Part F $-$ Example: Finding Two Forces $($Part I$)$

Two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in $($ Figure $1)$ has abrass $m=10kg$ and is being pulled by a force $F$ on a table with coefficient of static friction ${}_s=0\mathrm{.}3.$ Four forces act on it:

The applied force $F($directed $=30$ above the horizontal$).$

The force of gravity $F_g=mg($directly down, where $g=9\mathrm{.}8\frac{m}{s^2}).$

The normal force $N($directly up$)$

The force of static friction $f_s($directly left, opposing any potential motion$).$

If we want to find the size of the force necessary to just barely overcome static friction $($in which case $f_s={}_{s}N),$ we use the condition that the sum of the forces in both directions must be $0.$ Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as$.$

Fcos$-{}_{s}N=0$

Fsin$+N-mg=0$

In order to find the magnitude of force $F,$ we have to solve a system of two equabions with both $F$ and the normal force $N$ unknown. Use the methods we have learned to find an expression for $F$ in terms of $m,g,,$ and ${}_s($no $N).$

Express your answer in terms of $m,g,,$ and ${}_s.$