Answer:
hello the diagram attached to your question is missing attached below is the missing diagram
answer : Ffb = 1.9381 KN
Fdb = 2.6 KN
Explanation:
Given data
M = 125-kg
Determine The force on member FB
∅ = tan^-1 ( 3/1) = 71.56°
∑ Me = 0
= 125 * 9.81 * 3 - Ffb sin∅ * 2 = 0
hence : Ffb = 1.9381 KN
also
∑ Ex = 0
Ex = Ffb cos ( 71.56 ) = 1.9381 * cos 71.56 = 0.6132 KN
Determine the Force on member DB
∅ = tan^-1 ( 1/1 ) = 45°
∑ Mc = 0
Ex * 3 = Fdb * sin 45 * 1
Hence Fdb = 2.6 KN
attached below is the free body diagram
