A 10 kg block sits on a surface (μₛ = 0.4, μₖ = 0.3, g = 10 m/s²). Normal force is 100 N, so maximum static friction is 40 N. You apply a 25 N horizontal force. What is the actual friction force on the block?
A25 N — static friction adjusts to balance the applied force exactly, since 25 N < 40 N
B40 N — static friction always acts at its maximum value μₛN
C30 N — kinetic friction applies because the block is in contact with the surface
D0 N — friction only activates when the block is moving
Static friction is a reaction force: it matches the applied force to maintain equilibrium, up to its maximum μₛN. With 25 N applied and a 40 N maximum, the block stays stationary with friction = 25 N. Option B is the classic mistake — μₛN is the ceiling, not the default value. Kinetic friction (option C) only applies once sliding begins, and it remains constant at μₖN regardless of the applied force.
Question 2 Multiple Choice
A wooden crate is being pushed across a concrete floor at constant velocity. To find the friction force opposing its motion, which coefficient should you use?
AKinetic friction coefficient μₖ — the crate is sliding, so kinetic friction applies
BStatic friction coefficient μₛ — constant velocity means equilibrium, so static friction is balancing the push
CEither — they are approximately equal for most surfaces
DThe average of μₛ and μₖ — transitional motion uses a blended value
Once surfaces are in relative sliding motion, kinetic friction applies — always. The fact that velocity is constant means the net force is zero, but that is a consequence of the applied push equaling μₖN, not an indication that static friction is at work. Static friction only applies when surfaces are not sliding. μₛ > μₖ for essentially all real material pairs, so using μₛ would give an overestimate.
Question 3 True / False
Static friction typically equals μₛN whenever the surfaces in contact are stationary.
TTrue
FFalse
Answer: False
Static friction is a variable reaction force in the range 0 ≤ f_s ≤ μₛN. It equals exactly what equilibrium demands — no more, no less. When no horizontal force acts on a stationary block, static friction is zero. As applied force increases, static friction increases to match it. Only at the moment of impending motion does it reach μₛN. Treating μₛN as the default value leads to incorrect free-body diagrams.
Question 4 True / False
It requires more force to start sliding two surfaces against each other than to sustain that sliding, because μₛ > μₖ.
TTrue
FFalse
Answer: True
This asymmetry is fundamental to friction analysis. The maximum static friction force (μₛN) exceeds the kinetic friction force (μₖN) because μₛ > μₖ for all common surface pairs. In practice, this produces the familiar 'snap': you push harder and harder until the object breaks loose, then it suddenly accelerates — because the force you were applying (just above μₛN) now exceeds the smaller kinetic resistance (μₖN). ABS brakes exploit this by keeping tires in the static regime.
Question 5 Short Answer
Explain why the friction force on a stationary block increases as you push harder, up to a point, and then suddenly drops when the block starts moving.
Think about your answer, then reveal below.
Model answer: Before the block moves, static friction is a reaction force that adjusts to maintain equilibrium — it exactly cancels the applied force, so the net force stays zero. When the applied force reaches μₛN (the maximum static friction), equilibrium can no longer be maintained and sliding begins. Once sliding, friction switches to kinetic friction f_k = μₖN, which is fixed and smaller than μₛN. Since the applied force now exceeds f_k, the net force is nonzero and the block accelerates.
The key is the distinction between static friction as a variable reaction force (0 to μₛN) and kinetic friction as a fixed value (μₖN). The 'snap' moment — when the block breaks loose and suddenly moves faster — occurs precisely because μₛ > μₖ. There is no smooth transition; once the static limit is exceeded, the friction force drops discontinuously from μₛN to μₖN.