Engineering Board Exam Reviewer · Engineering Mechanics · hard
A vertical load of 525 N hangs from joint A, which is supported by two members: AB, horizontal, and AC, inclined at 30° above the horizontal. Using the method of joints, what is the force in member AB?
- 303.11
- 262.5
- 1050
- 909.33
Answer
D. 909.33
Solution
First, Fy = 0 at joint A gives F(AC) = 525 / (30°) = 1050 N. Then Fx = 0: F(AB) = F(AC) · (30°) = 1050 × (30°) = 909.33 N.
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