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

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

Show that for the object $\left|{\mu }_k\right|<\left|{\mu }_s\right|$.

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

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

${\overrightarrow{F}}_g={\overrightarrow{F}}_n=mg$

${\overrightarrow{F}}_n=20·10$

${\overrightarrow{F}}_n=200\mathrm{N}$

• At the threshold of motion

  • For applied force, $F$

$\left|F\right|=\left|F_{f,s}\right|=\left|{\mu }_s{\overrightarrow{F}}_n\right|=50\mathrm{N}$

• Moving at constant velocity

  • For applied force, $F$

$\left|F\right|=\left|F_{f,k}\right|=\left|{\mu }_k{\overrightarrow{F}}_n\right|=40\mathrm{N}$

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

• At the threshold of motion

$50=\left|{\mu }_s\right|·200$

$\left|{\mu }_s\right|=\frac{50}{200}$

$\left|{\mu }_s\right|=0.25$

• Moving at constant velocity

$40=\left|{\mu }_k\right|·200$

$\left|{\mu }_k\right|=\frac{40}{200}$

$\left|{\mu }_k\right|=0.20$

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

$0.25>0.20$

$\left|{\mu }_s\right|>\left|{\mu }_k\right|$