SME
Biomechanics
Joint forces
Use this for “A hand holds a 50 N ball 35 cm from the elbow; the 15 N forearm acts at 15 cm and the biceps at 4 cm. Find the biceps and elbow forces.”
Diagram
Moments about the elbow: FM dM = L dL + Wf df.
FM
=
1,975
4
=
493.8
N.
FJ = 493.8 − 50 − 15 = 428.8 N.
Arrows to one scale; the shortest are lengthened to be seen.
This is an example. Type your own values to start a new problem.
FM (N)
×
dM (cm)
=
L (N)
×
dL (cm)
+
Wf (N)
×
df (cm)
FJ (N)
=
FM (N)
−
Wf (N)
−
L (N)
Formulas
FM
×
dM
=
L
×
dL
+
Wf
×
df
494 × 4 = 50 × 35 + 15 × 15
FJ
=
FM
−
Wf
−
L
429 = 494 − 15 − 50
Assumptions
•
The forearm is level and still: the moments about the elbow add to zero.
•
The muscle, the weights and the joint force are vertical.
•
FJ > 0 means the upper arm pushes down on the forearm.
Step-by-step
WE KNOW
L = 50 N, dL = 35 cm, Wf = 15 N, df = 15 cm, dM = 4 cm
FIND
muscle force (FM), joint force (FJ)
Step 1 · Find muscle force (FM)
Use FM × dM = L × dL + Wf × df
Moments about the elbow balance: the loads’ moments over the muscle’s arm.
FM
=
L
× dL + Wf × df
dM
FM
=
50 × 35 + 15 × 15
4
FM
=
1,750 + 225
4
FM
=
1,975
4
FM = 494 N
Step 2 · Find joint force (FJ)
Use FJ = FM − Wf − L
Up and down balance: the muscle pulls up, the weights and the elbow push down.
FJ = 494 − 15 − 50
FJ = 429 N
Check
Put the numbers back in. Both sides should match.
494 × 4 = 50 × 35 + 15 × 15 ✓
429 = 494 − 15 − 50 ✓
Topic check
See what you know: 5 questions on this topic
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