If we have an object on a frictionless inclined bench, why does the minimum force applied to keep it steady have to be parallel to the incline (or perpendicular to the reaction force)?
In other words, why does the angle θ in the image have to be 0 in order for the force to be minimum?
Consider we know the weight of the object and the reaction of the incline.
The minimum force is in the direction the object would move without that force. Clearly the object would move along the inclined plain, down and right in your diagram.
Put another way, its the component of the force along the direction of travel that matters. More mathematically, it's the dot product of the force with the acceleration unit vector, with negative values contributing to cancelling the acceleration, and positive values enhancing it. The dot product is -1 when the force is exactly opposite the acceleration direction. A sideways force would do nothing, and has a dot product of 0. A force pushing down the incline has a dot product of 1, and just like the math says, would make the object go down the incline faster.
The force you're applying would have an effect that counteracts the body's tendency to move. That effect will be calculated by [applied force] $\cos \theta$.
If $\theta$ is not zero, $\cos{\theta}$ would be less than 1, so you would have to apply a larger force to produce the same result. For example, 10 degrees for $\theta$ would equate to a force parallel to the plane of 0.9848 times the applied force. You would need a larger applied force (approximately 1.015 times) to produce the same effect.
When $\theta = 0$, $\cos{\theta} = 1$.
Since I posed the question in comment, here is how to treat the friction case, and of course at the end it reduces to the expected friction free result.
Assume the applied force has components $f_x$ parallel (up) the plane and a normal component $f_y$ downward into the plane:
The normal force between block and plane is:
$$f_n = W\cos(t) + f_y$$
The force $f_x$ required to hold against sliding, assuming static coulomb friction with coefficient $u$ is
$$f_x = W\sin(t) - u f_n = W\sin(t) - u (W\cos(t) + f_y)$$
The magnitude of the applied force is then
$$f_{mag} = \sqrt{f_x^2+f_y^2}$$
without showing all the messy steps, setting $d f_{mag} / d f_y == 0$ we can solve for $f_y$:
$$f_y = u W\sin(t)\cdot\dfrac{1 - u /\tan(t)}{1 + u^2}$$
and
$$f_x = W\sin(t) ( 1 - u\cdot\dfrac{1/\tan(t) + u}{1 + u^2})$$
For the friction free case ($u=0$) this results in the expected $f_x=W\sin(t),f_y=0$ (ie parallel to the plane) but with friction angling the force toward the plane reduces the required force magnitude.
