For the problem attached in the picture, draw the fully dimensioned FBD for each body and determine the number of unknowns. Calculate all reactions. Assume friction at all contact surfaces. The diagram depicts a horizontal beam attached to a vertical support. The horizontal beam has a total length of 0.8 meters, divided into two segments: 0.3 meters and 0.5 meters. A load labeled "A" is suspended at the 0.3-meter point of the beam, while another load labeled "B" is suspended at the 0.5-meter point. The vertical support is 0.5 meters tall. The thickness of the beam is specified as 50 mm, and its mass is given as 30 kg/m.The diagram helps illustrate the mechanics of load distribution along a beam and can be useful in understanding concepts such as torque and equilibrium.

The free body diagram of each body is shown below: Here, Rh,RBv,RBH and RA are the reaction forces at point C, point B and Point A, PA and PB are load acting on it. The weight of the body is acting at the centroid of the vertical downward direction, so it is calculated as, w=30 kg/m×0.5 m×9.81 m/s2=147.15 N Here, m is the mass of the bar and it is given 30 kg/m and g is the acceleration due to gravity. The point load acting at point A can be calculated as, PA=30 kg/m×0.3×9.81 m/s2=88.29 N The point load acting at point B can be calculated as, PB=30 kg/m×0.5 m×9.81 m/s2=147.15 NConsider the equilibrium condition of entire system along horizontally, ∑FH=0RBH=0 Consider the equilibrium condition of vertical bar along vertically, ∑Fv=0Rh=wRh=150 N Consider the equilibrium condition of entire system along vertically, PA+PB+Rh=RA+RBv88.29 N+147.15 N+147.15 N=RA+RBv382.59=RA+RBv..........1 Take moment at point B is given as, ∑MB=0PB×0.25 m+PA×0.65 m+Rh×0.8 m=RA×0.5 m147.15 N×0.25 m+90 N×0.65 m+147.15 N×0.8 m==RA×0.5 mRA=426 N Substitute the value in equation 1, RA+RBV=382.59 NRBV=-43.41 N Negative sign indicates that it acts in downward direction.