Imagine you are using a wrench to loosen a nut. The idea is to turn the nut by applying a
force to the end of the wrench. If you push or pull the wrench directly toward or away
from the nut, it should be obvious from experience that no progress will be made in
turning the nut. The important thing is the component of force perpendicular to the
wrench. It is this component of force which will cause the nut to turn. For example see
the following picture.

PICT

In the picture a force, F is applied at the end of a wrench represented by the position
vector R and the angle between these two is θ. Then the tendency to turn will be

|R |

|F ⊥|

=

|R |

|F |

sinθ, which you recognize as the magnitude of the cross product of R
and F. If there were just one force acting at one point whose position vector is R,
perhaps this would be sufficient, but what if there are numerous forces acting at many
different points with neither the position vectors nor the force vectors in the same plane;
what then? To keep track of this sort of thing, define for each R and F, the torque
vector

τ ≡ R × F.

This is also called the moment of the force, F. That way, if there are several forces acting
at several points the total torque can be obtained by simply adding up the torques
associated with the different forces and positions.

Example 14.5.2Suppose R_{1} = 2i−j+3k,R_{2} = i+2j−6k meters and at the pointsdetermined by these vectors there are forces, F_{1} = i −j+2k and F_{2} = i − 5j + kNewtons respectively. Find the total torque about the origin produced by these forcesacting at the given points.

It is necessary to take R_{1}× F_{1} + R_{2}× F_{2}. Thus the total torque equals

|| i j k || || i j k ||
||2 − 1 3 ||+ ||1 2 − 6 ||= − 27i − 8j− 8k Newton meters
||1 − 1 2 || ||1 − 5 1 ||

Example 14.5.3Find if possible a single force vector F which if applied at thepointi + j + kwill produce the same torque as the above two forces acting at thegiven points.

This is fairly routine. The problem is to find F = F_{1}i + F_{2}j + F_{3}k which produces the
above torque vector. Therefore,

| |
|| i j k ||
|| 1 1 1 ||= − 27i− 8j− 8k
| F1 F2 F3 |

which reduces to

(F3 − F2)

i+

(F1 − F3)

j+

(F2 − F1)

k = − 27i − 8j − 8k. This
amounts to solving the system of three equations in three unknowns, F_{1},F_{2}, and
F_{3},

F3 − F2 = − 27,F1 − F3 = − 8,F2 − F1 = − 8

However, there is no solution to these three equations. (Why?) Therefore no single force
acting at the point i + j + k will produce the given torque.