The question is: Determine the minimal F necessary to keep the block from moving if the coefficient of friction is $\mu=.3$.
I know the horizontal force to the left is $cos (70)(50)=17.1 lbs$
My calculation for the friction force is $\mu(Fnormal)=\vert(.3)(sin (70)(50)-100\vert$. This frictional force is greater than the force to the left so I would calculate that the minimal force F to prevent the block from moving would be 0, since friction alone prevents it from moving.
Is this physics logic correct? I'm using forces to left and right rather than using the angle of $110$. I realize that the $cos(110)$ is negative, but I get around this by calling it a Force to the left opposite the forces to the right.
Best Answer
Your Physics is basically correct, but you are missing a bracket ... $$\mu(Fnormal)=\vert(.3)(50 \sin (70)-100)\vert \approx 15.9 $$