Mgsin θ = μ gcos θ, and sinθ=μcosθ, μ=tanθ.
A. If the external force F is vertically downward, and (F+mg)sinθ=μ(F+mg)cosθ, the resultant force on the whole body is zero, and B still drops at a constant speed. From the overall research, there is no static friction on the ground along the external force in the horizontal direction, so A is correct.
B, if the external force F inclines to the lower left, let the included angle between F and the inclined plane be α. The downward force of an object along the inclined plane is Fcosα+mgsinθ, and the sliding friction is μ(Fsinα+mgcosθ), in which the sizes of Fcosα and μFsinα are uncertain. If Fcosα=μFsinα,
B is still moving at a constant speed. If fcos α < μ fsinα, B slows down and slides down. If fcos α > μ fsinα, B accelerates the decline. For the inclined plane, if the pressure of B on the inclined plane is n and the sliding friction force is f, then f = μ n. Through orthogonal decomposition,
The horizontal component of f is fcosθ, the horizontal component of n is Nsinθ, and the horizontal component of n is right. If f=μN=Ntanθ, fcosθ=Nsinθ is obtained, and the ground obtained from the equilibrium condition has no static friction.
You can get c and d in the same way. No matter what direction F takes in the vertical plane, there is no static friction on the ground. So, C is wrong and D is right.
So choose AD.