Biomedical Engineering Reference
In-Depth Information
M
Q
at P
= −12.5 N × 0.175 m
= −2.188 Nm
Q
P
M
P
at P
= 10 N × 0.35 m
= 3.5 Nm
P
S
Net moment
= 1.312 Nm
Q
P
S
1 cm = 0.05 m
1 cm = 5 N
= 0.0625 Nm
FIGUre 1.7
boomerang (Problem 1.4).
ANSWER:
The boomerang is in translational but not rotational equilibrium. Since
the net moment is positive, it would tend to rotate in a counterclockwise
fashion.
Free body diagrams
The boomerang in Figure 1.7 seems much simpler to understand than the
patient's leg in Figure 1.5. This is because the boomerang is shown as a
free body diagram
;
that is, it is shown free of all other objects, but with
all forces acting on it depicted. Similarly, to simplify the understanding
of loading applied to the patient's leg and foot, Figure 1.8 is a free body
diagram of the depiction in Figure 1.5. There are five forces shown, two
of which are unknown. Solving for two unknown forces requires the
construction of at least two equations of equilibrium. Forces that are
R
´
-R
49 N
W
F
W
L
= 9.8 N
= 29.4 N
W
F
= 1 kg
W
L
= 3 kg
1 cm = 50 N
1 cm = 10 cm
FIGUre 1.8
free body diagram of leg and foot.
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