Solution:

Let's consider a block of weight w placed on a horizontal table. A pushing force is applied to the block at an angle θ to the horizontal. The angle of friction is given as ∅.

To determine the magnitude of the force required to move the block, we need to analyze the forces acting on the block.

1. Normal Force (N):
The weight of the block w acts vertically downwards. According to Newton's third law, the table exerts an equal and opposite force called the normal force (N) on the block in the upward direction. The normal force can be calculated as N = w.

2. Frictional Force (Ff):
The frictional force acts parallel to the surface and opposes the motion of the block. The magnitude of the frictional force can be given by Ff = μN, where μ is the coefficient of friction. In this case, the angle of friction is ∅, so μ = tan(∅). Therefore, Ff = tan(∅)N.

3. Applied Force (F):
The pushing force F is applied at an angle θ to the horizontal. We need to resolve this force into its horizontal and vertical components.

- The horizontal component of the applied force, Fx = Fcos(θ).
- The vertical component of the applied force, Fy = Fsin(θ).

4. Equilibrium in the Horizontal Direction:
For the block to move, the applied force must overcome the frictional force. So, Fx = Fcos(θ) = Ff.

5. Equilibrium in the Vertical Direction:
In the vertical direction, the block is not moving vertically, so the sum of the vertical forces must be zero. This gives us Fy + N - w = 0.

Now, let's substitute the values of Fx, Ff, and N into the equilibrium equations:

Fcos(θ) = tan(∅)N
Fcos(θ) = tan(∅)w

Solving for F, we get:

F = (w * tan(∅)) / cos(θ)

Hence, the magnitude of the force required to move the block is equal to (w * tan(∅)) / cos(θ), which matches the correct answer.