# Weight and kinetic friction relationship

### Static and kinetic friction example (video) | Khan Academy

The coefficient of friction (static or kinetic) is a measure of how difficult Example 1 - A box of mass kg travels at constant velocity down an. The force of friction is a force that resists motion when two objects are in contact. If you look at Some common values of coefficients of kinetic and static friction. To determine the coefficient of kinetic friction between two selected surfaces by To cause motion, you need to slightly increase m, the hanging mass, by

In some places, especially Alaska in the winter, you must keep friction in your mind constantly as you drive, in order to avoid an accident.

You have to limit your speed in order to be able to stop at a reasonable distance, and to negotiate curves. This is why the two second rule "travel at a speed so that two seconds pass between the moment the car in front of you passes a landmark and the moment you pass the same landmark" is not valid for high speeds; your stopping distance increases exponentially as you go faster.

A higher coefficient of friction decreases your stopping distance. It is better, therefore, for your tire to be using static friction rather than kinetic friction.

If the tire is rolling along so that the surface touching the ground is never sliding, then static friction is acting to slow the car. If the wheels are locked and sliding, then kinetic friction is acting to slow the car.

In order to utilize static friction when you need to stop quickly, there are several options.

- Static and kinetic friction example
- Relationship between friction and weight

You can attempt to apply just enough brake to stay within the static range of friction and not too much to lock the tires.

This is the best option, in terms of stopping you the quickest, but it can be difficult to be that precise with the brake. It can be especially difficult if you are about to hit a moose.

Another option is pumping the brake, which has the effect of alternating the use of kinetic and static friction as the wheels lock and unlock.

The box is moving at a constant velocity, so that means the acceleration is zero. Solving for the kinetic force of friction gives: The coefficient of kinetic friction can be found from the normal force and the frictional force: This is actually a relatively large value for the coefficient, so it's not easy to move this box along this ramp.

Example 2 Consider another example involving an inclined plane, only this time there will be two boxes involved.

This is quite a challenging example, so don't be too intimidated if it looks tricky. Start by seeing if you agree with the free-body diagrams; if you understand those, you've made an important step in learning some physics.

Box 1, a wooden box, has a mass of 8. Box 2, a cardboard box, sits on top of box 1. It has a mass of 1.

### Kinetic Friction Formula

The coefficient of kinetic friction between the two boxes is 0. The two boxes are linked by a rope which passes over a pulley at the top of the incline, as shown in the diagram. The inclined plane is at an angle of What is the acceleration of each box?

The diagram for the situation looks like this: The next step is to draw a free-body diagram of each box in turn. To draw these, it helps to think about which way the boxes will accelerate. The two boxes are tied together, and the heavier box will win The pulley, by the way, simply changes the direction of the tension force.

### Static & Kinetic Friction

We're assuming that the pulley is massless and frictionless, so both boxes feel the same tension force. It's important to know the direction of the acceleration or, if you don't know, to guess and apply what you figured out or your guess consistently to both boxes. The free-body diagram of box 1 is relatively complicated, with a total of 6 forces appearing.

The free-body diagram for box 2 is a little easier to deal with, having 4 forces, so that's a good place to start.

## Kinetic Friction Formula

For box 2 - start by summing the forces in the y-direction, where there is no acceleration: This can be solved to give the normal force: Now find the net force in the x direction, where there is an acceleration up the slope: There are two unknowns in this equation, the tension T and ax, the acceleration. We can at least solve for T in terms of ax, like this: Now move on to box 1. Going back to the free-body diagram and summing forces in the y-direction gives: This equation can be solved to give the normal force associated with the interaction between the inclined plane and box 1: Things are a little more complicated in the x-direction, but adding up the forces gives: Moving the acceleration terms to the left side gives: