Question

Consider the following picture:

F21Ь 가 F1 lb

F₁ = 118 lb

F₂ = 202 lb

α₁ = 31 deg

α₂ = 32 deg

1) Find the magnitude of the resultant force.

2) Calculate the directional cosine cos(α)

3) Calculate the directional cosine cos(β)

4) Calculate the directional cosine cos(γ).

Answer 1

From the vector diagram, we can write

                                    $\small \vec{F}_{1}= 118 \,\, \hat{i} \,\, (in \,\, lb)$

                               $\small \vec{F}_{2}= F_{2} cos \, \alpha_{1} sin \alpha_{2}\,\, \hat{i}+ F_{2} cos \, \alpha_{1} cos \alpha_{2}\,\, \hat{j} + F_{2} sin \, \alpha_{1} \,\, \hat{k}$

                                         $\small =202 \times cos \,31^{\circ} sin \, 32^{\circ} \,\, \hat{i}+ 202\, cos \, 31^{\circ} cos \, 32^{\circ}\,\, \hat{j} + 202\, sin \, 31^{\circ} \,\, \hat{k}$

                                         $\small =91.75 \,\, \hat{i}+ 146.84\,\, \hat{j} + 104.04 \,\, \hat{k}$     (in lb)

So,

                           $\small \vec{F}= \vec{F}_{1}+ \vec{F}_{2}=118\,\, \hat{i}+ \left ( 91.75 \,\, \hat{i}+ 146.84\,\, \hat{j} + 104.04 \,\, \hat{k} \right )$

                                            $\small = 209.75 \,\, \hat{i}+ 146.84\,\, \hat{j} + 104.04 \,\, \hat{k}$

(a) The magnitude of the resultant force is

                         $\small F = \left | \vec{F} \right |=\sqrt{(209.75)^{2}+(146.84)^{2}+(104.04)^{2}}=276.4\,\, lb$

(b) The direction cosine,

                         $\small cos \, \alpha = \frac{F_{x}}{F} = \frac{209.75}{276.4} = 0.76$

(c) The direction cosine,

                         $\small cos \, \beta = \frac{F_{y}}{F} = \frac{146.84}{276.4} = 0.53$

(d) The direction cosine,

                         $\small cos \, \gamma = \frac{F_{z}}{F} = \frac{104.04}{276.4} = 0.38$

For any doubt please comment.