- #1
bardia sepehrnia
- 28
- 4
I isolated the member ABC and drew the free body diagram:
α is then calculated using inverse tan: Tan-1=(6.25+15)/50=23.03
Then force of member BD on the joint can be found by sum of all moments around point A.
Then Ax is calculated which is equal to BD×Cos(α)=235.2×Cos(23.03) Ax=216.48
Then I draw free body diagram of the block with joint A:
Then P=Ax=216.48.
I cannot to the second part of this question and calculate the force on the member BC!?
I can calculate Ay in first free body diagram however would that at all help me find the force on Member BC? Is the resultant force of vectors Ay and Ax the force acting on member AB?