A particle of mass $m$ rests on a horizontal floor with which it has a coefficient of static friction $\mu$. It is desired to make the body move by applying the minimum possible force $F$. Find the magnitude of $F$ and the direction in which it has to be applied.

Step 1: Set up the forces acting on the particle.

Let the force $F$ be applied at an angle $\theta$ above the horizontal.

$$F_x = F\cos\theta$$ $$F_y = F\sin\theta$$

Step 2: Establish vertical equilibrium.

Balance the forces in the vertical direction to find the normal reaction force $N$.

$$N + F\sin\theta = mg$$ $$N = mg - F\sin\theta$$

Step 3: Relate the horizontal force to the maximum static friction.

To make the body move, the horizontal component of the force must equal the limiting static friction $f = \mu N$.

$$F\cos\theta = \mu(mg - F\sin\theta)$$

Step 4: Express the force $F$ as a function of $\theta$.

Rearrange the equation to solve for $F$.

$$F\cos\theta + \mu F\sin\theta = \mu mg$$ $$F(\cos\theta + \mu\sin\theta) = \mu mg$$ $$F = \frac{\mu mg}{\cos\theta + \mu\sin\theta}$$

Step 5: Maximize the denominator to minimize the force.

To find the minimum force $F_{\min}$, maximize the denominator $D(\theta) = \cos\theta + \mu\sin\theta$ by setting its derivative with respect to $\theta$ to zero.

$$\frac{dD}{d\theta} = -\sin\theta + \mu\cos\theta = 0$$ $$\tan\theta = \mu$$ $$\theta = \tan^{-1}\mu$$

Step 6: Calculate the maximum value of the denominator.

Using the trigonometric identity for the maximum value of $a\cos\theta + b\sin\theta$:

$$D_{\max} = \sqrt{1+\mu^2}$$

Step 7: Determine the minimum force magnitude.

Substitute the maximum value of the denominator back into the expression for $F$.

$$F_{\min} = \frac{\mu mg}{\sqrt{1+\mu^2}}$$

Answer: The minimum force magnitude is $F_{\min} = \dfrac{\mu mg}{\sqrt{1+\mu^2}}$ applied at an angle $\theta = \tan^{-1}\mu$ above the horizontal.