Torques - Biomechanics of the Human Body - page 25

lever arm 2 d

axis of rotation

F

direction of rotation

Fig. 2.2 A force F of the same intensity as that of Fig. 2.1 is now applied at the opposite extremity
to the axis. The torque will be twice that of Fig. 2.1 because the lever arm was doubled

F

axis of rotation

Fig. 2.3 A force F of the same magnitude as that of Fig. 2.2 is now applied at the end of the bar,
but perpendicular, and the torque is null because the lever arm is zero, since the line of action,
imaginary over the force vector, passes through the axis of rotation

2 d

lever arm x

F

Fig. 2.4 A force F of equal intensity but different direction to that of Fig. 2.2 is applied at the edge
of the bar. In this case, the lever arm x ⊥ , which must be perpendicular to the line of action of force,
is shorter than that of Fig. 2.2 and, as a consequence, the magnitude of the torque will be less

to the line of action of force F will be shorter, i.e., x ⊥ < 2 d ⊥ . If we want to rotate

the bar with the same ease, we have to apply a force with a larger magnitude. In the
case of Fig. 2.4 , the torque is M F ¼

Fx ⊥ .
We obtain the same result, decomposing the force F in its orthogonal components
(Fig. 2.5 ), as presented in Chap. 1 . The intensity of the component F y can be