Now that F_{BE} is known, another cut can be made to find F_{JB}, F_{JD}, and F_{IJ}. To find F_{JD} and F_{JB}, moment about L can be used with the sum of the vertical forces (notation JL mean length of JL)
ΣM_{L} = 0
= JL (2kN) - F_{JD}(BD) - F_{JB}(JL)
= 2(2) - F_{JD}(1.414) - F_{JB}(2)
ΣF_{y} = 0
= F_{JD} + F_{JB} + 5.167 - 2
-7.308(sin45)
F_{JD} and F_{JB} can be found by solving the above system of equations to give
F_{JD }= -0.8327(10^{-3}) kN (very small)
F_{JB} = 2.000 kN
F_{JD} is found to be 0.8 newton which is negligible and is because of the rounding errors.
Summation of forces in x direction can be written in order to find the value of F_{IJ}
ΣF_{x }= 0
= F_{EB}(cos45) + F_{JD}(cos45) + F_{IJ}
= -7.308 (0.707) + 0 (0.707) + F_{IJ}
F_{IJ }= 5.167 kN
F_{IJ} can also be found by taking the moment about joint B
ΣM_{B }= 0
= L(JL) - F_{JD}(BD) - F_{IJ}(BJ)
= 5.167(2) + 0 (1.414) - F_{IJ}(2)
F_{IJ }= 5.167 kN |