## Fundamentals of Physics Extended (10th Edition)

${F}^{>}_{3}$=-$7i$-$12j$ $N$
Let $F^{>}_{net}$ be the net force which produces acceleration so $F^{>}_{net}$=$F^{>}_{1}$+$F^{>}_{2}$+$F^{>}_{3}$ As we know that $F^{>}_{net}$=$ma^{>}$ Hence $ma^{>}$=$F^{>}_{1}$+$F^{>}_{2}$+$F^{>}_{3}$ $ma^{>}$-$F^{>}_{1}$-$F^{>}_{2}$=$F^{>}_{3}$ or $F^{>}_{3}$=$ma^{>}$-$F^{>}_{1}$-$F^{>}_{2}$ Putting the values, you will get $F^{>}_{3}$=$2(-8i+6j)-(3i+16j)-(-12i+8j)$ $F^{>}_{3}$=$-16i+12j-3i-16j+12i-8j$ $F^{>}_{3}$=$-7i-12j$ $N$