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# Biomechanics

Question No: 1

Fx = m x g

W1 = 35lbs

W2 = 45lbs

W3 = 55lbs

At angle = θ

g = 32.17405 lb (pound force)

Reactive force at knee = F2

F2 = w2 x g

= 45 lbs x 32.17

F2 = 1447 at knee point

1. All Assumptions.
2. Seated leg is in the position described as above.
3. W1 is the weight of leg at joint point while force acting at this point is considered to be F1. Rx and Ry are the resolved components.
4. W2 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, F2.  Rx2 and R y2 are the resolved components.  Force acting at knee point is product of weight of leg and g.
5. W3   weight at ground level point. Rx3 and Ry3 are the resolved components along x-axis and y-axis.

b)

Reactive force

∑Fx = 0

Rx1 = 0

∑Fy = 0

Ry1 = Fapplied + w1

∑M0   = 0

Ry2 = m2 g + Ry1

Ry2 = m2 g + Ry1

Muscle Moment in terms of θ = M x a sin θ = (b W1 – cW2) Cos (θ)

So

M =

Muscle moment in terms of θ.

Rx = W Cos (θ)

R y = W Sin (θ) – W2 – W3

c)

Using θ = 90, 30, 60, 0

At θ = 90 At knee point

Ry2 = m2 g + Ry1

Ry2 = m2 g + Ry1

M =

Fx = F Cos (θ)

Fy = F Sin (θ)

F = 1447

Fx = F Cos (θ) = 1447 x 0 = 1447

Fy = F Sin (θ) = 1447 x1 = 1447

M =  = 0

At θ = 60

Fx = F Cos (θ)

Fy = F Sin (θ)

F = 1447

Fx = F Cos (60) = 1447 x 0.5 = 723.5

Fy = F Sin (θ) = 1447 x0.866 = 1253.0

M =

=

At θ = 30

Fx = F Cos (30)

Fy = F Sin (30)

F = 1447

Fx = F Cos (30) = 1447 x 0.866 = 1253.5

Fy = F Sin (30) = 1447 x0.5 = 723.0

M =

=

At θ = 0

Fx = F Cos (0)

Fy = F Sin (0)

F = 1447

Fx = F Cos (30) = 1447 x 1 = 1447.5

Fy = F Sin (30) = 1447 x0 = 0

M =

=

Question No: 2

Reaction forces

∑Fx = 0

Rx1 = 0

∑Fy = 0

∑Fx = ma

a)

∑F x  = R ax  + R ka

∑Fx    = ma
∑Fy = may

∑Fy = Ry – Wf  + Ray  = may

Muscle Moment in terms of θ = M x a sin θ = (b W1 – cW2) 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 all F= ma To calculate acceleration total force acting on knee point should be given Reaction force at ankle Fax Mass of leg is not given ∑ Fy = R y – W f  + Ray  = may 