Math Problem Statement

Solve problems related to forces and energy usage during exercises, involving stretching muscles, lifting weights, and using a Nautilus machine.

Solution

Let's solve the problems from the uploaded image step by step. I'll go through each part carefully:


(a) Find Young’s modulus of the muscle tissue

Given:

  • Muscle length L=20cm=0.2mL = 20 \, \text{cm} = 0.2 \, \text{m}
  • Extension ΔL=3cm=0.03m\Delta L = 3 \, \text{cm} = 0.03 \, \text{m}
  • Force required when relaxed F=25.0NF = 25.0 \, \text{N}
  • Force required when tensed F=500NF = 500 \, \text{N}
  • Cross-sectional area A=50cm2=5×103m2A = 50 \, \text{cm}^2 = 5 \times 10^{-3} \, \text{m}^2

Young’s modulus formula is: E=StressStrain=FLAΔLE = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L}

  1. For relaxed muscles: Erelaxed=25.0×0.25×103×0.03E_{\text{relaxed}} = \frac{25.0 \times 0.2}{5 \times 10^{-3} \times 0.03}

  2. For tensed muscles: Etensed=500×0.25×103×0.03E_{\text{tensed}} = \frac{500 \times 0.2}{5 \times 10^{-3} \times 0.03}


(b) Find FForearm|\vec{F}_{\text{Forearm}}| and FTendon|\vec{F}_{\text{Tendon}}|:

Given:

  • Weight of the dumbbell: Fg=200NF_g = 200 \, \text{N}
  • Angle between FTendon\vec{F}_{\text{Tendon}} and horizontal: θ=80\theta = 80^\circ
  • The forces involved: FTendon|\vec{F}_{\text{Tendon}}|, FForearm|\vec{F}_{\text{Forearm}}|, and FgF_g.

Using torque equilibrium about the elbow joint: Sum of torques=0\text{Sum of torques} = 0 Fg×40cm=FTendon×4cm×cos(80)F_g \times 40 \, \text{cm} = |\vec{F}_{\text{Tendon}}| \times 4 \, \text{cm} \times \cos(80^\circ)

Solve for FTendon|\vec{F}_{\text{Tendon}}|. Then use force equilibrium (vertical): FForearm=Fg+FTendonsin(80)|\vec{F}_{\text{Forearm}}| = F_g + |\vec{F}_{\text{Tendon}}| \sin(80^\circ)


(c) Find FDeltoidF_{\text{Deltoid}}:

Given:

  • Cable tension: FNaut=40.0NF_{\text{Naut}} = 40.0 \, \text{N}
  • Cable angle with forearm: 120120^\circ
  • Deltoid attachment: 15.0cm15.0 \, \text{cm}
  • Forearm length: 60.0cm60.0 \, \text{cm}
  • Angle between deltoid and humerus: 170170^\circ

Using torque equilibrium about the shoulder joint: FNaut×60.0cos(120)=FDeltoid×15.0cos(170)F_{\text{Naut}} \times 60.0 \cos(120^\circ) = F_{\text{Deltoid}} \times 15.0 \cos(170^\circ)

Solve for FDeltoidF_{\text{Deltoid}}.


Would you like a detailed step-by-step calculation for each part or assistance with anything specific?

Related Questions:

  1. How does Young's modulus relate to the mechanical properties of a material?
  2. What is the significance of torque in biomechanics problems?
  3. Can you explain why static equilibrium conditions involve both torque and force balance?
  4. How do the angles of forces influence the magnitude of resultant forces in mechanics?
  5. Why is the deltoid muscle force higher than the cable tension in part (c)?

Tip:

Always check units carefully when plugging values into formulas to avoid calculation errors!

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Math Problem Analysis

Mathematical Concepts

Stress and Strain
Young's Modulus
Torque
Static Equilibrium
Force Analysis

Formulas

E = (F * L) / (A * ΔL)
ΣTorque = 0
ΣForce = 0
F_tendon = (F_g * d) / (d_tendon * cos(θ))
F_forearm = F_g + F_tendon * sin(θ)

Theorems

Static Equilibrium Theorem
Young's Modulus and Elasticity
Torque and Rotational Equilibrium

Suitable Grade Level

Undergraduate (Physics/Engineering)