Forces of 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

\(F_k\) = force of kinetic friction [N,Newtons]
pointed opposite the direction of motion

\(F_N\) = normal force [N,Newtons]

\(\mu _k\) = 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

\(F_{max}\) = maximum force of static friction [N,Newtons]
pointed opposite the direction of potential motion

\(F_N\) = normal force [N,Newtons]

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


Fs F=? 0.42kg 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 coefficient 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 coefficient on wikipedia
Find the force of kinetic friction.
Find how long it will take the cup to come to a stop.
solution $$F_{k}= \mu_{k}F_{N}$$ $$F_{N}=mg$$ $$F_{k}= \mu_{k}mg$$ $$F_{k}= (0.4)(0.42)(9.8)$$ $$F_{k}= 1.64N$$
$$F=ma$$ $$\frac{F}{m}=a$$ $$\frac{1.64}{0.42}=a$$ $$3.90\small\frac{m}{s^{2}}=\normalsize a$$
$$a=-3.90 \quad \Delta t=? \quad v_{i}=3.0 \quad v_{f}=0$$ $$v_{f} = v_{i}+a \Delta t$$ $$\frac{v_{f} - v_{i}}{a}=\Delta t$$ $$\frac{0 - 3}{-3.90}=\Delta t$$ $$0.77s=\Delta t$$
10kg Fs F=? 10kg Fg Example: A 10kg wood block is at rest on top of another 10kg wood block which is resting on a concrete slab. How much force will it take to overcome the static friction between the ground and the lower box? engineeringtoolbox
solution

Include both masses in the total mass: 10kg + 10kg = 20kg

$$F_g = mg$$ $$F_g = (20)(9.8)$$ $$F_g = 196N$$

Since the block isn't accelerating in the vertical the normal force equals the force of gravity for both blocks.

$$F_s = \mu_s F_N$$ $$F_s = \mu_s 196$$

The coefficient of static friction for concrete and wood is 0.62.

$$F_s = (0.62) (196)$$ $$F_s = 122N$$
Back