If a pushing force at an angle α with horizontal is applied on a block of mass m placed on a horizontal table and angle of friction is β, then what is the minimum magnitude of force required to move?

To find the minimum magnitude of the force required to move the block, we need to consider the forces acting on the block when a force is applied at an angle.

Given:
Mass of the block = m
Angle of the applied force with horizontal = α
Angle of friction = β (the angle of the frictional force relative to the normal force)
Coefficient of friction = μ (not explicitly given, but it is assumed to be related to the angle of friction)
Approach:
Force Components:

The pushing force has two components:
Horizontal component: F * cos(α)
Vertical component: F * sin(α)
Normal Force: The normal force (N) is the reaction force that balances the vertical forces. The vertical forces acting on the block are:

The weight of the block: mg (acting downward)
The vertical component of the applied force: F * sin(α)
Therefore, the normal force is: N = mg - F * sin(α)

Frictional Force: The frictional force (f) is given by: f = μ * N Since the angle of friction is β, the coefficient of friction (μ) can be written as: μ = tan(β)

Thus, the frictional force becomes: f = tan(β) * N = tan(β) * (mg - F * sin(α))

Condition for Motion: For the block to start moving, the horizontal component of the applied force must overcome the frictional force. This gives the condition: F * cos(α) = f

Substituting the expression for f, we get: F * cos(α) = tan(β) * (mg - F * sin(α))

Solve for F: Rearranging the equation: F * cos(α) + F * tan(β) * sin(α) = tan(β) * mg

Factor out F: F * (cos(α) + tan(β) * sin(α)) = tan(β) * mg

Solving for F: F = (tan(β) * mg) / (cos(α) + tan(β) * sin(α))

Final Answer:
The minimum magnitude of the force required to move the block is:

F = (tan(β) * mg) / (cos(α) + tan(β) * sin(α))