Static Friction: Definition, Coefficient & Equation (w/ Examples)

Static friction is a force that must be ​overcome​ for something to get going. For example, someone can push on a stationary object like a heavy couch without it moving. But, if they push harder or enlist a strong friend's help, it will overcome the friction force and move.

While the couch is still, the ​force of static friction is balancing the applied force of the push​. Therefore, ​the force of static friction increases in a linear fashion with the applied force acting in the opposite direction​, until it reaches a maximum value and the object just begins to move. After that, the object no longer experiences resistance from static friction, but from kinetic friction.

The static friction is usually a larger frictional force than kinetic friction – it's harder to start pushing a couch along the floor than to keep it going.

Coefficient of Static Friction

Static friction results from molecular interactions between the object and the surface it is on. Thus, different surfaces provide different amounts of static friction.

The coefficient of friction that describes this difference in static friction for different surfaces is ​μs.​ It can be found in a table, like the one linked with this article, or calculated experimentally.

Equation for Static Friction


  • Fs​ = force of static friction in newtons (N)
  • μs ​= coefficient of static friction (no units)
  • FN ​= normal force between the surfaces in newtons (N)

Maximum static friction is achieved when the inequality becomes an equality, at which point a different force of friction takes over as the object begins to move. (The force of kinetic, or sliding friction, has a different coefficient associated with it called the coefficient of kinetic friction and denoted ​μk .)

Example Calculation With Static Friction

A child tries to push a 10-kg rubber box horizontally along a rubber floor. The static friction coefficient is 1.16. What is the maximum force the child can use ​without​ the box moving at all?

[insert a free body diagram showing the applied, frictional, gravitational and normal forces on the still box]

First, note that the net force is 0 and find the normal force of the surface on the box. Since the box is not moving, this force must be equal in magnitude to the gravitational force acting in the opposite direction. Recall that ​Fg = mg​ where ​Fg​ is the force of gravity, ​m​ is the mass of the object and ​g​ is the acceleration due to gravity on Earth.


F_N=F_g=10\times 9.8 = 98\text{ N}

Then, solve for Fs with the equation above:

F_s=\mu_s\times F_N=1.16\times 98 = 113.68\text{ N}

This is the maximum static frictional force that will oppose the box's motion. Therefore, it is also the maximum amount of force the child can apply without the box moving.

Note that, so long as the child is applying any force ​less than the maximum value of static friction​, the box still won't move!

Static Friction on Inclined Planes

Static friction doesn't only oppose applied forces. It keeps objects from sliding down hills or other tilted surfaces, resisting the pull of gravity.

On an angle, the same equation applies but trigonometry is needed to resolve the force vectors into their horizontal and vertical components.

Consider a 2-kg book resting on an inclined plane at 20 degrees. For the book to remain still, the ​forces parallel to the inclined plane must be balanced​. As the diagram shows, the force of static friction is parallel to the plane in the upwards direction; the opposing downwards force is from gravity – in this case though, ​only the horizontal component of the gravitational force​ is balancing static friction.

By drawing a right triangle off the force of gravity in order to resolve its components, and doing a little geometry to find that the angle in this triangle is equal to the angle of incline of the plane, the ​horizontal component of the gravitational force​ (the component parallel to the plane) is then:

F_{g,x}=mg\sin{\theta}=2\times 9.8\times\sin{20}=6.7\text{ N}

This must be equal to the force of static friction holding the book in place.

Another value possible to find in this analysis is the coefficient of static friction. The normal force is ​perpendicular​ to the surface on which the book rests. So this force must be ​balanced with the vertical component​ of gravity's force:

F_{g,y}=mg\cos{\theta}=2\times 9.8\times\cos{20}=18.4\text{ N}

Then, rearranging the equation for static friction:


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