SME
Biomechanics · Problem type
Hip force on one leg
Use this for “On one leg, how hard does the hip joint push when the abductors are half as far out as the body’s weight?”
Diagram
Moments about the hip joint: Fab dab = (5/6)W dW.
Fab
=
5,833
5
=
1,167
N.
FJ = 1,167 + 583.3 = 1,750 N.
FJ
W
=
2.5.
Arrows to one scale; the shortest are lengthened to be seen.
This is an example. Type your own values to start a new problem.
Fab (N)
×
dab (cm)
=
5
6
×
W (N)
×
dW (cm)
FJ (N)
=
Fab (N)
+
5
6
×
W (N)
FJ/W
Joint force in body weights
calculated
Formulas
Fab
×
dab
=
(
5
6
)
×
W
×
dW
1,167
×
5
=
(
5
6
)
×
700
×
10
FJ
=
Fab
+
(
5
6
)
×
W
1,750
=
1,167
+
(
5
6
)
×
700
FJ/
W
=
FJ
W
2.5
=
1,750
700
Assumptions
•
The stance leg is 1/6 of body weight, so (5/6)W acts at the body’s center.
•
The pelvis is level and still: the moments about the hip joint add to zero.
•
The forces are vertical; the joint pushes up into the pelvis.
Step-by-step
WE KNOW
W = 700 N, dab = 5 cm, dW = 10 cm
FIND
abductor force (Fab), joint force (FJ), joint force in body weights (FJ/W)
Step 1 · Find abductor force (Fab)
Use Fab × dab = (5 ÷ 6) × W × dW
Moments about the hip joint balance: the body’s moment over the abductors’ arm.
Fab
=
(5 ÷ 6) ×
W
× dW
dab
Fab
=
(5 ÷ 6) × 700 × 10
5
Fab
=
0.8333
×
700
×
10
5
Fab
=
583.31
×
10
5
Fab
=
5,833.1
5
Fab = 1,167 N
Step 2 · Find joint force (FJ)
Use FJ = Fab + (5 ÷ 6) × W
The joint pushes up on both downward pulls: the abductors and the body.
FJ
=
1,167
+
(
5
6
)
×
700
FJ = 1,167 + 0.8333 × 700
FJ = 1,167 + 583.31
FJ = 1,750 N
Step 3 · Find joint force in body weights (FJ/W)
Use FJ/W = FJ ÷ W
The joint force in body weights.
FJ/
W
=
1,750
700
FJ/
W
=
2.5
Check
Put the numbers back in. Both sides should match.
1,167
×
5
=
(
5
6
)
×
700
×
10
✓
1,750
=
1,167
+
(
5
6
)
×
700
✓
2.5
=
1,750
700
✓
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