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# Friction forces

## What is the molecular origin of friction?

### * At small speeds friction is nearly independent of speed.

(text above adapted from ABE Advanced Level Physics, Derived from College Physics by OpenStax College)

The molecular origin of friction

http://www.drcruzan.com/Friction.html

http://archive.cnx.org/contents/904e502e-c9ec-4759-b7d1-3827fd3ab274@1/5-2-friction

https://www.quora.com/At-an-atomic-level-what-is-the-difference-between-static-and-kinetic-friction

Excerpted from:

### Friction is proportional to the force pressing the surfaces together.

 Drawing the frictional force on a free body diagram. Notice that the frictional force’s vector is (1) parallel to the force, (2) opposite the direction of motion -or intended motion.
Tension

### Tension exists in any body that is pulled by to opposing forces.  Typically we talk about ropes and chains as being in tension but any body can be in put in tension.

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A block of mass m = 0.75 Kg is pulled along a table at constant velocity. If it takes 2.0 N of force to maintain constant velocity, calculate the coefficient of friction for this system.

Solution: First draw a diagram of the situation. The 0.75 Kg mass exerts a force (F = mg) on the table, and the normal force is equal to it. I have not bothered to make these two forces of opposite sign but, strictly speaking, they are opposite vector forces. In these cases of constant velocity, there is no acceleration, so all forces must be balanced. Therefore it doesn’t matter how fast the block is moving, only that it is moving at constant velocity. Then the friction force Ff is equal to the pulling force of 2.0 N (but again of opposite sign, which I’ll ignore).

Now it’s a simple matter of rearranging the friction equation and using the normal and friction forces to calculate the coefficient of friction. Notice that in calculating μ, the units cancel.

Example 2

A 4.5 Kg mass is pulled up a 30˚ ramp at constant velocity by a falling 8.0 Kg mass, as shown. Calculate the coefficient of friction.

Solution: Here’s the diagram.

All of the relevant calculations are shown above. Notice that this is another of those problems that makes μ easy to find because the force up the ramp (22.0 N) is equal to the frictional force. The normal force of 38.2 N gives the result:

## Learning Standards

2016 Massachusetts Science and Technology/Engineering Curriculum Framework

3-PS2-1. Provide evidence to explain the effect of multiple forces, including friction, on an
object. Include balanced forces that do not change the motion of the object and
unbalanced forces that do change the motion of the object.

HS-PS2-1. Analyze data to support the claim that Newton’s second law of motion is a
mathematical model describing change in motion (the acceleration) of objects when
acted on by a net force… Forces can include contact forces, including friction.

A FRAMEWORK FOR K-12 SCIENCE EDUCATION: Practices, Crosscutting Concepts, and Core Ideas
PS2.B: TYPES OF INTERACTIONS
What underlying forces explain the variety of interactions observed?

All forces between objects arise from a few types of interactions… Any two objects in contact also exert forces on each other that are electromagnetic in origin. These forces result from deformations of the objects’ substructures and the electric charges of the particles that form those substructures (e.g., a table supporting a book, friction forces).

Massachusetts Science and Technology/Engineering Curriculum Framework (2006)

1. Motion and Forces. Central Concept: Newton’s laws of motion and gravitation describe and predict the motion of most objects.
1.6 Distinguish qualitatively between static and kinetic friction, and describe their effects on the motion of objects.