First, I need to explain my interpretation of the diagram. Then we can start to produce the FBDs applicable to the problem.
The bar ABC, as I understand it, has a weight derived from two parts: AB and BC, which have the same length=1.5ft. BC has a constant density ρ = 100lb/ft, so the weight of BC=150lb. But I understand AB to have less weight because the density is apparently variable, being 0 at A and 100lb/ft at B. If we take a distance x from A, the density at distance x, ρx=100x/1.5=200x/3. For example, at the midpoint of AB,
x=0.75ft and ρ0.75=50lb/ft.
If we consider the weight of a very small length δx of AB at a distance x from A, its weight would be xρx=200xδx/3, making the weight of AB (WAB) the sum of all these small lengths.
So WAB=0∫1.5200xdx/3=(200/3)0∫1.5xdx=(100/3)[x2]01.5=225/3=75lb (as you might expect, being half the weight of BC). The weight of the bar ABC would therefore be 150+75=225lb. This would represent the magnitude of a force in the y direction. As a vector it is -225lb weight (gravity converts this mass into weight, so we need to multiply this by g=32ft/s2 to convert mass into force).
I assume that AD and BD are weightless strings subject to tension. They need to counterbalance the weight of the bar plus the 200lb force applied perpendicularly to AD. Now we can begin to produce FBDs.
The relevant forces are:
The 200lb force, the tension U in the upper part of AD, the tension L in the lower part of AD, the weight of the bar 225g directed downwards, and T the downward tension in BD. We neglect any forces attributed to the fastening points on the wall at C and D. All these forces are in equilibrium and we can resolve them into x and y components.
The attachment AD is inclined to ABC at an angle θ=sin-1(⅘) because sinθ=CD/AD=4/5. Similarly sinφ=CD/BD, and tanφ=4/1.5=8/3.
The perpendicular 200lb pushing on AD affects the tensions U and L, which are already affected by the weight of the bar ABC. To me, 200lb represents a mass (not a weight) which needs to be converted into a force so that we are dealing entirely in forces. As a force, then, I assume that it is 200g, just as the downward force on ABC is 225g. In the absence of the force from 200lb, U and L would combine as one, affected only by the weight of the bar. When calculating the tensions U and L we need to add/subtract the effect of the force due to the 200lb mass. Since this mass doesn't touch BD, T is entirely due to the weight of the bar ABC.
Consider first the horizontal components of force. The weight of the bar has no horizontal component.
More to follow in due course...