Static Friction Force Formula (College Board AP® Physics 1: Algebra-Based)

Study Guide

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Written by: Ann Howell

Reviewed by: Caroline Carroll

Static friction force equation

The maximum magnitude of the static friction force exerted on an object is the product of the normal force the surface exerts on the object and the coefficient of static friction
The possible magnitudes of static friction is given by the inequality:

$|{\vec{F}}_{f, s}| ⩽ |μ_{s} {\vec{F}}_{n}|$

Where:
- $|{\vec{F}}_{f, s}| =$ magnitude of static friction, measured in $N$
- $μ_{s} =$ coefficient of static friction
- ${\vec{F}}_{n} =$ normal reaction force between the two contact surfaces, measured in $N$
- $⩽$ means less than or equal to

Coefficient of kinetic vs static friction

The coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces
- As explained in the Static friction study guide, the force of static friction is greater than the force of kinetic friction
- We can rearrange this inequality to show that the coefficient of static friction is greater

$|F_{f, k}| < |F_{f, s}|$

$|μ_{k} {\vec{F}}_{n}| < |μ_{s} {\vec{F}}_{n}|$

$|μ_{k}| < |μ_{s}|$

Worked Example

An object of mass $20 kg$ is at rest on a rough horizontal surface. When a force of $50 N$ is applied, the object is on the point of its threshold of motion and will not move, when a force of $40 N$ is applied, this force is enough to keep it moving at a constant velocity.

Show that for the object $|μ_{k}| < |μ_{s}|$ .

Answer:

Step 1: Draw a free body diagram of the two scenarios

When the object is at its threshold of motion, the magnitude of the static frictional force is equal to the magnitude of the applied force
When the object is moving at a constant velocity, the kinetic frictional force is equal to the applied force

A block at rest with 50N applied force, and in motion with 40N applied force, depicting static and kinetic friction thresholds.

Step 2: Determine the magnitudes of friction and the normal force in each scenario

Normal force is created due to the gravitational force acting on the object
- The normal force is the same in each scenario

${\vec{F}}_{g} = {\vec{F}}_{n} = m g$

${\vec{F}}_{n} = 20 \cdot 10$

${\vec{F}}_{n} = 200 N$

At the threshold of motion
- For applied force, $F$

$|F| = |F_{f, s}| = |μ_{s} {\vec{F}}_{n}| = 50 N$

Moving at constant velocity
- For applied force, $F$

$|F| = |F_{f, k}| = |μ_{k} {\vec{F}}_{n}| = 40 N$

Step 3: Calculate the magnitude of the kinetic and static friction

At the threshold of motion

$50 = |μ_{s}| \cdot 200$

$|μ_{s}| = \frac{50}{200}$

$|μ_{s}| = 0.25$

Moving at constant velocity

$40 = |μ_{k}| \cdot 200$

$|μ_{k}| = \frac{40}{200}$

$|μ_{k}| = 0.20$

Step 4: Compare the magnitudes of the kinetic and static friction

$0.25 > 0.20$

$|μ_{s}| > |μ_{k}|$

Last updated: 19 October 2024

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