Comprehensive Study Notes on Newtonian Forces, Friction, Elasticity, and Biomechanics

Fundamental Definition and Vectorial Nature of Forces

  • Physical Definition of Force:

    • A force is a quantitative physical vector quantity that expresses and measures the interaction between two physical systems.
    • A force never exists in isolation; it always represents a mutual interaction acting between two bodies.
    • Every day physical actions—such as pushing, pulling, compressing, or lifting an object—involve the exertion of a force.
  • Vector Characteristics of a Force:

    • Magnitude (intensity or modulus).
    • Direction (the line along which the force acts).
    • Sense (the orientation along the direction arrow).
    • Point of application (the specific point on the body where the force is exerted).
  • Primary Effects of Forces:

    • Static Effects: Cause elastic or plastic deformation in a body (e.g., a bone subjected to a mechanical load, or the force required for a syringe needle to penetrate human skin).
    • Dynamic Effects: Alter the state of motion of a body by producing an acceleration.

Syringe needle force application on arm

Dynamic Nature of Force: Refuting Common Misconceptions

  • Historical Misconception vs. Newtonian Physics:

    • Common intuition (historically formalized by Aristotle) suggested that force is directly proportional to velocity (\text{Force} \n\n\propto v)—asserting that a continuous force is required to maintain constant speed, and that an object stops simply because the driving force ceases.
    • Experimental physics established by Galileo Galilei and Sir Isaac Newton disproved this notion. Friction is the real agent that causes moving objects to decelerate and come to rest.
    • In the absence of resistive forces, no net force is required to keep a body moving at a constant velocity.
  • Fundamental Relationship:

    • A net force is required to change the velocity of a body (i.e., to cause acceleration or deceleration).
    • The correct mathematical relation is:   Forceacceleration\text{Force} \propto \text{acceleration}F=ma\vec{F} = m \mathbf{a}
    • Whenever a change in the state of motion of a body is observed, a net force must be identified as the underlying cause.

Fundamental Classification of Forces

  • Action-at-a-Distance Forces:

    • Forces that act without physical contact between objects. Many of these forces are conservative.
    • Gravitational Force: Universal attractive interaction between masses.
    • Electrostatic Force: Interaction between electric charges.
  • Contact Forces:

    • Forces that result from direct macroscopic contact between physical surfaces or bodies.
    • Friction Force: Dissipative force opposing relative sliding or motion.
    • Tension Force: Force transmitted through stretched ropes, cords, or cables.
    • Fluid Pressure: Force exerted by fluids per unit surface area.

Gravitational Force and Weight Force

  • Newton's Law of Universal Gravitation:

    • Any two point masses m1m_1 and m2m_2 separated by a distance rr attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them:   F=G×m1×m2r2F = G \times \frac{m_1 \times m_2}{r^2}
    • Universal Gravitational Constant: G=6.67428×1011m3kg1s2G = 6.67428 \times 10^{-11}\,\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}.
  • Gravitational Acceleration at Earth's Surface:

    • Setting m1=MEarthm_1 = M_{\text{Earth}} (mass of the Earth) and r=REarthr = R_{\text{Earth}} (mean radius of the Earth):   F=(G×MEarthREarth2)×m=g×mF = \left(G \times \frac{M_{\text{Earth}}}{R_{\text{Earth}}^2}\right) \times m = g \times m
    • Acceleration due to gravity at Earth's surface:   g=G×MEarthREarth2=9.81m/s2g = \frac{G \times M_{\text{Earth}}}{R_{\text{Earth}}^2} = 9.81\,\text{m/s}^2
  • Weight Force Definition:

    • The weight force Fg\vec{F}_g (or P\vec{P}) acting on a mass mm is the gravitational attraction exerted on it by Earth:   Fg=P=mg\vec{F}_g = \vec{P} = m\mathbf{g}
    • Direction and Sense: Pointed vertically downward, directed toward the center of Earth.
    • Geoid Effect: Earth is an oblate spheroid (geoid) rather than a perfect sphere. The Earth's radius RR is larger at the Equator than at the Poles. Consequently, gravitational acceleration gg is slightly smaller at the Equator compared to the Poles.
  • Units and Dimensional Analysis:

    • SI Unit of Force: Newton (N\text{N}).   1N=1kg×1m/s21\,\text{N} = 1\,\text{kg} \times 1\,\text{m/s}^2
    • A mass of 1kg1\,\text{kg} subjected to standard gravity (g9.81m/s210m/s2g \approx 9.81\,\text{m/s}^2 \approx 10\,\text{m/s}^2) experiences a weight force of approximately 9.81N9.81\,\text{N} (or 10N\approx 10\,\text{N}).
    • Practical System Unit: Kilogram-force (kgp\text{kgp} or kgf\text{kg}_f):   1kgp=1kg×g=9.81N10N1\,\text{kgp} = 1\,\text{kg} \times g = 9.81\,\text{N} \approx 10\,\text{N}
  • Distinction Between Mass and Weight:

    • Mass (mm): An intrinsic scalar property of matter, measured in kilograms (kg\text{kg}), which remains constant everywhere in the universe.
    • Weight (Fg\vec{F}_g): A force dependent on local gravitational acceleration gg.
    • Free-Fall Kinematics (in vacuum with y0=Hy_0 = H and v0=0v_0 = 0):v(t)=g×tv(t) = -g \times ty(t)=Hg2×t2y(t) = H - \frac{g}{2} \times t^2
    • Earth vs. Moon Comparison:
    • On Earth (g9.81m/s2g \approx 9.81\,\text{m/s}^2): A mass m=1kgm = 1\,\text{kg} has Fg=9.8NF_g = 9.8\,\text{N}; a mass m=10kgm = 10\,\text{kg} has Fg=98NF_g = 98\,\text{N}.
    • On the Moon: Gravitational acceleration is roughly one-sixth of Earth's gravity (gMoon16gEarthg_{\text{Moon}} \approx \frac{1}{6} g_{\text{Earth}}). A body's weight on the Moon drops to one-sixth of its Earth weight, while its mass remains entirely unchanged.
  • Numerical Examples and Order-of-Magnitude Calculations:

    • Example 1 (Weight Estimation):
    • Problem: A backpack has a mass m=8kgm = 8\,\text{kg}. What is its approximate weight?
    • Calculation: Fg=m×a=8kg×9.81m/s28×10=80NF_g = m \times a = 8\,\text{kg} \times 9.81\,\text{m/s}^2 \approx 8 \times 10 = 80\,\text{N}.
    • Example 2 (Gravitational Attraction Between Small Spheres):
    • Problem: Two spheres of mass m1=m2=5kgm_1 = m_2 = 5\,\text{kg} are placed at a distance r=0.5mr = 0.5\,\text{m}. Estimate the order of magnitude of their mutual gravitational force (G=6.7×1011Nm2kg2G = 6.7 \times 10^{-11}\,\text{N}\,\text{m}^2\,\text{kg}^{-2}).
    • Calculation:     F=G×m1×m2r2(6.7×1011)×5×5(0.5)2=6.7×1011×250.25=6.7×109N108NF = G \times \frac{m_1 \times m_2}{r^2} \approx \left(6.7 \times 10^{-11}\right) \times \frac{5 \times 5}{(0.5)^2} = 6.7 \times 10^{-11} \times \frac{25}{0.25} = 6.7 \times 10^{-9}\,\text{N} \approx 10^{-8}\,\text{N}
    • Resulting order of magnitude: 108N10^{-8}\,\text{N}.

Electrostatic Force and Relative Strength Comparison

  • Coulomb's Law:

    • Calculates the magnitude of the electrostatic attraction or repulsion between two point charges q1q_1 and q2q_2 separated by distance dd:   F=k×q1×q2d2F = k \times \frac{q_1 \times q_2}{d^2}
    • Units: FF in Newtons (N\text{N}), charges in Coulombs (C\text{C}), distance in meters (m\text{m}).
    • Electrostatic Proportionality Constant:k=14πε0=μ0×c24π=8.99×109Nm2C2k = \frac{1}{4\pi \varepsilon_0} = \frac{\mu_0 \times c^2}{4\pi} = 8.99 \times 10^9\,\text{N}\,\text{m}^2\,\text{C}^{-2}
  • Comparison Between Electrostatic and Gravitational Constants:

    • k=8.99×109Nm2C2k = 8.99 \times 10^9\,\text{N}\,\text{m}^2\,\text{C}^{-2}
    • G=6.7×1011Nm2kg2G = 6.7 \times 10^{-11}\,\text{N}\,\text{m}^2\,\text{kg}^{-2}
    • Since kGk \gg G, electrostatic forces completely dominate over gravitational forces at atomic and molecular scales. In the presence of electrostatic interactions, gravitational forces between charged particles are negligible.

Normal Contact Force (Constraint Force)

  • Definition and Mechanism:

    • The normal force FN\vec{F}_N (or N\vec{N}) is the contact force exerted perpendicular (normal) to the surface of contact.
    • It acts as a mechanical constraint force that prevents a body from penetrating or falling through a solid support surface.
  • Equilibrium on a Horizontal Plane:

    • For a stationary body resting on a flat horizontal plane (a=0\mathbf{a} = 0):   FN+Fg=0\vec{F}_N + \vec{F}_g = 0FNmg=0    FN=mgF_N - m\mathbf{g} = 0 \implies F_N = m\mathbf{g}

Free body diagram of normal force on horizontal table

Friction Forces Between Solid Surfaces

  • Microscopic Origin:

    • Friction arises from electrostatic interactions and physical interlocking between microscopic irregularities (asperities) of contacting materials.
    • Biological systems utilize special lubricants to reduce friction (e.g., synovial fluid in human joint cavities).
  • Classification of Friction Forces:

    • Static Friction (FsF_s): Prevents the initiation of relative sliding motion between stationary contacting surfaces. It increases proportionally with the applied force up to a maximum threshold value, at which point the static bond breaks.
    • Kinetic (Dynamic) Sliding Friction (FkF_k): Resists continuous relative sliding motion between two surfaces in contact.
    • Rolling Friction (Dynamic Volvente): Resists the rolling motion of spherical or cylindrical objects (such as wheels, spheres, or wheelchair castors) across a surface. Rolling friction is generally much weaker than kinetic sliding friction because the instantaneous surface area of contact is substantially smaller.
  • General Mathematical Formulation:Fa=μa×F=μa×FNF_a = \mu_a \times F_{\perp} = \mu_a \times F_N

    • FaF_a: Friction force acting parallel to the surface, directly opposing motion or the attempted direction of motion.
    • μa\mu_a: Dimensionless coefficient of friction characteristic of the pair of materials.
    • FF_{\perp} (or FNF_N): Normal pressing force acting perpendicular to the contact plane.
  • Direction and Vector Orientation Clarification:

    • The normal force FNF_N acts perpendicular to the surface.
    • Friction (FsF_s or FkF_k) acts parallel to the surface, oriented opposite to the motion (or impending motion).
    • The equation Fa=μa×FNF_a = \mu_a \times F_N scales the magnitude only; the two forces are mutually perpendicular vectors.
  • Properties of Dynamic Friction:

    • Independent of the macroscopically visible surface contact area.
    • Independent of relative sliding speed over ordinary ranges.
    • Directly proportional to the normal force FNF_N
    • Coefficient relationship: For any given pair of materials, the kinetic friction coefficient μk\mu_k is strictly smaller than the static friction coefficient μs\mu_s (μk<μs\mu_k < \mu_s).
  • Self-Adjusting Nature of Static Friction:

    • The maximum threshold of static friction is Fs,max=μs×FNF_{s,\max} = \mu_s \times F_N.
    • If the applied pushing force FappliedF_{\text{applied}} is less than Fs,maxF_{s,\max}, the static friction force adapts to exactly match the applied force: Fs=FappliedF_s = F_{\text{applied}}, keeping the body at rest.
    • Motion starts only when Fapplied>Fs,maxF_{\text{applied}} > F_{s,\max}. Once motion begins, the opposing force drops immediately to the kinetic friction value Fk=μk×FNF_k = \mu_k \times F_N

Friction force versus applied force graph

  • Standard Coefficients of Friction (μs\mu_s and μk\mu_k):
Surface PairStatic Coefficient (μs\mu_s)Dynamic/Kinetic Coefficient (μk\mu_k)
Wood on wood0.40.40.20.2
Ice on ice0.10.10.030.03
Metal on metal (lubricated)0.150.150.070.07
Steel on steel (dry)0.70.70.60.6
Rubber on dry concrete1.01.00.80.8
Rubber on wet concrete0.70.70.50.5
Rubber on other solid surfaces141\text{--}411
Teflon on Teflon in air0.040.040.040.04
Teflon on steel in air0.040.040.040.04
Lubricated ball bearings<0.01<0.01<0.01<0.01
Synovial joints in human bodies0.010.010.010.01
  • Practical Worked Examples:
    • Example 1 (Threshold force required to initiate movement of 1kg1\,\text{kg} mass):
    • For Wood on Wood (m=1kgm = 1\,\text{kg}, μs=0.4\mu_s = 0.4):       Fthreshold=μs×mg=0.4×1kg×9.8m/s24NF_{\text{threshold}} = \mu_s \times m\mathbf{g} = 0.4 \times 1\,\text{kg} \times 9.8\,\text{m/s}^2 \approx 4\,\text{N}
    • For Ice on Ice (m=1kgm = 1\,\text{kg}, μs=0.1\mu_s = 0.1):       Fthreshold=μs×mg=0.1×1kg×9.8m/s21NF_{\text{threshold}} = \mu_s \times m\mathbf{g} = 0.1 \times 1\,\text{kg} \times 9.8\,\text{m/s}^2 \approx 1\,\text{N}
    • Example 2 (Static Friction Adaptation):
    • A block of mass m=10kgm = 10\,\text{kg} rests on a horizontal plane (N=100NN = 100\,\text{N}). The static coefficient is μs=0.4\mu_s = 0.4. A horizontal force of 20N20\,\text{N} is applied.
    • Maximum potential static friction: Fs,max=0.4×10kg×10m/s2=40NF_{s,\max} = 0.4 \times 10\,\text{kg} \times 10\,\text{m/s}^2 = 40\,\text{N}.
    • Since 20N<40N20\,\text{N} < 40\,\text{N}, the block remains motionless, and the active static friction force is exactly 20N20\,\text{N}.
    • Example 3 (Crate Motion Thresholds):
    • A crate weighs 200N200\,\text{N} on a horizontal floor with μs=0.40\mu_s = 0.40 and μk=0.30\mu_k = 0.30.
    • Minimum force to start motion: Fs,max=μs×FN=0.40×200N=80NF_{s,\max} = \mu_s \times F_N = 0.40 \times 200\,\text{N} = 80\,\text{N}.
    • Force required to maintain constant sliding velocity: Fk=μk×FN=0.30×200N=60NF_k = \mu_k \times F_N = 0.30 \times 200\,\text{N} = 60\,\text{N}.

Analysis of Motion on an Inclined Plane

  • Force Decomposition:

    • Consider an object of mass mm on a plane inclined at an angle θ\theta to the horizontal:
    • Perpendicular component of gravity: F=mg×cos(θ)F_{\perp} = m\mathbf{g} \times \cos(\theta), balanced by the normal force FN=mg×cos(θ)F_N = m\mathbf{g} \times \cos(\theta).
    • Parallel component of gravity (driving force down the slope): F=mg×sin(θ)F_{\parallel} = m\mathbf{g} \times \sin(\theta).
    • Opposing static friction force: FaF_a.
  • Static Equilibrium on an Inclined Plane:

    • Condition to prevent sliding down the ramp:   Fa=F=mg×sin(θ)F_a = F_{\parallel} = m\mathbf{g} \times \sin(\theta)
    • Maximum available static friction force:   Fs,max=μs×FN=μs×mg×cos(θ)F_{s,\max} = \mu_s \times F_N = \mu_s \times m\mathbf{g} \times \cos(\theta)
    • Critical threshold condition for slipping:   mg×sin(θ)=μs×mg×cos(θ)    μs=sin(θ)cos(θ)=tan(θ)m\mathbf{g} \times \sin(\theta) = \mu_s \times m\mathbf{g} \times \cos(\theta) \implies \mu_s = \frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta)
  • Worked Inclined Plane Problems:

    • Problem 1 (Actual static friction calculation):
    • A 10kg10\,\text{kg} block rests on a 3030^\circ slope with μs=0.8\mu_s = 0.8. What is the actual static friction force acting on it?
    • F=mg×sin(30)10kg×10m/s2×0.5=50NF_{\parallel} = m\mathbf{g} \times \sin(30^\circ) \approx 10\,\text{kg} \times 10\,\text{m/s}^2 \times 0.5 = 50\,\text{N}.
    • Fs,max=μs×mg×cos(30)=0.8×100×0.866=69.3NF_{s,\max} = \mu_s \times m\mathbf{g} \times \cos(30^\circ) = 0.8 \times 100 \times 0.866 = 69.3\,\text{N}.
    • Since 50N<69.3N50\,\text{N} < 69.3\,\text{N}, the block does not slide, and the active static friction force is 50N50\,\text{N}.
    • Problem 2 (Minimum coefficient for equilibrium):
    • What minimum static friction coefficient μs\mu_s prevents a block from sliding down a 3030^\circ incline?
    • μs=tan(30)=sin(30)cos(30)=0.50.8660.58\mu_s = \tan(30^\circ) = \frac{\sin(30^\circ)}{\cos(30^\circ)} = \frac{0.5}{0.866} \approx 0.58

Viscous Friction in Biological and Fluid Dynamics

  • Low Velocity Fluid Resistance (Stokes' Law):

    • Describes the viscous drag force FaF_a acting on a small spherical body moving at low velocity through a laminar fluid:   Fa=6π×η×r×vF_a = 6\pi \times \eta \times r \times v
    • rr: Radius of the sphere (m).
    • η\eta: Dynamic viscosity of the fluid (Nsm2\text{N}\,\text{s}\,\text{m}^{-2} or Pas\text{Pa}\,\text{s}).
    • vv: Velocity relative to the fluid (m/s\text{m/s}).
    • Dimensional Verification: (Nsm2)×m×(m/s)=N\left(\text{N}\,\text{s}\,\text{m}^{-2}\right) \times \text{m} \times \left(\text{m/s}\right) = \text{N}.
  • High Velocity Aerodynamic Drag:

    • When object velocity is high and turbulence occurs, resistive force scales quadratically with speed:   Fa=12×cx×ρ×S×v2F_a = \frac{1}{2} \times c_x \times \rho \times S \times v^2
    • cxc_x: Aerodynamic drag coefficient (dimensionless shape-dependent index).
    • ρ\rho: Fluid density (kg/m3\text{kg/m}^3).
    • SS: Cross-sectional frontal area (m2\text{m}^2).
    • vv: Velocity relative to the fluid (m/s\text{m/s}).
    • Dimensional Verification: (kg/m3)×m2×(m2/s2)=kgm/s2=N\left(\text{kg/m}^3\right) \times \text{m}^2 \times \left(\text{m}^2/\text{s}^2\right) = \text{kg}\,\text{m/s}^2 = \text{N}.

Tension Forces and Pulley Systems

  • Ideal Rope Properties:

    • Massless (m0m \approx 0).
    • Inextensible (fixed length under stress).
    • Incompressible (only transmits tensile pulling forces).
    • Transmits mechanical force along its entire length from one end to the other, causing linked bodies to move together with identical acceleration magnitude aa
  • Ideal Pulleys:

    • A massless, fixed pivot point around which an ideal rope redirects its vector direction while keeping tension magnitude TT identical across all segments on both sides.
    • Clinical Application: Mechanical traction of injured human limbs via arrangements of cords and pulleys. The pulley changes the direction of pulling force without altering its magnitude (P=TP = T).
  • Centripetal Tension in Rotational Motion:

    • Tension in a tethered system provides the required centripetal acceleration aca_c to maintain circular orbital motion:   T=m×acT = m \times a_c

Elastic Forces, Hooke's Law, and Harmonic Motion

  • Hooke's Law for Ideal Springs:

    • When an elastic body or spring undergoes displacement Δx=xx0\Delta x = x - x_0 from its resting length x0x_0, a restoring force arises that opposes deformation:   Fel=k×Δx=k×(xl0)F_{\text{el}} = -k \times \Delta x = -k \times (x - l_0)
    • kk: Elastic spring constant (N/m\text{N/m}).
    • The negative sign signifies that the restoring force acts opposite to displacement.
  • Differential Equation of Simple Harmonic Motion:

    • Applying Newton's Second Law to a spring-mass system (l0=0l_0 = 0):   m×a=k×xm \times a = -k \times x
    • Expressing acceleration as the second derivative of position with respect to time (a=d2xdt2a = \frac{d^2 x}{d t^2}):   d2xdt2=(km)×x=ω2×x\frac{d^2 x}{d t^2} = -\left(\frac{k}{m}\right) \times x = -\omega^2 \times x
    • Solutions to this differential equation take the general sinusoidal form of simple harmonic motion:   x(t)=A×sin(ωt+φ)x(t) = A \times \sin(\omega t + \varphi)
    • Angular frequency ω\omega, frequency ff, and period TT:   ω=2πf=2πT=km\omega = 2\pi f = \frac{2\pi}{T} = \sqrt{\frac{k}{m}}
  • Numerical Practice Example:

    • Problem: An ideal spring with elastic constant k=200N/mk = 200\,\text{N/m} is stretched by Δx=3cm\Delta x = 3\,\text{cm} (0.03m0.03\,\text{m}). Calculate the magnitude of the restoring force.
    • Calculation: F=k×x=200N/m×3×102m=6NF = k \times x = 200\,\text{N/m} \times 3 \times 10^{-2}\,\text{m} = 6\,\text{N}.

Continuum Mechanics and Biomechanical Elasticity of Tissues

  • Stress and Strain Definitions:

    • Stress (Specific Load, τ\tau): The deforming force applied per unit cross-sectional area:     τ=FS[N/m2 or Pa]\tau = \frac{F}{S} \quad \left[\text{N/m}^2 \text{ or } \text{Pa}\right]
    • Strain (Relative Deformation, γ\gamma): Fractional change in length relative to initial resting length:     γ=Δll0[dimensionless]\gamma = \frac{\Delta l}{l_0} \quad \left[\text{dimensionless}\right]
  • Deformation Phases of Biological Materials:

    • Elastic Region: Linear behavior where stress is directly proportional to strain. Upon removing the load, the body returns completely to its original shape.
    • Inelastic / Plastic Region: Occurs once stress exceeds the critical yield point (carico specifico di snervamento). Permanent irreversible deformation remains after removing the load.
    • Fracture / Breaking Point: The ultimate terminal stress Σ\Sigma at which material structural failure occurs.
  • Young's Modulus (EE):

    • Within the linear elastic regime, Hooke's generalized law holds:   τγ=E    τ=E×γ\frac{\tau}{\gamma} = E \implies \tau = E \times \gamma
    • Young's modulus EE (Pa\text{Pa} or N/m2\text{N/m}^2) quantifies material stiffness and depends strictly on the material composition and type of stress (tensile vs. compressive).
  • Transverse Contraction and Poisson's Ratio (PP):

    • When a continuous body is stretched longitudinally, it narrows laterally. The fractional lateral dimensional change Δs\Delta s relates to longitudinal strain Δl\Delta l via Poisson's ratio PP:   Δs=P×Δl\Delta s = P \times \Delta l
  • Mechanical Properties of Biological Tissues vs. Biomaterials:

MaterialUltimate Breaking Stress Σ\Sigma (MPa\text{MPa})Young's Modulus EE (GPa\text{GPa})Poisson's Ratio PP
Human Tendon70700.40.40.40.4
Human Skin880.50.50.50.5
Human Bone13013017170.40.4
Steel400850400\text{--}8502002000.30.3
Titanium500900500\text{--}9001001000.30.3
  • Biomechanical Behavior of Human Femur Under Stress:
    • The human femur obeys Hooke's linear relation near the origin (dashed regime).
    • Tension vs. Compression Asymmetry: Young's modulus for tensile loading is approximately half of that observed for compressive loading (Etension12EcompressionE_{\text{tension}} \approx \frac{1}{2} E_{\text{compression}}). This is reflected by differing initial linear slopes.
    • Ultimate breaking stress Σ\Sigma also differs significantly between tension and compression.

Stress strain relationship for human femur under tension and compression

  • Torsional Moments and Bone Spiral Fractures:
    • Internal torsional moment opposes external applied torque T=FR\mathbf{T} = \mathbf{F} \wedge \mathbf{R} to maintain section equilibrium.
    • Excessive rotational torque applied across long bones (e.g., the tibia) produces classic helical torsional fractures, known as spiral fractures.

Radiograph showing spiral fracture of the tibia

Muscle Forces and Musculoskeletal Mechanics

  • Muscular Force Significance:

    • Muscle forces represent the primary internal active forces in physiological and biomechanical applications.
  • Structural Hierarchy of Striated Skeletal Muscle:

    • Whole Muscle \n\n\rightarrow Muscular Fiber Bundles (muscle cells) \n\n\rightarrow Individual Muscle Fiber (Myofiber) \n\n\rightarrow Myofibril \n\n\rightarrow Sarcomere (the functional contractile unit bounded by Z-lines, featuring H-zone, I-bands, and A-bands containing sliding actin and myosin filaments).

Levels of structural organization in skeletal muscle

  • Musculoskeletal Levers and Antagonistic Action:
    • Skeletal muscles attach to bones via collagenous tendons across biological joints.
    • Muscle contraction exerts tensile forces that generate rotational moments around joint axes.
    • Movements are coordinated in opposing pairs: agonist muscles (which produce a specific joint rotation) work opposite antagonist muscles (which produce reverse rotation).

Summary of Key Mechanical Force Equations

  • Universal Gravitation:F=G×m1×m2r2F = G \times \frac{m_1 \times m_2}{r^2}

  • Weight Force:P=mgP = m\mathbf{g}

  • Friction Between Solid Surfaces:Fa=μa×FNF_a = \mu_a \times F_N

  • Viscous Friction in Fluids:

    • Low velocity (Stokes): Fa=6π×η×r×vF_a = 6\pi \times \eta \times r \times v
    • High velocity: Fa=12×cx×ρ×S×v2F_a = \frac{1}{2} \times c_x \times \rho \times S \times v^2
  • Hooke's Law for Springs:F=k×ΔxF = -k \times \Delta x

  • Generalized Elasticity Law for Continuous Media:τγ=E\frac{\tau}{\gamma} = EF/SΔl/l0=E\frac{F/S}{\Delta l / l_0} = E