 Now we're going to look at a slightly complicated example for free-body diagrams and in this question We are given a beam That is pinned, but that pin connection is welded So it actually behaves like a cantilever or fixed boundary condition And on this beam Which has a mass of 200 kilograms and the center of gravity position is shown here There is also a man with a mass of 80 kilograms standing on the beam and He is pulling on this rope here with a force of 300 Newtons And that rope wraps around a pulley attaches itself to the beam again and that pulley itself is attached to the ground and We are asked to calculate the force and the moment That it will occur at the welded pin in order for this system to be in static equilibrium so we're going to analyze this system by First making a free-body diagram, so we'll make a free-body diagram. We will cut the beam Here from free it from this support and we'll have to put the reactions of the welded pin there And we're going to cut here through the rope here at the bottom Okay, I've written down here the givens from the problem description So the mass of the beam is 200 kilograms the mass of the man is 80 kilograms The force the man is pulling on the rope is 300 Newtons and gravity is 9.81 meters per second square so now we're going to start by drawing our free-body diagram and We have to first establish a coordinate system Then we will draw our body So we cut the beam at point a where the pin was and at the bottom of the pulley and the man is still on the beam Now we have to consider the loads acting on the system We have the mass times gravity of the beam itself We have the mass of the man times gravity itself Then we have a force in this rope of the pulley and because it's a rope it has to be a tensile force and We have at our pin connection at a It is fixed in translation in the x and y so we will have a reaction a x and a y and It also because it's welded the pin can't rotate so we will get a moment at point a in the z direction now if we look at this we have a x a y m a z and f as Unknowns because the tension and he's pulling on the rope with 300 Newtons f is not 300 Newtons So we have four unknowns But only three equations So what can we do to? To bring down the number of unknowns Well, if you know how pulleys work you can actually already do the relation, but let's Look at cutting away the pulley separating it and Coming up with two free-body diagrams where we have tension in this rope So we know that tension is 300 Newtons and the tension will remain constant as it goes around the pulley because it's frictionless Now you can prove this to yourself if you take some of the moments for this free-body diagram around this point we get this tension Times the moment arm, which is the radius so tension times radius creates a clockwise and Then this tension times the radius creates a counterclockwise and so that sums to zero So t has to be equal to t in a frictionless pulley Okay, so now if we reanalyze this system We see that we now only have three unknowns a x a y and our moment a z Okay The masses are known and the tension is known as 300 Newtons, so we can now solve Now first I will look at some of the forces and we'll do some of the forces in the x direction And I'll use the right direction as my positive reference and here we see that our only force in the x direction is the Horizontal reaction force a x that is possible at the built-in condition But because we have no other external forces in the horizontal direction that is zero If I now look at some of the forces in the y direction and take upwards as my positive reference I have a y acting upwards and one times g acting downwards so negative M2 times g acting downwards so negative and then to T forces acting downwards Now all of these are known so I can rearrange and get a y is equal to m1 plus m2 times Acceleration due to gravity plus 2t I can then sub in the values here And I haven't written units here, but you can see I have everything in base units So 200 kilograms 80 kilograms 9.81 meters per second squared This will give me newtons and 2 times 300 newtons So the units are compatible. I can add those and I'll get 3346.8 newtons or about 3.3 kilo newtons Now if I do some of the moments I'm going to pick point a as my reference point and Counterclockwise as my positive reference direction Now because I pick point a both a x and a y pass through a And do not contribute to the moment equilibrium So my first moment will be ma z which is in the counterclockwise Thus positive in my equilibrium equation If I go along the beam the next force I encounter is m1 times g and it will have a moment arm of 1200 millimeters and it will cause a clock wise or negative moment and you see here I've transformed the 1200 millimeters into meters to keep everything in base units So that my mass times gravity times meters will give me Newton meters Now if I continue to travel along the beam, I will hit the person and There is a mass 2 times gravity with a moment arm of 1.2 meters plus 0.6 meters Which will be clockwise thus negative moment Then I have the same for t And the same for the second t which is at 1.2 plus 0.6 plus 0.3 meters Now of course, this is an equilibrium equation. So all of that has to sum to zero So I can rearrange this all of these are negative So the moment becomes the sum of all of these as positives and here I've substituted in all the numbers and again, I haven't written the units here But I've ensured that everything is in base units. So my result will come out in Newton meters So I'll get 4.937 times 10 to the 3 Newton meters which is 4.9 kilo Newton meters Now did we actually answer the question? If we look back here, we have components So in a way, we've we've answered it But the question asked us for the reaction force and the reaction moment So to be a little bit more Complete in our answer We should either draw a point a and show the resultant force and the resultant moment with an indication of its directions Or we can give it in vector formulation