You are sliding a block with mass m up a ramp inclined at an angle of $\theta$ with respect to the horizontal where the coefficient of static friction between the block and the ramp is $\mu_s$. What is the maximum horizontal force $F$ you can apply to the ramp so that the block does not slide up the ramp, assuming $\theta > 90 - \theta_s$, where $\theta_s$ is the angle between the ramp and horizontal when the force of static friction is maximum (so that $\tan \theta_s = \mu_s$).
Let $W$ be the weight and $f_s$ be the force of static friction. Take the positive x-axis to be down the ramp and the positive y-axis to be perpendicular to the ramp and pointing above the ramp. I can use Newton's first law to get the following equations:
$W\sin \theta - F\cos \theta + f_s = 0\tag{1}$ $-W\cos\theta - F\sin\theta + N = 0\tag{2}$
The following image illustrates how these equations were deduced:
Solving the equations gives $f_s = F\cos\theta - W\sin\theta \leq f_s^{\max} = \mu_s N = \mu_s (W\cos \theta + F\sin\theta).$ So $F(\cos\theta - \mu_s \sin\theta) \leq W(\mu_s \cos\theta +\sin\theta) \Rightarrow F \leq \frac{W(\mu_s + \tan\theta)}{1 - \mu_s\tan\theta} =: F_\max$. But since $\theta > 90-\theta_s,$ $F_\max$ is negative, so by the diagram, it points to the right. So does this mean that any horizontal force applied to the left will not cause the block to slide up? I tried drawing another free body diagram but I still don't understand why any horizontal force applied to the left would not cause the block to slip in this case, though this may be wrong.
let's look at this from a different perspective for a second...
$\theta$ is the the inclination of the ramp. $\theta\gt90-\theta_s$. if $\theta_s\lt45$ then $\theta\gt\theta_s$. However, if $\theta_s\gt45$ then $\theta\lt\theta_s$ until the point where $\theta=\theta_s$ and then it becomes $\theta\gt\theta_s$.
obviously it becomes easier to move the object up the ramp the smaller the angle gets so we're only interested in the $\theta\gt\theta_s$ case but this case happens for $\theta_s$ bigger or smaller than 45. since $\theta\gt\theta_s$ then $\tan\theta\gt\tan\theta_s=\mu_s$ which leads us to $1-\mu_s\tan\theta>0$ when $\mu_s\tan\theta<1$ this of course can still happen when $\theta$ isn't too big or $\mu_s$ is sufficiently small or both.
what you're saying though isn't all wrong... but I believe you either mean
or your image is mirrored...
but why is it true? it's simply because by applying horizontal force to the right you're pulling the object away from the ramp in the direction of it sliding down (even though it's not sliding nor moving down) so technically that is the same as the object is not sliding up.