Biophysics: Friction, Translational Motion, and Energy in Physical Activity

Friction

Introduction to Friction

Friction is a fundamental force in physics that resists the relative motion of two surfaces in contact. It plays a crucial role in everyday phenomena and biological systems, such as walking and joint movement.

• Definition: Friction is the resistance to the sliding or moving of one surface over another.

• Applied Force: To move an object along a surface, the applied force must overcome the frictional force.

• Types of Friction:

  • Static friction – acts when an object is at rest.

  • Kinetic friction – acts when an object is moving.

Frictional Force and Coefficient of Friction

The magnitude of frictional force depends on the nature of the surfaces and the force pressing them together.

• Coefficient of Friction (μ): A dimensionless quantity representing the frictional property of surfaces.

• Normal Force (Fn): The force perpendicular to the surfaces in contact.

• Frictional Force Equation: $F_f = F_n$

• Static Friction: $F_{f,s} = _s F_n$

• Kinetic Friction: $F_{f,k} = _k F_n$

Table: Coefficients of Friction for Various Surfaces

Surfaces	μs (Static)	μk (Kinetic)
Leather on oak	0.6	0.5
Rubber on dry concrete	0.9	0.7
Steel on ice	0.02	0.01
Dry bone on bone	0.3	—
Bone on joint, lubricated	0.01	0.003

Additional info: The table above compares static and kinetic coefficients for common biological and engineering surfaces.

Standing at an Incline

Forces Acting on a Person Standing on an Inclined Plane

When a person stands on an inclined surface, several forces act on the body, determining whether the person will slide or remain stationary.

• Normal Force (Fn): $F_n = W cos$

• Static Friction Force (Ff): $F_f = 1c_s F_n = 1c_s W cos$

• Component of Weight Down the Incline (Fp): $F_p = W sin$

Condition for Sliding

The person will begin to slide when the downslope component of weight exceeds the maximum static friction force.

• Sliding Condition: $F_p 3e F_f$

• At the threshold: $F_p = F_f$

• Solving for the angle: $W sin = 1c_s W cos$ $tan = 1c_s$ For $1c_s = 0.6$, $tan = 0.6 = 31^$

Example

If a person of weight W stands on an oak board with $1c_s = 0.6$, the maximum angle before sliding is $31^$.

Friction at the Hip Joint

Biomechanics of Walking and Joint Forces

During walking, the body's weight is supported by one leg at a time, and the forces on the hip joint can be much greater than the body weight due to the position of the center of gravity.

• Force on the Joint: The force on the hip joint is greater than the body weight because the center of gravity is not directly above the joint.

• Magnitude: Depending on walking speed, the force can be several times the body weight.

• Frictional Wear: Large forces during motion can cause frictional wear at the joints, which is minimized by smooth cartilage and synovial fluid lubrication.

Example

In a typical step, the force on the hip joint may reach 2–3 times the body weight, especially during fast walking or running.

Translational Motion

Kinematics of Uniform Acceleration

Translational motion describes the movement of an object in a straight line, often analyzed under uniform acceleration.

• Velocity under Uniform Acceleration: $v = v_0 + a t$ $a = frac{v - v_0}{t}$

• Average Velocity: $v_{av} = frac{v + v_0}{2}$

• Distance Traversed: $S = v_{av} t = v_0 t + frac{a t^2}{2}$

• Alternative Equation: $v^2 = v_0^2 + 2 a s$

Example

If a person accelerates from rest ($v_0 = 0$) at $a = 2 m/s^2$ for $t = 3 s$, the final velocity is $v = 0 + 2 imes 3 = 6 m/s$.

Energy Consumed in Physical Activity

Conversion of Chemical Energy to Mechanical Work

Physical activity requires energy, which is supplied by the chemical energy in food. Muscles convert this energy into mechanical work, but the process is not highly efficient.

• Efficiency: Muscular efficiency in converting food energy to work is less than 20%.

• Energy Calculation Example:

  • For a 70 kg person jumping up 60 cm (0.6 m) once per second for 10 minutes:

  • Work per jump: $W = m g h = 70 imes 9.8 imes 0.6 = 411.6 J$

  • Total jumps: $10 min imes 60 s/min = 600 jumps$

  • Total work: $411.6 imes 600 = 247,000 J$

  • Food energy required (20% efficiency): $E_{food} = frac{247,000}{0.2} = 1,235,000 J$

Example

To sustain this activity, the person must consume enough food to provide approximately 1.23 MJ of energy.

Additional info: 1 calorie = 4.184 J; thus, 1.23 MJ = 294,000 calories.