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Anarel [89]
3 years ago
5

You are to assess the biomechanics of a male’s arm using his bicep to hold a 20 kg object in his hand. The upper arm is perpendi

cular to the ground while the forearm is parallel to the ground. The male’s forearm is 26.5 cm long and weighs 1.42 kg. The man’s hand weighs 0.54 kg and is 19 cm long. The center of masses for the forearm and hand are 43% and 50% of the extremity lengths from the proximal joint, respectively.
You may assume that only the bicep generates the force, the angle between the bicep and forearm is 75 degrees, and the bicep insertion is 5cm from the elbow joint.

a) Draw a complete free body diagram of the above described activity. (5 pts)

b) Calculate the elbow moment due to external load and body segments weight. (5 pts)

c) Calculate the elbow contact force and force exerted by the bicep. (5 pts)

Engineering
1 answer:
cestrela7 [59]3 years ago
5 0

Answer:

Explanation:

The detailed analysis, with free body diagram and step by step calculations with appropriate substitution is as shown in the attached files.

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3 years ago
A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 × 106 psi) and an original diameter of 3.7
Keith_Richards [23]

Answer:

the maximum length of specimen before deformation is found to be 235.6 mm

Explanation:

First, we need to find the stress on the cylinder.

Stress = σ = P/A

where,

P = Load = 2000 N

A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4

A = 1.0752 x 10^-5 m²

σ = 2000 N/1.0752 x 10^-5 m²

σ = 186 MPa

Now, we find the strain (∈):

Elastic Modulus = Stress / Strain

E = σ / ∈

∈ = σ / E

∈ = 186 x 10^6 Pa/107 x 10^9 Pa

∈ = 1.74 x 10^-3 mm/mm

Now, we find the original length.

∈ = Elongation/Original Length

Original Length = Elongation/∈

Original Length = 0.41 mm/1.74 x 10^-3

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5 0
3 years ago
can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.​
marishachu [46]

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

Sum of moments about the x axis:

∑τx = Iα

Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

∑τy = Iα

My = 0 Nm

Sum of moments about the z axis:

∑τz = Iα

Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0

Mz = -30.8 Nm

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Answer:

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