Question

Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If the coefficient of static friction between the brick and the floor is 0.4, what is the minimum value of F required to start the brick to move?

Answer 1

Let $F_{1}=F_{2}=F$.

Normal force equals (using Newton's third law) $N=mg+F\sin30^{o}-F\sin37^{o}$.

$F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o})$, but $F_{f}\leq F(\cos30^o+\cos37^o)$ for all $F_{f}$ (in order to start moving the break). Therefore $F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o})$, solving for $F$: $F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}$