In this problem. On the topic of equilibrium and elasticity, we have shown a uniform beam which has a weight of 60 newtons and a length of 3.2 m. It is hinged at its lower end and a horizontal force of magnitude five newtons acts the up end. The beam is held vertical by a cable, making 25 degrees with the ground and is attached to the beam at a height of two m. We want to calculate firstly the tension in the cable and next the force on the beam from the hinge. Now, if we choose an access through the hinge perpendicular to the plane of the figure and take talks, that would cause a clockwise rotation to be positive. Then we require the net talk to vanish since the beam is in equilibrium and this net talk fl sign of 90 degrees minus T. H. Sign of 65 degrees must equal to zero here the length of the beam L is given to be 3.2 m. And the height at which the cable attach is is H. And age is attached at two m. And we note that the weight of the beam does not enter in this equation, since this line of action is directed towards the hinge. So if we take if as 50 newtons has given in the problem statement, we can rearrange the equation above and solve for T. And we get T. Is equal to F times L over H. Sign of 65 degrees. If we put our values into this, we get this to be 50 newtons times three 0.2 m, divided by the height two m times the sine of 65 degrees. This gives us detention T. To be 88 newton's and that was the solution for part A. In part B. We want to find the force on the beam from the hinge in unit vector notation. So to find the components of the force F. P. Will call it. We have to balance the X and Y components, so we know that some of the X components of this force was equal to zero, which means if P X is equal to T. Timely co sign of 25 degrees minus Mhm Force F, which is purely horizontal. Similarly, we have these some the vertical forces was also equal to zero for equilibrium. In this case, that's if P Y must equal to T. Sign 25 degrees plus the weight of the rod W. And that weight of the beam is 60 newtons. So we therefore find that the hinge force components, when we calculate uh F. P. X, which is 30 newton's, and this points to the right. And if P Y is 97 newtons pointing upward. So when you infect a notation, we can write this force that the hinge applies to the beam. F p is simply 30 newton's long, unit factor i less 97 newtons times unit vector J.