Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 5 - Force and Motion-I - Problems: 8

Answer

${F}^{>}_{3}$=-$7i$-$12j$ $N$

Work Step by Step

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 $F^{>}_{3}$=$2(-8i+6j)-(3i+16j)-(-12i+8j)$ $F^{>}_{3}$=$-16i+12j-3i-16j+12i-8j$ $F^{>}_{3}$=$-7i-12j$ $N$
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