Biomedical Engineering Reference
In-Depth Information
F
x
F
y
lever arm 2
d
F
Fig. 2.5
Decomposition of force
F
in its orthogonal components. The torque due to the compo-
nent
F
x
is zero because the lever arm is zero and the torque due to
F
y
is given by
F
y
(2
d
⊥
)
θ
between
F
and the vertical line, applying the law of
calculated if we know the angle
trigonometry:
F
y
¼
F
cos
θ
. The resultant torque about
θ
is then given by
M
F
¼
F
x
0
F
y
2
d
⊥
ð
Þ¼
F
cos
θ
ð
2
d
⊥
Þ¼
Fx
⊥
:
The torque due to the component
F
x
is zero and, therefore, can produce no
rotation because the line of action of the vector
F
x
goes through the pivot point
O. Figure
2.4
shows that since the angle between 2
d
⊥
and
x
⊥
is also
θ
, we can write
that cos
x
⊥
.
Example 2.1
In the exercise of lateral lifting of an arm, one object with a 2 kg mass
is held by the hand, as can be seen in the figure of Example 2.1. The length of the
arm + forearm + center of the hand is 70 cm. The axis of rotation is located at the
shoulder. Find the torque exerted by the weight of the object for each situation in
which the angle between the arm and the body is (a) 30
downward and (b) 90
.
θ
(2
d
⊥
)
¼
W
90°
30°