Engineering Board Exam Reviewer · Engineering Mechanics · medium
A vertical load of 500 N hangs from joint A, which is supported by two members: AB, horizontal, and AC, inclined at 25° above the horizontal. Using the method of joints, what is the force in member AB?
- 233.15
- 1072.25
- 211.31
- 1183.1
Answer
B. 1072.25
Solution
First, Fy = 0 at joint A gives F(AC) = 500 / (25°) = 1183.1 N. Then Fx = 0: F(AB) = F(AC) · (25°) = 1183.1 × (25°) = 1072.25 N.
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