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 100\text{ 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 50\text{ N}.

Then we have the normal reaction force R = 100\text{ 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 30\degree. 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 = W\cos 30°.

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

So, parallel to F, we have T + \text{Thrust} = F + W\sin 30°.

By F_{max} = \textcolor{red}{\mu} R, the limit of friction is F_{max} = 5092.23\text{ 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 + W\sin 30° = 12442.23\text{ N}.