Biomedical Engineering Reference
In-Depth Information
Numeric
.
M
MP
at
P
=× ×=
−
398015
.
.
−
441
.
Nm
A
at
=+ ××=
59801
49
44149049
.
.
+
.
Nm
Nm
B
M
=−
.
+
.
=+
.
net
PROBLEM 1.9
A bone plate is placed in a bending press (Figure 1.14a), and a load of
150 N is applied. What are the magnitudes of the resultant forces at
A
and
B
?
Assume that the plate has a mass of 0.5 kg (a very heavy plate!)
and draw a free body diagram.
ANSWER:
Graphic.
Represent the plate by a line (since there are no horizontal
components) and construct the appropriate moment “boxes” for moments
about point
A
(Figure 1.14b). Partition into squares of 0.625 N ∙ m each.
The 150 N force contributes -18 squares, and the plate contributes about
1 square. Therefore, the force at
B
must contribute +19 squares. Thus,
B
must point up and have a magnitude of (19 × 0.625)/0.2 or about 60 N.
Constructing 150 + 4.9 =
A
+
B
yields
A
≈ 95 N.
Numeric
.
M
net at
A
= -150 × 0.075 - 4.9 × 0.1 +
B
× 0.2
Therefore:
B
= 58.7 N
Since:
A
+
B
= 150 + 4.9,
A
= 96.2 N
Note:
If the mass of the plate is neglected,
B
= 60 N and
A
= 90 N;
their values are the inverse ratio of their lever arms measured to the point
of action of the 150 N force.
(a)
(b)
150 N
150 N
18
~1
4.9 N
A
B
A
= ?
B
= ?
19
Free body
diagram and
moment equilibrium (at
A
)
B
60 N
A
95 N
Force
equilibrium
A
150 N
=
=
B
1 cm = 50 N
1 cm = 5 cm
= 0.625 Nm
4.9 N
FIGUre 1.14
bone plate in bending press (Problem 1.9).
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