Question #e1fc3

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Question #e1fc3

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Answer 1 · 2017-04-14T17:21:48

It would be $120^{0}$ $120^0$

Explanation:

Given the forces are equal, it would be reasonable to assume they have the same angle between them as in the diagram below:

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If we then consider components of forces in the x direction:

$F_{2} {cos 30}^{0} - F_{1} {cos 90}^{0} - F_{3} {cos 30}^{0}$ $F_2cos 30^0-F_1cos90^0-F_3cos30^0$
$\Rightarrow F_{2} {cos 30}^{0} - F_{3} {cos 30}^{0}$ $=>F_2cos 30^0-F_3cos30^0$

Since $F_{1} = F_{2} = F_{3}$ $F_1=F_2=F_3$ then the above equates to nil

Similarly in the y direction:

$F_{1} - F_{2} {sin 30}^{0} - F_{3} {sin 30}^{0}$ $F_1 -F_2sin 30^0-F_3sin30^0$
$\Rightarrow F_{1} - 0.5 \cdot F_{2} - 0.5 \cdot F_{3} = 0$ $=>F_1-0.5*F_2-0.5*F_3=0$

Hence no resultant force

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