A body of mass M is placed on a rough horizontal surface having a...

The force exerted by the surface on the body is the resultant contact force, which is the vector sum of the normal reaction $N$ and the force of static friction $f$.

$F = \sqrt{N^2 + f^2}$

Since the body is in vertical equilibrium on a horizontal surface, the normal reaction is equal to the weight of the body.

$N = Mg$

The body remains stationary under the applied horizontal force, so the static friction $f$ is self-adjusting. Its value ranges from zero to the limiting friction.

$0 \leq f \leq \mu N$

$0 \leq f \leq \mu Mg$

Substituting the minimum and maximum values of $f$ into the expression for $F$:

When $f = 0$, F_\min = \sqrt{(Mg)^2 + 0} = Mg

When $f = \mu Mg$, F_\max = \sqrt{(Mg)^2 + (\mu Mg)^2} = Mg\sqrt{1 + \mu^2}

Therefore, the total force $F$ exerted by the surface satisfies the condition:

$Mg \leq F \leq Mg\sqrt{1 + \mu^2}$