Question

multiple choice question
in the relationship ( f_{f, \text { static }} leq mu_{s} f_{n} ), what is the physical meaning of the ( leq ) sign in the expression?

the static friction force is a theoretical value that can never be reached.
the static friction force will always be slightly lower than ( mu_{s} f_{n} ).
all frictional values must be less than the coefficient of static friction values.
the static friction force can increase up to, but cannot exceed, the calculated value ( mu_{s} f_{n} ).

Explanation:

Static friction force \(F_{f,\text{static}}\) adjusts according to the applied force. It can increase up to the maximum value \(\mu_sF_N\) (when motion is about to start). Once the applied force exceeds \(\mu_sF_N\), kinetic friction takes over. The inequality \(F_{f,\text{static}}\leq\mu_sF_N\) means \(F_{f,\text{static}}\) can reach \(\mu_sF_N\) (at the verge of motion) but not exceed it.

• The first option is wrong because \(F_{f,\text{static}}=\mu_sF_N\) is reached at the point of impending motion.
• The second option is incorrect as \(F_{f,\text{static}}\) can equal \(\mu_sF_N\) (not just be slightly lower).
• The third option misinterprets the relationship; it's about the static - friction force and \(\mu_sF_N\), not all frictional values and \(\mu_s\).

Answer:

The static friction force can increase up to, but cannot exceed, the calculated value \(\mu_sF_N\).