = 0.30$ m from the pivot, at an angle $\theta = 60^\circ$ to the bar. Find the magnitude of the torque about the pivot.
Method 1, full formula.
$$\tau = rF\sin\theta = (0.30)(50)\sin 60^\circ = (0.30)(50)(0.866) \approx 13.0\,\text{N}\cdot\text{m}.$$Method 2, lever-arm form. The perpendicular distance from the pivot to the force's line of action is
$$r_\perp = r\sin\theta = (0.30)(0.866) = 0.260\,\text{m},$$and so
$$\tau = r_\perp F = (0.260)(50) = 13.0\,\text{N}\cdot\text{m}.$$Method 3, perpendicular-component form. The component of the force perpendicular to the bar is
$$F_\perp = F\sin\theta = (50)(0.866) = 43.3\,\text{N},$$and so
$$\tau = rF_\perp = (0.30)(43.3) = 13.0\,\text{N}\cdot\text{m}.$$Three forms, three routes, one answer. Use whichever fits the figure in front of you.
Units: $r$ in meters, $F$ in newtons, $\tau$ in newton-meters ($\text{N}\cdot\text{m}$). Torque has the same SI unit as energy ($1\,\text{J} = 1\,\text{N}\cdot\text{m}$), but the two are different quantities. Always write torque in $\text{N}\cdot\text{m}$, never $\text{J}$.
When more than one force acts on an object, what rotates it is the net torque: the signed sum of the individual torques. To take a signed sum you have to pick a positive direction first. Throughout this course, CCW (counter-clockwise) is positive.
$$\Sigma\tau = \tau_1 + \tau_2 + \tau_3 + \dots$$Each $\tau_i$ gets its own sign: $+$ if that force would rotate the object CCW about the chosen axis, $-$ if CW. Once every torque has a sign, add them like ordinary signed numbers.
Worked example. A horizontal beam is pivoted at its center. A $6$ N weight hangs at $1.0$ m to the left of the pivot. A $3$ N weight hangs at $2.0$ m to the right of the pivot. What is the net torque about the pivot?
Left weight. A downward force on the left side rotates the beam CCW about the pivot. With CCW positive, this torque is positive:
$$\tau_L = +(1.0\,\text{m})(6\,\text{N}) = +6\,\text{N}\cdot\text{m}.$$Right weight. A downward force on the right side rotates the beam CW. With CCW positive, this torque is negative:
$$\tau_R = -(2.0\,\text{m})(3\,\text{N}) = -6\,\text{N}\cdot\text{m}.$$Sum.
$$\Sigma\tau = \tau_L + \tau_R = (+6) + (-6) = 0\,\text{N}\cdot\text{m}.$$Net torque is zero. A bigger force at a shorter arm exactly cancels a smaller force at a longer arm. The beam balances.
Two more points the lab below makes concrete. First, the axis matters: the same forces give a different $\Sigma\tau$ about a different axis. "The torque" with no axis named is not a number. Second, the sign convention is yours to pick, but use it consistently: once CCW is positive, every torque in the problem follows that rule.
Three traps in this topic. Each one is a real student sentence; each fix is the rule that catches it.
The intuition is fair: push harder, more should happen. For straight-line motion, that's right; double the force and you double the acceleration. For rotation, it's wrong. A $50$ N push right at a door's hinge gives zero torque about the hinge, no matter how hard you push. A $5$ N push at the doorknob beats it cleanly. Force alone is half a torque; the other half is the lever arm.
Fix. Compute $\tau = rF\sin\theta$ in full. If you find yourself comparing torques by comparing forces, stop and check the lever arms. Big force at a tiny lever arm can produce a smaller torque than a small force at a long lever arm.
This works when the force happens to be perpendicular to the bar ($\theta = 90^\circ$, so $\sin\theta = 1$). Textbook diagrams are often drawn that way, which trains the habit. The trap fires the moment the force is at any other angle: a $50$ N pull at $30^\circ$ to a $0.30$ m bar gives $\tau = (0.30)(50)\sin 30^\circ = 7.5\,\text{N}\cdot\text{m}$, not $15\,\text{N}\cdot\text{m}$. Half the force was wasted pulling along the bar, where it can't twist.
Fix. Always include the $\sin\theta$ factor unless you can see at a glance that the force is perpendicular to $\vec r$. If you'd rather skip the angle, use the lever-arm form $\tau = r_\perp F$ (where $r_\perp$ is the perpendicular distance from the axis to the force's line of action), or the perpendicular-component form $\tau = rF_\perp$.
Said about which axis? Without a named axis, "the torque" of a force is not a number. The same $40$ N force at the end of a $2.0$ m wrench gives $80\,\text{N}\cdot\text{m}$ about the other end of the wrench, $40\,\text{N}\cdot\text{m}$ about the midpoint, and $0$ about any point on the force's own line of action. Torque is a property of force and axis, not of the force alone.
Fix. Always state the axis. "The torque of $\vec F$ about point P" is a number. "The torque of $\vec F$" is not. When the axis changes, every torque in the problem changes too; recompute from scratch.