Forces of Friction

Friction Forces

Crash Course youtube video on Friction

We are going to focus on the two types of dry friction: static and kinetic. Dry friction is a complex interaction, but we have equations that are a good approximation for most situations.


Kinetic Friction

Kinetic means motion. Kinetic friction is a force that occurs when two surfaces in contact slide against each other. Kinetic friction is primarily caused by chemical bonding between surfaces, however, in many cases roughness is dominant.

$$F_{k}=\mu_{k} F_{N}$$

Fn Fk v

Fk = force of kinetic friction [N,Newtons]
pointed opposite the direction of motion

FN = normal force [N,Newtons]

μk = mu, coefficient of friction [no units]
μ is different for every pair of surfaces, but you can look the values up: wikipedia engineeringtoolbox




Static Friction

Static means not moving. Static friction is friction between two or more solid objects that are not moving relative to each other. For example, static friction can prevent an object from sliding down a sloped surface. We can estimate the maximum static friction force. Up to that point the friction force will match any applied forces to keep the object stationary.

$$F_{max}=\mu_{s} F_{N}$$

Fn Fs

Fmax = force of static friction [N,Newtons]
pointed opposite the direction of potential motion

FN = normal force [N,Newtons]

μs = mu, coefficient of friction [no units]
μ is different for every pair of surfaces, but you can look the values up: wikipedia, engineeringtoolbox





Example: You place a 0.42kg glass IKEA calls POKAL on a flat copper pan. How much horizontal force will you have to apply to get the glass to move?

  • Look up the coeffcient on wikipedia

    solution $$F_{max}= \mu_{s}F_{N}$$ $$F_{N}=mg$$ $$F_{max}= \mu_{s}mg$$ $$F_{max}= (0.68)(0.42)(9.8)$$ $$F_{max}= 2.80N$$




    Example: You slide the 0.42kg POKAL glass cup on a glass table at 3.0 m/s. This time it doesn't fall over.

  • Look up the coeffcient on wikipedia
  • Find the force of kinetic friction.
  • Find how long it will take the cup to come to a stop.

    solution $$F_{k}= \mu_{k}F_{N}$$ $$F_{N}=mg$$ $$F_{k}= \mu_{k}mg$$ $$F_{k}= (0.4)(0.42)(9.8)$$ $$F_{k}= 1.64N$$
    $$F=ma$$ $$\frac{F}{m}=a$$ $$\frac{1.64}{0.42}=a$$ $$0.69\small\frac{m}{s^{2}}=\normalsize a$$

    \(a=-0.69\), \(\Delta t=?\), \(v_{i}=3.0\), \(v_{f}=0\)

    $$v_{f} = v_{i}+a \Delta t$$ $$\frac{v_{f} - v_{i}}{a}=\Delta t$$ $$\frac{0 - 3}{-0.69}=\Delta t$$ $$4.35s=\Delta t$$





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