Question
Download Solution PDFThe force in member AB is ______. (take angle BAC as 60° and angle BCA as 30°)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Given:
∠BAC = 60°
∠BCA = 30°
Let FAB = compressive
FBC = compressive
FAB = ?
Considering joint B,
∑FH = 0
FAB cos 60° = FBC cos 30°
\(F_{AB} = \sqrt{3} \ F_{BC}\)
∑FV = 0
FAB sin 60° + FBC sin 30° = 10
\(\dfrac{\sqrt{3}}{2} F_{AB} + \dfrac{F_{AB}}{\sqrt{3}} \times \dfrac{1}{2} = 10\)
Multiplying both sides by \(\sqrt {3}\)
\(3 F_{AB} + F_{AB} = 20\sqrt{3}\)
\(4F_{AB} = 20\sqrt{3}\)
\(F_{AB} = 5 \sqrt{3}\)
Compressive (As direction assumed comp)
Last updated on May 28, 2025
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