Let F, F_N and f denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero.
(a) F > F_N
(b) F > f
(c) F_N > f
(d) F_N − f < F < F_N + f.

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Solution

(a) F > F_N
(b) F > f
(d) F_N − f < F < F_N + f

The system is in equilibrium condition when F = f.
Hence, the net horizontal force is zero.

f = μF_N
F > F_N
f = F_Nand 0 ≤ μ ≤ 1
Therefore, we can say that F > f. So the net horizontal force is nonzero.

F > f, and so the net horizontal force is zero.

F_N > f $\Rightarrow$ F_N > μF_N $\Rightarrow$ μ < 1

Here, the given relation between F and f i.e

F > f and f = μF_N will not be satisfied So it cannot be said that the net horizontal force is zero or nonzero.

F_N − f < F < F_N + f
∵ f = μF_N
$\frac{f}{μ} - f < F < \frac{f}{μ} + f f (\frac{1 - μ}{μ}) < F < f (\frac{1 + μ}{μ})$
For the above relation, we can say that F ≠ f and so the net horizontal force is nonzero.

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Let F, FN and f denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero. (a) F > FN (b) F > f (c) FN > f (d) FN − f < F < FN + f.

Let F, F_N and f denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero.
(a) F > F_N
(b) F > f
(c) F_N > f
(d) F_N − f < F < F_N + f.