Friction

Fundamentals

Example: Here’s a box on a flat surface with two people pushing and pulling in the same direction.

Let’s now say that the weight of the box, $W$, is $100text{ N}$, and that the friction is limiting (i.e. applying any more force in $T$ or Thrust will cause the box to accelerate). Say, also, that the friction $F$ is $50text{ N}$.

Then we have the normal reaction force $R = 100text{ N}$.

Since $F = textcolor{red}{mu} R$, the coefficient of friction, $textcolor{red}{mu} = dfrac{50}{100} = 0.5$.

Friction on a Tilted Plane

Example: Here’s a $1500 text{ kg}$ car being towed up an incline at an angle of $30degree$. Say the slope has a coefficient of friction $textcolor{red}{mu} = 0.4$. What would be the required combined force of the tension in the rope, $T$, and the thrust from the car? Assume the car is travelling at constant velocity up the hill.

First, adapt this system by rotating it, to visualise the perpendicular forces a little more easily.

Perpendicular to $F$, we have $R = Wcos 30°$.

$W = 1500 times 9.8 = 14700text{ N}$, so $R = 12730.57text{ N}$

So, parallel to $F$, we have $T + text{Thrust} = F + Wsin 30°$.

By $F_{max} = textcolor{red}{mu} R$, the limit of friction is $F_{max} = 5092.23text{ N}$

Since we have $W$ and $F_{max}$, we can see that the minimum thrust and tension required to move the car up the slope is $5092.23 + Wsin 30° = 12442.23text{ N}$.