Suppose F , F_N , and f represent the magnitudes of the contact...

The contact force $F$ is the vector sum of the normal force $F_N$ and the friction force $f$. Since the normal force and friction are perpendicular to each other, the magnitude of the contact force is given by $F = \sqrt{F_N^2 + f^2}$.

Since neither $F_N$ nor $f$ is zero, we have $F^2 = F_N^2 + f^2 > f^2 \Rightarrow F > f$.

Similarly, $F^2 = F_N^2 + f^2 > F_N^2 \Rightarrow F > F_N$.

Squaring the sum of the magnitudes gives $(F_N + f)^2 = F_N^2 + f^2 + 2F_N f$. Since $2F_N f > 0$, we get $(F_N + f)^2 > F^2 \Rightarrow F < F_N + f$.

Squaring the difference gives $(F_N - f)^2 = F_N^2 + f^2 - 2F_N f$. Since $2F_N f > 0$, we get $(F_N - f)^2 < F^2 \Rightarrow F > |F_N - f| \ge F_N - f$.

Combining these inequalities yields $F_N - f < F < F_N + f$.

The friction force $f$ can be greater than the normal force $F_N$ if the coefficient of friction $\mu > 1$. Thus, $F_N > f$ is not necessarily true.