To solve this problem, we analyze the forces acting on the block at the moment it is about to move up the incline. We use a coordinate system with the x-axis parallel to the incline (upwards) and the y-axis perpendicular to it.

Let the mass of the block be $m$ and the acceleration due to gravity be $g$. The forces acting on the block are:

1. Gravitational force, $mg$, acting vertically downwards.
2. Applied force, $F$, at an angle $\beta$ to the incline.
3. Normal force, $N$, perpendicular to the incline.
4. Maximum static friction force, $f_s$, acting down the incline.

We apply Newton's first law (condition for equilibrium) along the x and y axes.

1. Force Balance Equations Sum of forces perpendicular to the incline (y-direction):

   $$ \Sigma F_y = N + F\sin\beta - mg\cos\theta = 0 $$ $$ N = mg\cos\theta - F\sin\beta $$

   The friction force is at its maximum value, $f_s = \mu N$.

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

   Sum of forces parallel to the incline (x-direction):

   $$ \Sigma F_x = F\cos\beta - mg\sin\theta - f_s = 0 $$ $$ F\cos\beta - mg\sin\theta - \mu (mg\cos\theta - F\sin\beta) = 0 $$

2. Derive the Expression for F We rearrange the equation to solve for $F$:

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

3. Minimize the Force F The problem states that the coefficient of friction is $\mu = \tan\varphi$. Substituting this into the expression for $F$:

   $$ F = mg \frac{\sin\theta + (\tan\varphi)\cos\theta}{\cos\beta + (\tan\varphi)\sin\beta} $$

   By writing $\tan\varphi = \frac{\sin\varphi}{\cos\varphi}$ and multiplying the numerator and denominator by $\cos\varphi$, we get:

   $$ F = mg \frac{\sin\theta\cos\varphi + \sin\varphi\cos\theta}{\cos\beta\cos\varphi + \sin\beta\sin\varphi} $$

   Using the trigonometric angle sum and difference identities:

   $\sin(A+B) = \sin A \cos B + \cos A \sin B$ $\cos(A-B) = \cos A \cos B + \sin A \sin B$

   The expression for $F$ simplifies to:

   $$ F = mg \frac{\sin(\theta+\varphi)}{\cos(\beta-\varphi)} $$

   To minimize the force $F$, the denominator $\cos(\beta-\varphi)$ must be maximized. The maximum value of the cosine function is 1. This occurs when its argument is zero.

   $$ \cos(\beta-\varphi) = 1 \implies \beta - \varphi = 0 $$

   Therefore, the angle $\beta$ that minimizes the force $F$ is:

   $$ \beta = \varphi $$

(Alternative Calculus Method) To minimize $F$, we can maximize the denominator $D(\beta) = \cos\beta + \mu\sin\beta$. Taking the derivative with respect to $\beta$ and setting it to zero:

Since $\mu = \tan\varphi$, we find that $\tan\beta = \tan\varphi$, which gives $\beta = \varphi$.