Determine the magnitude and coordinate direction angles of the resultant force. Set FB = 630 N, FC = 520 N and FD = 750 N, and x = 3 m and z = 3.5 m.

Question

Determine the magnitude and coordinate direction angles of the resultant force. Set [latex]F_B[/latex] = 630 N, [latex]F_C[/latex] = 520 N and [latex]F_D[/latex] = 750 N, and x = 3 m and z = 3.5 m.

Accepted Answer

Force Vectors: The unit vectors [latex]u _B, u _C ext {, and } u _D ext { of } F _B, F _C ext {, and } F _D[/latex]must be determined first. From Fig. a,

[latex]u _B=\frac{ r _B}{r_B}=\frac{(-3-0) i +(0-6) j +(4.5-2.5) k }{\sqrt{(-3-0)^2+(0-6)^2+(4.5-2.5)^2}}=-\frac{3}{7} i -\frac{6}{7} j +\frac{2}{7} k[/latex]

[latex]u _C=\frac{ r _C}{r_C}=\frac{(2-0) i +(0-6) j +(4-2.5) k }{\sqrt{(2-0)^2+(0-6)^2+(4-2.5)^2}}=\frac{4}{13} i -\frac{12}{13} j +\frac{3}{13} k[/latex]

[latex]u _D=\frac{ r _D}{r_D}=\frac{(3-0) i +(0-6) j +(-3.5-2.5) k }{\sqrt{(0-3)^2+(0-6)^2+(-3.5-2.5)^2}}=\frac{1}{3} i -\frac{2}{3} j -\frac{2}{3} k[/latex]

Thus, the force vectors [latex]F _B, F _C ext {, and } F _D[/latex] are given by

[latex]F _B=F_B u _B=630\left(-\frac{3}{7} i -\frac{6}{7} j +\frac{2}{7} k ight)=\{-270 i -540 j +180 k \} N[/latex]

[latex]F _C=F_C u _C=520\left(\frac{4}{13} i -\frac{12}{13} j +\frac{3}{13} k ight)=\{160 i -480 j +120 k \} N[/latex]

[latex]F _D=F_D u _D=750\left(\frac{1}{3} i -\frac{2}{3} j -\frac{2}{3} k ight)=\{250 i -500 j -500 k \} N[/latex]

Resultant Force:

[latex]\begin{aligned}F _R &= F _B+ F _C+ F _D=(-270 i -540 j +180 k )+(160 i -480 j +120 k )+(250 i -500 j -500 k ) \&=[140 i -1520 j -200 k ] N\end{aligned}[/latex]

The magnitude of [latex]F _R[/latex] is

[latex]\begin{array}{c}{{F_{R}=\sqrt{\left(F_{R} ight)_{x}^{2}+\left(F_{R} ight)_{y}^{2}+\left(F_{R} ight)_{x}^{2}}}}\ {{=\sqrt{140^{2}+\left(-1520 ight)^{2}+\left(-200 ight)^{2}}=1.539.48\,\mathrm{N}=1.54\,\mathrm{kN}}}\end{array}[/latex]

The coordinate direction angles of [latex]F _R[/latex] are

[latex]\alpha=\cos ^{-1}\left[\frac{\left(F_R ight)_x}{F_R} ight]=\cos ^{-1}\left(\frac{140}{1539.48} ight)=84.8^{\circ}[/latex]

[latex]\beta=\cos ^{-1}\left[\frac{\left(F_R ight)_y}{F_R} ight]=\cos ^{-1}\left(\frac{-1520}{1539.48} ight)=171^{\circ}[/latex]

[latex]\gamma=\cos ^{-1}\left[\frac{\left(F_R ight)_z}{F_R} ight]=\cos ^{-1}\left(\frac{-200}{1539.48} ight)=97.5^{\circ}[/latex]

Question 2.108: Determine the magnitude and coordinate direction angles of t...

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