Engineering Board Exam Reviewer · Engineering Mechanics · medium
A vertical load of 575 N hangs from joint A, which is supported by two members: AB, horizontal, and AC, inclined at 40° above the horizontal. Using the method of joints, what is the force in member AB?
- 482.48
- 369.6
- 894.54
- 685.26
Answer
D. 685.26
Solution
First, Fy = 0 at joint A gives F(AC) = 575 / (40°) = 894.54 N. Then Fx = 0: F(AB) = F(AC) · (40°) = 894.54 × (40°) = 685.26 N.
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