College Physics I: BIIG problem-solving method

Friction

• Friction is a force that opposes relative motion between systems in contact.
• Three type of friction:  Static friction, Kinetic friction, and Rolling friction.

Static Friction

• When objects are stationary, static friction can act between them.
• Magnitude of static friction fs is

            f_s   ≤   μ_s N

where,  μ_s is the coefficient of static friction and N is the magnitude of the normal force.

• Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit.
• Once the applied force exceeds f_s(max), the object will move. Thus

            f_s(max)   =   μ_s  N

Kinetic Friction

• If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on ice.
• Magnitude of kinetic friction f_k is

            f_k   =   μ_k  N

where, μ_k is the coefficient of kinetic friction.

• The coefficients of kinetic friction are less than their static counterparts.
• Problem (E5.1):  A skier with a mass of 62 kg is sliding down a snowy slope of 25°. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N.                                ( 0.082 )

Drag Forces

• The force of drag on an object when it is moving in a fluid (either a gas or a liquid).
• Like friction, the drag force always opposes the motion of an object.
• The drag force F_D is given by

 F_D   =   C ρ A v²

where, C is the drag coefficient, A is the area of the object facing the fluid, and ρ is the density of the fluid.

• Drag force F_D is found to be proportional to the square of the speed of the object.

Terminal Velocity

• For a falling object, the downward force of gravity remains constant.
• However, as the object’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero.
• A zero net force means that there is no acceleration, as given by Newton’s second law.
• At this point, the object’s velocity remains constant and we say that the object has reached his terminal velocity ( v_t ).
• The terminal velocity is given by

            v   =   √ ( 2 m g / ρ C A )

The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air.

• Problem (E5.2):  Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position with an area approximately A = 0.70 m² and a drag coefficient of approximately C = 1.0. Assume the density of air is ρ = 1.21 kg/m³.                                                                                                       ( 44 m/s )

Stoke’s Law

• For fluids, the drag force is proportional just to the velocity. This relationship is given by Stokes’ law, which states that

            F_s   =   6 π r η v

where,  r is the radius of the object, η is the viscosity of the fluid, and v is the object’s velocity.

Elasticity

• Forces affect the motion of an object (such as friction and drag) and they also affect an object’s shape.
• A change in shape due to the application of a force is a deformation.
• For small deformations, the object returns to its original shape when the force is removed—that is, the deformation is elastic. The size of deformation is proportional to the force.
• Hooke’s law is given by

            F  =   k ΔL

where,  ΔL is the amount of deformation produced by the force F, and k is a proportionality constant (spring constant) that depends on the shape and composition of the object and the direction of the force.

Elastic Modulus

• A change in length ΔL is produced when a force is applied to a wire or rod parallel to its length L₀, either stretching it (a tension) or compressing it.

            ΔL  =   ( 1 / Y )  ( F / A )  L₀

where, ΔL is the change in length, F the applied force, Y is a factor, called the elastic modulus or Young’s modulus, A is the cross-sectional area, and  L₀ is the original length.

• The change in length is proportional to the original length L₀ and inversely proportional to the cross-sectional area of the wire or rod.
• Tensile strength: Materials with a large Y are said to have a large tensile strength because they deform less for a given tension or compression.
• Problem (E5.3):  Suspension cables are used to carry gondolas at ski resorts. Consider a suspension cable that includes an unsupported span of 3 km. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of 5.6 cm and the maximum tension it can withstand is 3.0×10⁶ N.  For steel, use Y = 210×10⁹ N/m².                                                                                             ( 20 m ) 
• Problem (E5.4):  Calculate the change in length of the upper leg bone (the femur) with young’s modulus of 9×10⁹ N/m² when a 70.0 kg man supports 62.0 kg of his mass on it, assuming the bone to be equivalent to a uniform rod that is 40.0 cm long and 2.00 cm in radius.                                     ( 0.02 mm )  

Stress-Strain Equation

• The stress-strain equation is given by

F / A   =   Y  ( ΔL / L₀ )

• Stress: The ratio of force to area,

            F / A

is defined as stress measured in N/m².

• Strain: The ratio of the change in length to length,

            ΔL / L₀

is defined as strain (a unitless quantity).

• Rearranging in the form of Hooke’s law, we get

            F   =   ( Y A / L₀ )  ΔL

The constant of proportionality is

            k   =   Y A / L₀ 

is the spring constant.

• In general, force and the deformation it causes are proportional for small deformations - applies to changes in length, sideways bending, and changes in volume.

Shear Modulus

• Shearing forces are applied perpendicular to the length L₀ and parallel to the area A, producing a deformation Δx. Here the force is a sideways stress or shearing force!
• The expression for shear deformation is

                  Δx   =   ( 1 / S )  ( F / A )  L₀ 

where,  S is the shear modulus and F             is the force applied perpendicular to L₀ and parallel to the cross-sectional area A.

• For example, it is easier to bend a long thin pencil (small A) than a short thick one, and both are more easily bent than similar steel rods (large S).
• Shear moduli are less than Young’s moduli for most materials.

Bulk Modulus

• An object will be compressed in all directions if inward forces are applied evenly on all its surfaces.
• It is relatively easy to compress gases and extremely difficult to compress liquids and solids.
• The expression for bulk modulus is

ΔV   =   ( 1 / B )  ( F / A )  V₀ 

where,  B is the bulk modulus and V₀  is the original volume, and  F / A is the force per unit area applied uniformly inward on all surfaces.

• Problem (E5.6):  Calculate the fractional decrease in volume (ΔV / V₀ ) for seawater at 5.00 km depth, where the force per unit area is 5.00×10⁷ N/m². The bulk modulus for water is 2.2×10⁹ N/m².   ( 2.3 % )

BIIG: Problems & Solutions