The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center.

Find (b) the horizontal and vertical components of the force ex-erted on the beam at the wall.

Video transcript

Hi everyone today we are going to find the horizontal and vertical forces exerted By the Hinge on the rot at the wall. Okay, so we know that this is going to be a horizontal uniform broad with a mass of two kg Which is supported by a hinge on the barrier at one end and the light string at its other end as shown in this image with the length of the Rod being four m like. So okay, now that we've identified all this, what is known or given in our problem statement, we want to start by making our free body diagram. So first we definitely know that we will have a weight or the weight off the rod which is just going to be m rod multiplied by G. And we know that it is a horizontal uniform rock. So because of that we know that the center of mass is going to be directly located Exactly in the middle of the Rod which is at a distance two m from the Hinge. Next we have the tension which is going to be the t. And as always we want to start by making the projection of our attention. This side is just going to be the vertical side. T. multiplied by sine data with data of 20° and horizontal side of the multiplied by co sign of data which is also going to be 20°. Next we will have what is being asked here which are the hinges forces P. Y. On the vertical direction and P X on the horizontal direction. Alright, so now that we've identified all the forces we want to know what is actually known and what is actually unknown. We can easily calculate the weight but we are still missing the tension P. Y. And P. X. So we have three different unknowns of forces. So how will we want to start to proceed with solving this problem First. What we want to do is to actually use the second equilibrium condition which is sigma, torque at a certain access, T. Z. Or Z equals to zero at equilibrium which the system is at. So the reason why I want to use this is because using the second equilibrium condition allows us to pick a pivot point at which we can eliminate all the forces exerted directly onto that different point itself as we know we have two different forces, P. Y and P X. Directly located at the hinge. So we want to use that as our perfect point here. That is going to be a different point and therefore we can neglect P. Y and P X. While we are using the second equilibrium condition equation. Okay, so recall that the torque itself is actually going to be calculated by multiplying the distance with the force which has to actually be perpendicular to one another. So in this case the weight we know is perpendicular to the distance and t multiplied by sine data is also perpendicular to the rod or the distance. So we are good on that. Okay, next we want to also recall that we have the convention for torques. So we have a convention where the torquing clockwise or this way clockwise is negative while the torquing counterclockwise is positive. And in this equilibrium situation we actually don't really care what we want to subtract it with, but because it's equal to zero. But for consistency purposes, I am going to stick to this convention. So now we want to start with the torque going counterclockwise, which is the multiplied by science data. So now t multiplied by science data and this force right here, this projected force is perpendicular. So we we just multiplied that by the distance which is from the pivot point or the hinge to this force is going to be L rod or the overall rod because it's at the end. We want to subtract this with the weight once applied by G and multiply this by the center of my distance to to hinge which is all rotten, multiplied by two. We want to equal this 20. Like. So, okay, now we know that we do not know what he is. We know what sign data and we know all the other things. So we can figure out what the is. Okay, I'm just going to start with uh plugging things in sign of 20 degrees, multiplied by l rod which is four m minus this with emerald times G is going to be two kg multiplied by 9.81 m per second squared multiplied by two m which is going to be L Rod four m divided by 22 m equal this to zero. And then we want to start rearranging things which is going to just be the equals to two kg multiplied by 9.81 m/s squared, multiplied by two m Divided by the city which is four m multiplied by sine of 20°. This will come out to be 28. Newton like. So Okay, so now we know what he is knowing what the is allows us to actually start solving for P. Y and P X. By utilizing the first equilibrium condition. So recall that the first ecuadorian condition is going to be sigma force equals to zero at a certain access that we want. In this case, I am just going to start with the X axis because it is simpler. Look at looking at this diagram here, we know that there are going to be two only two different forces on the X axis which is P X. And also the multiplied by cosine data. So in this case, I am just going to start with whatever is pointing to the right minus whatever is pointing to the left. But essentially it doesn't matter as well because it is at equilibrium. So we can immediately just calculate it. Okay, so we know what T. Is. We know what the coastline data is. So we can just immediately calculate this. The T. Is 28.7 Newton. And the data is going to be 20 degrees and this will turns out to be 26.9 newton like. So. Okay, next we want to also apply the other first equilibrium condition on the different axis, which is going to be the Y axis or the first he called direction. Okay, looking at this diagram right here we have three different forces acting on the Y direction. We have B Y. P. Multiplied by sine data and the weight or brought multiplied by G. So, I'm just going to do whatever is pointing to pointing up minus whatever is pointing down minus m. Broad Times cheap. It goes to zero like. So essentially it also doesn't matter because we are at equilibrium. Okay, so we have then py that will equal to uh and Rod, which is two kg, multiplied by 9.81 m per second square which is the g minus this by T. Sign data. Which is 28.7. Newton multiplied by sine of 20 degrees. Like. So, And then this will turn out to be P Y two B equals 29.8. Newton like. So. Okay, So we found that PX is going to be 26.9 and py is going to be 9.8. Newton which going to correspond to option C With a PX of 26.9 and py of 9.8. So that should be all for this particular problem. Let me know if you guys have any questions, but at the same time, if you guys are still confused, there are a lot more other videos on our website regarding this particular example, which you guys should check out. Thank you.