Biomedical Engineering Reference
In-Depth Information
while the force in the muscle is given by
F
M
=−
F
M
cos(
θ
)
e
x
+
F
M
sin(
θ
)
e
y
.
Notice that both the
x
- and
y
-component of the joint reaction force,
F
J
x
and
F
J
y
,
respectively, are unknown (the joint is modelled by a hinge), while for the muscle
only the magnitude of the muscle force
F
M
is unknown.
Application of the force balance in the
x
- and
y
-direction yields
F
J
x
−
F
M
cos(
θ
)
=
0
and
F
J
y
+
F
M
sin(
θ
)
−
W
−
W
0
=
0.
Force equilibrium supplies two equations for three unknowns, hence moment
equilibrium needs to be enforced as well. With the points A, B and C located
at
x
A
=
a
e
x
,
x
B
=
b
e
x
and
x
C
=
c
e
x
, respectively, moment equilibrium with
respect to point J requires:
aF
M
sin(
θ
)
−
bW
−
cW
0
=
0.
From this equation it follows that
bW
+
cW
0
a
sin(
F
M
=
.
θ
)
Hence, from the force balance in the
x
- and
y
-direction the joint reaction forces
can be computed immediately.
Suppose that
10
,
a
θ
=
=
0.1 [m],
b
=
0.25 [m],
c
=
0.6 [m],
and
W
=
5[N],
W
0
=
10 [N],
then
F
M
=
235 [N],
F
Jx
=
223 [N],
F
Jy
=−
58 [N].
Exercises
3.1
The position vectors of the points P, Q, R and S are given, respectively:
x
P
=
e
x
+
e
y
3
x
Q
=
4
e
x
+
2
e
y
