A fancy bathroom scale is calibrated in Newtons. This scale is put on a ramp, which is at a $40^\circ$ angle to the horizontal. A box is then put on the scale and the box-scale system is then pushed up the ramp by a horizontal force $F$. The system slides up the ramp at a constant speed. If the bathroom scale reads $R$ and the coefficient of static friction between the system and the ramp is $0.40$, what is $\frac{F}{R}$? Round to the nearest thousandth.

[i](Proposed by Ahaan Rungta)[/i]
1. **Identify the forces acting on the box-scale system:**
   - The gravitational force \( mg \) acting vertically downward.
   - The normal force \( R \) exerted by the ramp on the box, perpendicular to the ramp.
   - The frictional force \( f \) acting parallel to the ramp, opposing the motion.
   - The applied horizontal force \( F \).

2. **Resolve the gravitational force into components:**
   - The component parallel to the ramp: \( mg \sin \theta \).
   - The component perpendicular to the ramp: \( mg \cos \theta \).

3. **Set up the equilibrium conditions:**
   - Since the system slides up the ramp at a constant speed, the net force along the ramp must be zero.
   - The normal force \( R \) is equal to the perpendicular component of the gravitational force: 
     \[
     R = mg \cos \theta
     \]

4. **Frictional force:**
   - The frictional force \( f \) is given by \( f = \mu R \), where \( \mu \) is the coefficient of static friction.
   - Substituting \( R \):
     \[
     f = \mu mg \cos \theta
     \]

5. **Horizontal force \( F \):**
   - Resolve \( F \) into components parallel and perpendicular to the ramp:
     - The component of \( F \) parallel to the ramp: \( F \cos \theta \).
     - The component of \( F \) perpendicular to the ramp: \( F \sin \theta \).

6. **Equilibrium along the ramp:**
   - The net force along the ramp is zero:
     \[
     F \cos \theta - mg \sin \theta - f = 0
     \]
   - Substituting \( f \):
     \[
     F \cos \theta - mg \sin \theta - \mu mg \cos \theta = 0
     \]

7. **Solve for \( F \):**
   - Rearrange the equation:
     \[
     F \cos \theta = mg \sin \theta + \mu mg \cos \theta
     \]
   - Factor out \( mg \):
     \[
     F \cos \theta = mg (\sin \theta + \mu \cos \theta)
     \]
   - Divide both sides by \( \cos \theta \):
     \[
     F = mg \left( \frac{\sin \theta + \mu \cos \theta}{\cos \theta} \right)
     \]

8. **Express \( \frac{F}{R} \):**
   - Recall that \( R = mg \cos \theta \):
     \[
     \frac{F}{R} = \frac{mg \left( \frac{\sin \theta + \mu \cos \theta}{\cos \theta} \right)}{mg \cos \theta}
     \]
   - Simplify:
     \[
     \frac{F}{R} = \frac{\sin \theta + \mu \cos \theta}{\cos \theta}
     \]

9. **Substitute the given values:**
   - \( \theta = 40^\circ \)
   - \( \mu = 0.40 \)
     \[
     \frac{F}{R} = \frac{\sin 40^\circ + 0.40 \cos 40^\circ}{\cos 40^\circ}
     \]

10. **Calculate the trigonometric values:**
    - \( \sin 40^\circ \approx 0.6428 \)
    - \( \cos 40^\circ \approx 0.7660 \)
      \[
      \frac{F}{R} = \frac{0.6428 + 0.40 \times 0.7660}{0.7660}
      \]
      \[
      \frac{F}{R} = \frac{0.6428 + 0.3064}{0.7660}
      \]
      \[
      \frac{F}{R} = \frac{0.9492}{0.7660} \approx 1.239
      \]

The final answer is \(\boxed{1.239}\)