no slide titleclasses.engr.oregonstate.edu/mime/fall2018/ie545... · 2 second-class: fulcrum is at...
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First-class: Fulcrum is between the two loads, which is good for fine positional control.
Review: Skeletomuscular Levers
2
Second-class: Fulcrum is at the end; Force is exerted through a longer moment arm than the resistance.
Third-class: Fulcrum is at the end; Force is exerted through a shorter moment arm than the resistance
Review: In-Class Exercise
∑ FV = 0: RELBOW –WFOREAR AND HAND-WLOAD =0RELBOW -15.8N – 49N = 0RELBOW= 64.8N
∑ FH =0: N/A
∑ MA = 0: ME – MFOREAR AND HAND-MLOAD =0ME – 0.172m*15.8N-0.355m*49N=0ME =20.1NM
A
Basic for calculations of joint reaction forces and netmuscle moments throughout the body.
Link-Segment Model
Modeling components Anthropometric datao Segment lengtho Segment weighto Moment of inertia
Posture data External load: hand load Internal loado Segment weighto Muscle contraction force
Link-Segment Model
Basic for calculations of joint reaction forces and netmuscle moments throughout the body.
Link-Segment Model
Weight
Assumptions A fixed point mass, located at its center of mass. Joints are represented by simple hinge joints (not free to
translate, but free to rotate) Constant moment of inertiao I=mr2
Shape of the body does not change Constant segment length
Link-Segment Model
WBW
Low Back Pressure, FM?
Link-Segment Model
F
WBW
DBW
DF
L: Low back
FM DM
L
F: Hand load (external load)
WBW : Upper body weight (internal load)
FM : Low back muscle force (internal load)
ΣML = 0: MM - M BW - MF =0
FM = (WBW× DBW + F× DF)/ DM
FM × DM - WBW× DBW - F× DF = 0
θ
Low Back Pressure, Joint compression and shear forces?
Link-Segment Model
FCompression
FShear
L
θ
FCompression = FM + (WBW + F) * cosθ
L
θ
FM
WBW
F
FShear = (WBW + F) * sinθF
WBW
DBW
DF
FM DM
L
θ
In-class Exercise
Upper body weight, WBW = 300NHand load, F = 100NTrunk flexion angle, θ = 30 degreeDBW = 0.25 m, DF = 0.5 m, DM = 0.05 mPlease calculate the following variables:
1. Muscle force at the low back joint, FM2. Compression forces at the low back joint, FCompression3. Shear forces at the low back joint, FShear
F
WBW
DBW
DF
FM DM
θ
In-class Exercise 1. Muscle force at the low back joint, FM
F
WBW
DBW
DF
FM DM
θ
L: Low back
L
ΣML = 0:
MM - M BW - MF =0
FM = (WBW× DBW + F× DF)/ DM
FM × DM - WBW× DBW - F× DF = 0
FM = (300 × 0.25 + 100 × 0.5)/ 0.05=2500 N