Using Statics find the x and y components of force at the pin connections of member ABC shown in the figure.

Question

Please show all steps in the calculation and also any and all free body diagrams, Rod can you help. See several notes below figure.

Note #2 Members AD, BD, and ABC are all the same say steal rods and all weights are neglected, and as for the graduated down arrows they represent a compound load of 100 lb/ft for 1.5 ft from B to C and from A to B represents a compound load at 1/2  the base of 1.5 ft times the height of 100 lb/ft, so the force acting on A to B is 75 lb and from B to C is 150 lb. the center of gravity for the force acting on A to B is 1 ft from point A and for the force acting from B to C the center of gravity for that force is 2.25 ft from point A. Note #3 The x y axis starts at point A.  Note #1 from global analysis of the frame using the sum of the moments about point D we find that the normal force  at point C is 215.625 lb and using Free Body Diagrams of members AD and ABC at pin connections A and B, Ay is 50 lb and By is 275 lb but  Ax is supposedly 9.67 1b and Bx is 206 lb but how is that possible help me is Ax and Bx equal to those numbers.

Comments

I see you added some notes. I need to refresh my memory about statics in applied maths. I wasn't sure about the meaning of the graduated down arrows on ABC. I guess the line density decreases to zero at A from being 100lb/ft between B and C. One thing to note is that the diagram clearly indicates a 3-4-5 triangle which simplifies sines and cosines (sinDÂC=⅘), so that's a big clue. I'm not sure if AD and BD are meant to represent (weightless) strings attaching the various components. With your extra figures it should be possible to verify what I believe to be the total structure. I hope I can help...in due course!

Answer 1

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, ρ_x=100x/1.5=200x/3. For example, at the midpoint of AB,

 x=0.75ft and ρ_0.75=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 xρ_x=200xδx/3, making the weight of AB (W_AB) the sum of all these small lengths.

So W_AB=₀∫^1.5200xdx/3=(200/3)₀∫^1.5xdx=(100/3)[x²]₀^1.5=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...