Solution:

🔹 Step-by-step:

• Block of mass M is on a frictionless incline of angle $\theta$
• Force
  $F$ is applied via a pulley system, split into two components:
  • One acts up along the incline: F
  • One acts horizontally, which when resolved along the incline becomes: F$F \cos \theta$

Total upward force along incline = $F + F \cos \theta$

Downward component of weight =$Mg \sin \theta$

🔹 For the block to just start moving upward:

$$F+F\cos \theta =Mg\sin \theta$$

$$F(1+\cos \theta )=Mg\sin \theta$$

$$F = \frac{Mg \sin \theta}{1 + \cos \theta}$$

Now use the identity:

$$\frac{\sin \theta}{1 + \cos \theta} = \cot\left(\frac{\theta}{2}\right)$$

 Final Answer:

$$\boxed{F = Mg \cot\left(\frac{\theta}{2}\right)}$$