5.3 Torque

Learning objectives

5.3A—Identify the torques exerted on a rigid systemIdentify the torques exerted on a rigid system.• T orque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force.• The lever arm is the perpendicular distance from the axis of rotation to the line of action of the exerted force.5.3B—Describe the torques exerted on a rigid systemDescribe the torques exerted on a rigid system.• T orques can be described using force diagrams.- i. Force diagrams are similar to free-body diagrams and are used to analyze the torques exerted on a rigid system.- ii. Similar to free-body diagrams, force diagrams represent the relative magnitude and direction of the forces exerted on a rigid system. Force diagrams also depict the location at which those forces are exerted relative to the axis of rotation.• The magnitude of the torque exerted on a rigid system by a force is described by the following equation, where is the angle between the force vector and the position vector from the axis of rotation to the point of application of the force. BOUNDARY STATEMENT While AP Physics 1 expects students to mathematically manipulate the magnitude of torque using vector conventions, the direction of torque is beyond the scope of the course. TOPIC 5.4 Rotational Inertia

Locate the rotational effect

A force exerts torque about a specified axis only through the component perpendicular to the position vector from that axis to the force's application point. Torque describes the force's rotational effect about that axis.

Trace the geometry

Identify each torque in this order:

Name the axis of rotation.
Mark where the force is applied and its direction.
Extend the force's line of action.
Measure the lever arm: the perpendicular distance from the axis to that line.

Recognize zero and nonzero torque

Force geometry about the chosen axis	Lever arm	Torque?
Line of action passes through the axis	Zero	None
Force is parallel to the position vector	Zero	None
Force has a perpendicular component away from the axis	Nonzero	Yes

Apply the geometry to a door

Door comparison: for an axis through the hinges, a push at the handle perpendicular to the door has a large lever arm and produces torque. A push directed along the door toward the hinges has a line of action through the axis, so its lever arm and torque are zero.

Measure the correct distance

The lever arm is not automatically the distance from the axis to the application point. It is the perpendicular distance to the line of action. Also, only the force component perpendicular to the position vector contributes to torque.

Preserve force location

A torque force diagram must show each force's relative magnitude and direction and where it acts relative to the chosen axis. That location information is what an ordinary free-body diagram may omit.

Use either torque form

$\tau=r_{\perp}F=rF\sin\theta$

Symbol	Meaning
$r$	Distance from axis to application point
$\theta$	Angle between $\vec r$ and $\vec F$
$r_\perp=r\sin\theta$	Perpendicular lever arm
$F$	Force magnitude

Calculate an angled force

Angled force: a $30\,\text{N}$ force acts $0.40\,\text{m}$ from an axis at $60^\circ$ to the position vector.

$\tau=rF\sin\theta=(0.40\,\text{m})(30\,\text{N})\sin60^\circ=10.4\,\text{N}\,\text{m}$ .

Equivalently, $r_\perp=(0.40\,\text{m})\sin60^\circ=0.346\,\text{m}$ and $\tau=r_\perp F$ gives the same result.

Keep angle, direction, and units bounded

The angle is between the position vector and force vector, not automatically the angle drawn against a surface. AP Physics 1 requires manipulation of torque magnitude but not the vector direction of torque. Torque uses N·m; do not relabel it as joules.

Next Topic

—5.4 Rotational InertiaRead