A student wants to slide a steel 15 kg mass across a steel table. The static coefficient of friction is 0.74. How much force must the student apply in order to start the box moving? Use _g_ = 10 $\frac{m}{s^2}$. ### ==Answer== $\vec{F}_f = \mu_s \cdot \vec{F}_N$ Substitute $mg$ into $\vec{F}_N$ $\vec{F}_f = \mu_s \cdot mg$ $\vec{F}_f$ = 0.74 $\cdot$ 15 kg $\cdot$ 10 $\frac{m}{s^2}$ $\vec{F}_f$ = 111 N