Newtons Laws EX 20

For equilibrium, the sum of the three forces must be zero.

$F_{1}+F_{2}+F_{3}=0$
$F_{3}=-(F_{1}+F_{2})$

We will now solve the problem using components method:

Forces	x-component	y-component
$F_{1}$	30 cos(30°) = 8.7 N	30 sin(30°) = 5 N
$F_{2}$	12.8 cos(-70°) = 4.4 N	12.8 sin(-70°) = -12 N
$F_{3}$	${F_{3}}_{x}$	${F_{3}}_{y}$

We can write the equations for the second Newton Law:

$8.7+4.4+{F_{3}}_{x}=0$
$5-12+{F_{3}}_{y}=0$

We find that the force $F_{3}$ is (-13.1, 7). We can now compute the magnitude and the direction of this force: $F_{3}$ = 14.9 N and $\theta$ = 152°