The hoist supports the 125-kg engine. Determine the force the load creates in member DB and in member FB, which contains the hydraulic cylinder H.

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

how is this a question? in order for me to answer it i have to understand the question

explanation:

answer:

yes that is right have a good day