If FB = 560 N and FC = 700 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.

Question

If [latex]F_{B} [/latex] = 560 N and [latex]F_{C} [/latex] = 700 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.

Accepted Answer

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

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

[latex] \mathbf{u}_C=\frac{\mathbf{r}_C}{r_C}=\frac{(3-0) \mathbf{i}+(2-0) \mathbf{j}+(0-6) \mathbf{k}}{\sqrt{(3-0)^2+(2-0)^2+(0-6)^2}}=\frac{3}{7} \mathbf{i}+\frac{2}{7} \mathbf{j}-\frac{6}{7} \mathbf{k}[/latex]

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

[latex] \mathbf{F}_B=F_B \mathbf{u}_B=560\left(\frac{2}{7} \mathbf{i}-\frac{3}{7} \mathbf{j}-\frac{6}{7} \mathbf{k}\right)=\{160 \mathbf{i}-240 \mathbf{j}-480 \mathbf{k}\} \mathrm{N}[/latex]

[latex] \mathbf{F}_C=F_C \mathbf{u}_C=700\left(\frac{3}{7} \mathbf{i}+\frac{2}{7} \mathbf{j}-\frac{6}{7} \mathbf{k}\right)=\{300 \mathbf{i}+200 \mathbf{j}-600 \mathbf{k}\} \mathrm{N}[/latex]

Resultant Force:

[latex]\mathbf{F}_R=\mathbf{F}_B+\mathbf{F}_C=(160 \mathbf{i}-240 \mathbf{j}-480 \mathbf{k})+(300 \mathbf{i}+200 \mathbf{j}-600 \mathbf{k}) [/latex]

= {460 i - 40 j + 1080 k} N

The magnitude of [latex]\mathbf{F}_R[/latex] is

[latex]F_R=\sqrt{\left(F_R\right)_x^2+\left(F_R\right)_y^2+\left(F_R\right)_z^2}[/latex]

[latex] =\sqrt{(460)^2+(-40)^2+(-1080)^2}=1174.56 \mathrm{~N}=1.17 \mathrm{kN}[/latex]

The coordinate direction angles of [latex]\mathbf{F}_R [/latex] is

[latex] \alpha=\cos ^{-1}\left[\frac{\left(F_R\right)_x}{F_R}\right]=\cos ^{-1}\left(\frac{460}{1174.56}\right)=66.9^{\circ}[/latex]

[latex]\beta=\cos ^{-1}\left[\frac{\left(F_R\right)_y}{F_R}\right]=\cos ^{-1}\left(\frac{-40}{1174.56}\right)=92.0^{\circ} [/latex]

[latex] \gamma=\cos ^{-1}\left[\frac{\left(F_R\right)_z}{F_R}\right]=\cos ^{-1}\left(\frac{-1080}{1174.56}\right)=157^{\circ}[/latex]

Question 2.92: If FB = 560 N and FC = 700 N, determine the magnitude and co...

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