Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 3 - Equilibrium of a Particle - Section 3.4 - Three-Dimensional Force Systems - Fundamental Problems - Page 111: 7


$F_3=776N$ $F_1=466N$ $F_2=879N$

Work Step by Step

We can find the magnitude of the required forces as follows: The sum of the forces in the x-direction $\Sigma F_x=0$ $\implies (\frac{3}{5})F_3(\frac{3}{5})+600-F_2=0$.eq(1) The sum of the forces in the y-direction $\Sigma F_y=0$ $\implies (\frac{4}{5})F_1-(\frac{3}{5})F_3(\frac{4}{5})=0$.eq(2) and the sum of the forces in the z-direction $F_z=0$ $\implies (\frac{4}{5})F_3+(\frac{3}{5})F_1-900=0$..eq(3) From eq(2), we obtain: $5F_1-3F_3=0$ $\implies F_1=\frac{3}{5}F_3$..eq(4) From eq(3), we obtain: $4F_3+3F_1=4500$eq(5) We plug in the value of $F_1$ from eq(4) into eq(5) to obtain: $4F_3+3(\frac{3}{5}F_3)=4500$ This simplifies to: $F_3=776N$ We plug in the value of $F_3$ in eq(4) to obtain: $F_1=\frac{3}{5}(776)=466N$ From eq(1), we have $(\frac{9}{25})F_3+600-F_2=0$ $\implies (\frac{9}{25})(776)+600-F_2=0$ This simplifies to: $F_2=879N$
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