Linear Algebra: A Modern Introduction

Published by Cengage Learning
ISBN 10: 1285463242
ISBN 13: 978-1-28546-324-7

Chapter 1 - Vectors - 1.4 Applications - Exercises 1.4 - Page 54: 12

Answer

$5\sqrt{2}N$

Work Step by Step

Due to symmetry between the string, the tension applied between both is the same: $F$ force applied in vertical direction = force applied downward $10=2F\sin\sin45^\circ$ $F=\frac{10}{\sqrt{2}}=5\sqrt{2}$
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