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In order to calculate a torque, we always have a pivot P0 (where the axis is) and a force acting at another point P1. For example, in Figure 7-3, the pivot is at the crocodile's belly, and the force F acts at his snout at P1.

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Figure 7-3


The torque is defined by


where τ is the torque, r is the distance from P0 to P1, F is the size of the force, and φ is the angle between the direction of the force and the line P0 and P1. But is φ the big angle or the little angle? Well, it turns out it doesn't matter, since we are taking the sine, and the sines of supplementary angles are the same. Recall that
sin 0˚ = 0, ……….(4)
sin 90˚ = 1,
sin 180˚ = 0.
Also the convention is that
counterclockwise = positive torque
clockwise = negative torque.
Up until this chapter, we have drawn force vector arrows anywhere as long as the tail of the arrow sat on the object the force acted on. In doing torque problems, we must be more careful to put the arrows in the right place.

A large tarot card (the Fool) measuring 0.3 m by 0.4 m lies at the lower left in the first quadrant of the xy-plane, so that one corner is at (0.4 m, 0.3 m). See Figure 7-4. There is a force of 1.5 N in the y-direction located at point (0.4 m, 0.3 m). The pivot is at the origin. What is the torque?


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Figure 7-4


We can see that the force tends to turn the card counterclockwise, so the torque is positive. We can find the radius by the Pythagorean theorem:


The angle φ is shown in the two places in the diagram (corresponding angles with the parallel lines). The sine of φ can be obtained by looking at the portion of the diagram shown in Figure 7-5:


Putting all this together gives us


τ = rFsinφ
= (0.5 m)(1.5 N)(0.8)
= 0.6 Nm.

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Figure 7-5



A massless meter stick is supported by a fulcrum at the mark 0.3 m (point B). A mass A of 10 kg is sitting at the 0.1-m mark (point A). A mass C of 4 kg is sitting at the 0.8-m mark (point C). Consider the forces on the ruler (Figure 7-6).

  1. What is the torque due to the weight of A about point B?
  2. What is the torque due to the weight of C about point B?
  3. What is the torque due to the force of the fulcrum about point B?

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Figure 7-6

  1. In this case, we have r = 0.2 m, F = mg = 100 N, sinφ = 1, and the torque is counterclockwise, so
    τ = (0.2 m)(100 N)(1) = 20 Nm.
  2. In this case, the torque is clockwise, so we have
    τ = –(0.5 m)(40 N)(1) = –20 Nm.
  3. In this case, r = 0 m, so τ = 0 Nm.



A massless meter stick is hanging from the ceiling at the mark 0.3 m, point B. Mass A (10 kg) is hanging by string A (0.2 m long) connected to point A at the 0.1-m mark on the meter stick. Mass C (4 kg) is hanging by string C (0.3 m long) connected to point C at the 0.8-m mark on the meter stick. (See Figure 7-7.)

  1. What is the torque due to the weight of A about point B?
  2. What is the torque due to the weight of C about point B?
  3. What is the torque due to the force of the fulcrum about point B?

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Figure 7-7


This example looks exactly like the previous example; the strings are only a slight modification. If we apply the strict definition for torque (equation [3]), however, we will end up making an enormous effort, calculating rA and sinφA, and so on. Thankfully, there is an easier way.


Trick: The torque due to a force is not changed by moving the force vector to a new point, as long as (hat point lies on the line of the vector. That is, the new point must be on a line containing the old point and running in the same direction as the vector. We can think of this as sliding the vector along the direction it is already pointing until we have


In this example, this trick is the equivalent of sliding the force up the string to the meter stick. In fact, the example is exactly equivalent to Example 2.
  1. τ = 20 Nm.
  2. τ = –20 Nm.
  3. τ = 0 Nm.
Example 1, revisited solution: We can slide the 1.5-N force down the edge of the card to the x-axis. In this case we have r = 0.4 m and sinφ = 1, so we have
τ = (0.4 m)(1.5 N)(1) = 0.6 Nm
Once we move the force to a new point, so that φ = 90˚ (and sinφ = 1), the torque is especially easy to calculate. The line segment from the pivot to the new position of the force is called the lever arm. In Figure 7-8, the lever arm is the line segment .

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Figure 7-8

In drawing torque diagrams, it is helpful to keep the following principles in mind:
  1. Gravity acts at the center of mass.
  2. A string or rope exerts only a pulling force at the point of connection.
  3. When a stick or pole meets a wall or floor, the force acts at the point of contact.
    1. If the surface is frictionless, then there is only a normal force.
    2. If the stick is connected by a hinge, then the force is along the stick (either pushing or pulling).
    3. If the stick is connected to the wall or floor, then there are two forces, one normal and one parallel to the surface.

We will use these in Next Topic.

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