Determine the horizontal and vertical components of reaction on the member at the pin A, and the normal reaction at the roller B.
Take moment about A.
"\\Sigma M =O"
NBcos30(6)-NBsin30(2)-750(3)=0
5.19NB-NB-2250=0
NB=536.22 lb
equilibrium of forces along x-direction
"W\\Sigma Fx=0"
Ax=NBsin30
Ax=536.22sin30
Ax=268.11 lb
equilibrium of forces along y-direction
"W\\Sigma Fy=0"
Ay+NBCos(30)-750=0
Ay+536.22Cos(30)-750=0
Ay=285.61 lb
Comments
Leave a comment