Question No: 1
F_{x} = m x g
W_{1 }= 35lbs
W_{2 }= 45lbs
W_{3} = 55lbs
At angle = θ
g = 32.17405 lb (pound force)
Reactive force at knee = F_{2}
F_{2} = w_{2} x g
= 45 lbs x 32.17
F_{2 }= 1447 at knee point
- All Assumptions.
- Seated leg is in the position described as above.
- W_{1 }is the weight of leg at joint point while force acting at this point is considered to be F_{1}. R_{x }and R_{y }are the resolved components.
- W_{2 }is the weight of leg at knee point while force acting at this point is considered to be downward due to gravitational force acting at this point, F_{2}. R_{x2} and R _{y2 }are the resolved components. Force acting at knee point is product of weight of leg and g.
- W_{3 }weight at ground level point. R_{x3} and R_{y3 }are the resolved components along x-axis and y-axis.
b)
Reactive force
∑F_{x }= 0
R_{x1} = 0
∑F_{y }= 0
R_{y1} = F_{applied} + w_{1}
∑M_{0 }= 0
R_{y2} = m_{2 }g + R_{y1}
R_{y2} = m_{2 }g + R_{y1}
Muscle Moment in terms of θ = M x a sin θ = (b W_{1} – cW_{2}) Cos (θ)
So
M =
Muscle moment in terms of θ.
R_{x} = W Cos (θ)
R _{y} = W Sin (θ) – W_{2} – W_{3}
c)
Using θ = 90, 30, 60, 0
At θ = 90 At knee point
R_{y2} = m_{2 }g + R_{y1}
R_{y2} = m_{2 }g + R_{y1}
M =
F_{x }= F Cos (θ)
F_{y }= F Sin (θ)
F = 1447
F_{x }= F Cos (θ) = 1447 x 0 = 1447
F_{y }= F Sin (θ) = 1447 x1 = 1447
M = = 0
At θ = 60
F_{x }= F Cos (θ)
F_{y }= F Sin (θ)
F = 1447
F_{x} = F Cos (60) = 1447 x 0.5 = 723.5
F_{y} = F Sin (θ) = 1447 x0.866 = 1253.0
M =
=
At θ = 30
F_{x }= F Cos (30)
F_{y }= F Sin (30)
F = 1447
F_{x }= F Cos (30) = 1447 x 0.866 = 1253.5
F_{y }= F Sin (30) = 1447 x0.5 = 723.0
M =
=
At θ = 0
F_{x }= F Cos (0)
F_{y }= F Sin (0)
F = 1447
F_{x }= F Cos (30) = 1447 x 1 = 1447.5
F_{y }= F Sin (30) = 1447 x0 = 0
M =
=
Question No: 2
Reaction forces
∑F_{x }= 0
R_{x1} = 0
∑F_{y }= 0
∑F_{x }= ma
a)
∑F _{x } = R a_{x } + R k_{a}
∑F_{x
} _{ } =
ma
∑F_{y
}= ma_{y}
∑F_{y = }R_{y} – W_{f } + Ra_{y } = ma_{y}
Muscle Moment in terms of θ = M x a sin θ = (b W_{1} – cW_{2}) Cos (θ)
So
M =
- Length of total leg is missing to calculate moment of muscle total length of leg should be given. Only length of lower leg part is given so we can calculate moment around knee.
- If the muscle of leg under consideration is of female the mass and weight of female muscle will be different from mass of male muscle therefore the moment and forces around knee and ankle will be different due to different mass values.
Variable Description | Variable Name | Known/ unknown | Value of symbol |
Length of lower leg | L_{ ll} | Moment of muscle and force can be calculated | = 37.16 |
Acceleration of lower leg | a_{ll} | F= ma | To calculate acceleration total force acting on knee point should be given |
Reaction force at ankle | F_{ax} | Mass of leg is not given | ∑ F_{y = }R _{y} – W _{f } + Ra_{y } = ma_{y } |