Confusion regarding the direction of static friction and point of contact

Question

I'm doing a practice question for my Physics 1 class, and after solving this I believe the answer key is incorrect. However, I'd like to confirm this and make sure I'm not incorrectly deciding upon the direction of static friction in my free body diagram; it's not entirely clear to me:

A block is pushed up against a vertical wall by a force of 20N. The force is at an angle of 40 degrees above the horizontal. The coefficient of static friction between the block and the wall is 0.50. Find the maximum mass, in kg, that the force can prevent from sliding down.

Here is my free body diagram:

My equations:

$$20\sin(40)+\mu_sN=mg \tag{1}$$ $$N=20\cos(40) \tag{2}$$ $$m=\frac{20\sin(40)+(0.50)(20\cos(40))}{g} \tag{3}$$ $$m\approx2.1kg \tag{4}$$

My answer is m is approximately equal to $2.1$ kg, assuming the direction of static friction is upward. The answer key says that m is approximately equal to $0.52$ kg. I noticed that this would be the case if the direction of static friction were downward.

From a little bit of research, it's my understanding that the direction of static friction would be dependent upon whether the vertical component of the applied force $(F\sin(40))$ was greater or less than the weight of the object ($mg$). If $F \sin (40)$ is less $mg$, then the direction of static friction should be upward, and if greater, downward. Adding to the confusion is that the weight of the object, in this case, is an unknown. The problem is asking for the maximum mass.

So my question is, is there anyone that can provide more clarification on the concepts and this particular problem? What is the direction of static friction in this particular problem, and why?

Answer 1

Friction is always opposite to the resultant force as you pointed out. The modulus of the static force will be equal to the resultant force if this is less than $\mu N$.

So in your exercise if you want to prevent it from sliding down, you can assume than $mg > F_y$, where $F_y = 20sin(40)$. This would be the case if $m>\frac{F_y}{g} = 1.31$. And your answer would be correct.

However, if $F_y > mg$ (if there were no friction it would slide upwards) You can find the minimum mass that is necessary in order to "stop" $F$. So if you assume this:

$$F_y - m_{min}g - \mu N= 0$$ $$m_{min} = \frac{F_y-\mu N}{g}$$ $$m_{min} = 0.53$$

Which pretty much fit the answer key you have been given. In conclusion, to prevent it from sliding the mass should be between $0.53$ and $2.1$, but the maximum mass to prevent from sliding down is the one you calculated.