 Torque: An Introduction Printer Friendly Version
Basic Definition and Rotating Beams

Whenever a force is applied to a rigid body (a bar, a beam, a pole) it usually results in the rigid body rotating about an axis or pivot - that is, a torque has been applied.

In the following diagrams, we first see a uniform beam balanced on a knife-edge. This is accomplished by placing the knife-edge in the exact center directly under the beam's center of gravity. If the beam is shifted to either the right or left side of the knife-edge, it will no longer remain in equilibrium since the weight of the beam will be producing a torque. The beam will now be tilted. If a mass having the same weight of the beam is placed equidistant from the pivot on the opposite side, the beam will once again be brought back to equilibrium and return to its original level orientation. To calculate the magnitude of the torque produced by a force we use the formula or where

• F represents the magnitude of applied force
• r represents its moment arm
• represents the angle formed between F and r

If F and r are mutually perpendicular, sin 90º = 1 and .  The magnitude of the moment arm, r, is often written as since it, more often than not, represents a length along a beam. The moment arm is defined as the perpendicular distance from the line of action of the force to the pivot point. Since F is measured in newtons and is in meters, torque is measured in m nt. Alert! torque is not measured in joules, even though a joule equals a newton-meter.

Refer to the following information for the next three questions.

Now let's practice identifying the magnitude of some moment arms and the direction of the resulting torque on a horizontal beam. In each diagram, the green triangle represents the pivot point.
 A force F is applied to a bar midway between the pivot and the end of the bar. A force F now pulls downward on the end of the bar at an angle θ. A force F now pulls upward on the end of the bar at an angle θ. If you would like to practice more on calculating moment arms, torques, and the direction of rotation on horizontal beams, then use this accompanying worksheet.

Torque as a Vector Cross Product and the Right Hand Rule

Newton's Second Law, net F = ma, has an analogous rotational expression, net τ = Iα. It is this analogous expression that we are going to study in this lesson. Remember that we previously calculated torque as the product of a force and its moment arm. Torque is an example of a vector quantity formed by a cross product of two vectors r and F. One mathematical equation used to calculate this vector cross product is where

θ represents the angle between vectors r and F
and the double bars represent the magnitude, or norm, of vectors A and B

θ's location is determined by following this procedure: with r and F drawn head-to-tail, θ begins on r and terminates on F. The direction of the cross product is determined by a right hand rule (RHR). With your right hand, curl your fingers in the direction of θ. The extended thumb points in the direction of r x F.

Refer to the following information for the next question. For the diagram shown above, use the RHR to determine the direction of A x B?

Refer to the following information for the next three questions.

First, let's practice identifying the magnitude of some moment arms and the direction of the resulting torque on a circular wheel. In each diagram, the green dot represents the axis of revolution.
Each force has a magnitude of F and is being applied tangentially to the wheel. What is the moment arm of each force and in what direction would the wheel rotate?   Each force has a magnitude of F and is being applied tangentially to the wheel. What is the moment arm of each force and in what direction would the wheel rotate?   Each force has a magnitude of F and is being applied radially to the wheel. What is the moment arm of each force and in what direction would the wheel rotate?   Refer to the following information for the next three questions.

If a spool of thread were sitting on a horizontal table, which way would it move if it were to be pulled by each of the three strings shown below? You might want to take a moment and practice the CP workbook page on cams and spools.

 F1?

 F2?

 F3?

Determinants

Numerically, a cross product can also be calculated as a determinant. For torque, where rx, ry, and rz are the components of the radius vector r and Fx, Fy, and Fz are the components of the force vector F. The letters i, j, k represent the unit vectors along the x, y, and z axis.

i = (1, 0, 0)
j = (0, 1, 0)
k = (0, 0, 1)

Let's look at some numerical examples.

 Force F equals 100 N and is applied 25 cm from the pivot. What is the magnitude and direction of the torque it creates? Force F equals (100 N, 37º) and is applied at the end of the beam, 50 cm from the pivot. What is the magnitude and direction of the torque it creates? Force F equals (100 N, -37º) and is applied at the end of the beam, 50 cm from the pivot. What is the magnitude and direction of the torque it creates? What torque is produced by the weight of a 60-kg girl who has climbed 1 meter along a ladder that is leaning against a wall at an angle of 53º? Notice in the last example that using a determinant allows you to not need to calculate the angle between r and F - an area where some students might make a mistake. Related Documents