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°