We have already studied in school physics that a turning effect is produced when a nut is tightened with a spanner (Fig. 2.13). the turning effect increases when you push harder on the spanner. It also depends on the length of the spanner: the longer the handle of the spanner, the greater is the turning effect of an applied force.

The turning effect of a force is called its moment or torque and its magnitude is defined as the product of force F and the perpendicular distance from its line of action to the pivot which is the point O around which the body (spanner) rotates. This distance OP is called moment arm I. Thus the magnitude of torque represented by τ is

τ = IF                      ………………..                         (2.26)

Fig. 2.13: the nut is easy to turn with a spanner.

It is easier still if the spanner has a long handle.

 turn_with a_spanner

when the line of action of the applied force passes through the pivot point, the value of moment arm I = 0, so in this case torque is zero.

We now consider the torque due to a force F acting on a rigid body. Let the force F acts on rigid body at point P whose position vector relative to pivot O is r. the force F can be resolved into two rectangular components, F sin θ perpendicular to r and F cosθ along the direction of r (Fig. 2.14 a). the torque due to F cos θ about pivot O is zero as its line of action passes through point O. therefore, the magnitude of torque due to F is equal to the torque due to F sin θ only about O. it is given by

τ = (F sinθ) r = r F sin θ                    …………..                                                (2.27)


Fig. 2.14 (a)

alternatively the moment arm I is equal to the magnitude of the component of R perpendicular to the line of action of F as illustrated in Fig. 2.14 (b). thus

τ = (r sinθ) F = r F sin θ                    ………………                            (2.28)

perpendicular_to the_line of_action

Fig 2.14 (b)

where θ is the angle between r and F

from Eq. 2.27 and Eq. 2.28 it can be seen that the torque can be defined by the vector product of position vector r and the force F, so

τ = r x F

or                                                                            τ = (r F sin θ) n̂ ………………                                                              (2.29)

where (rF sin θ) is the magnitude of the torque. The direction of the torque represented by n̂ is perpendicular to the plane containing r and F given by right hand rule for the vector product of two vectors.

The SI unit for torque is newton metre (N m).

Just as force determines the linear acceleration produced in a body, the torque acting on a body determines its angular acceleration. Torque is the analogous of force for rotational motion. If the body is at rest or rotating with uniform angular velocity, the angular acceleration will be zero. In this case the torque acting on the body will be zero.

Point to ponder


Do you think the rider in the above figure is really in danger? What if people below were removed?

Example  2.6: the line of action of a force F passes through a point P of a body whose position vector in metre is î-2ĵ+k̂. if F = 2î -3 ĵ + 4k̂ (in Newton), Determine the torque about the point A whose position vector (in metre) is 2 Î + ĵ + k̂


The position vector of point solution

The position vector of pointsolution_2relative to O.

The position vector of P relative to A is

solution_3AP = (î – 2 ĵ + k̂) – (2 î + ĵ + k̂) = -î -3 ĵ

The torque about A = r x F

= (-Î -3 ĵ) x (2 Î – 3 ĵ + 4k̂)

= -12 Î +4 ĵ +9k̂ N m

Can you do?

can you do?

Stand with one arm and the side of one foot pressed against a wall. Can you raise the other leg side ways? If not, then why not