Rowan tries to slide the family refrigerator. He applies a 50 N force, and it does not slide.
He doubles the force to 100 N. Still it does not slide.
Twice the force produced exactly the same stillness. Something is matching his force, newton for newton.
Before anything slides, static friction is not a fixed number. It adjusts.
The refrigerator does not slide, so the forces on it balance. While Rowan applies 50 N, the static friction from the floor is exactly 50 N the other way.
While he applies 100 N and it still does not slide, the static friction has grown to exactly 100 N.
Here is the rule: static friction adjusts to match the applied force up to a maximum of μ_s N, and only beyond that maximum does sliding begin.
μ_s is the pair's coefficient of static friction (the subscript s says static), and N is the normal force, so the maximum is the product μ_s N.
Give the refrigerator numbers. For this pair, μ_s = 0.30, and the normal force is 700 N. The maximum is 0.30 × 700 = 210 N.
Any applied force up to that maximum is matched exactly: apply 150 N, get 150 N of static friction; apply 209 N, get 209 N.
The maximum static friction on a dresser is 180 N. Beth applies a 120 N horizontal force, and the dresser does not slide. How large is the static friction on it?
The dresser does not slide, so static friction matches the applied force.
The applied force is 120 N, below the 180 N maximum.
So the static friction is exactly 120 N, not 180 N.
120 N: static friction matches the applied force, not its own maximum.
The maximum static friction on a storage chest is 260 N. Imre applies a 90 N horizontal force, and the chest does not slide. Enter the static friction on the chest in N.
The maximum static friction on a washing machine is 300 N. Tilda applies a 240 N horizontal force, and it does not slide. Enter the static friction on the machine in N.
Past the maximum, the matching fails. At 220 N applied, the refrigerator begins to slide.
A fridge pushed gently or firmly without budging feels a friction force exactly matching each push.
Beth says: 'While a cabinet stays put, raising the applied force raises the static friction on it.' Is she correct?
For a toolbox on a workbench, μ_s = 0.50 and the normal force is 380 N. Calculate the maximum static friction.
Write down the values in the question:
μ_s = 0.50
N = 380 N
maximum static friction = ?
Write down the equation:
maximum static friction = μ_s N
Substitute in the values:
maximum = 0.50 × 380
Simplify:
maximum = 190 N
The maximum static friction on the toolbox is 190 N.
The maximum static friction on a planter box is 420 N. A mover applies a 350 N horizontal force, and it does not slide. Enter the static friction on the box in N.
The applied force on a crate rises past the maximum static friction μ_s N. What happens?
For a shipping crate on a warehouse floor, μ_s = 0.60 and the normal force is 260 N. Enter the maximum static friction in N.