A 9.00-m-long uniform beam is hinged to a vertical wall and held horizontally by a 5.00-m-long cable attached to the wall 4.00 m above the hinge (Fig. E11.17). The metal of this cable has a test strength of 1.00 kN, which means that it will break if the tension in it exceeds that amount.

(c) Find the horizontal and vertical compo-nents of the force the hinge exerts on the beam. Is the vertical component upward or downward?

Video transcript

Everyone today, we are going to calculate the horizontal and vertical components of the hinges force on the board at the maximum tension of the rod. So they also asked us to specify whether the vertical component is up or down for the hinge forces. And also assume that the center of mass of the system of the boxes and the board is located at the board center. Okay, so this is going to be a pretty complicated question. So we are just going to start it off with creating a list of what is given. So first we have the length of the board which is going to just be L board here. This is going to be 8.4 m, like, so, and then the board itself is used in stacking boxes and that it is actually held by a rod with a length of 7.5 m. And then the board is horizontal and the rod hooks at zero a 6.4 m vertically above the hinges, which is I'm just going to write that down as the l wall because essentially that's the length of the wall from the end of the hinges to the end of the rod up here. Okay, so those are all the dimensions that is listed in the question. And the next we are given the maximum time style strength of the rod, which I'm going to be representing that with an R. For the rod and then that is going to be 1500 newton. Okay, the way we want to actually tackle this problem is to first draw the free body diagram. So I'm just gonna draw that real quick for you guys because this picture right here is a bit too complicated. I feel like that is not horizontal. Okay. Um yeah, it's a bit too complicated. So it's gonna be easier for you guys to actually draw all the forces through this. Okay, so we know that this is going to be the rod and this is going to be the wall right here and then this is going to be the board like so, So we know that the line of the Rod is 7.5 m which is going to be here. And then the length of the wall is 6.4 m, which is also going to be here. And then from there we have the board, the length of the board itself, which is 8.4 m. Okay, so next, what we want to do is to identify all the forces. So obviously, first we will have detention detention force from the rock itself, which is just going to be our here. And then as always we want to do the projection of the uh force. This is going to be our multiplied by sine of theater for the vertical component. And then we will also have a horizontal component, which is going to just be, I'm gonna make this a little bit more thick. The horizontal component is going to be our multiplied by co sine theta like So okay, so this is going to be our Y. And this is going to be our R. X. Next we want to also draw the next force that we have which is going to be the weight. So the weight of the board itself, we are not sure whether it's going to be before the point where the rod and the board meets or after. So we can actually check this right after we finish this free body diagram. But I am just going to draw the weight here for simplicity purposes because it is just too complicated on that side. Okay, so this is going to be the weight here and then next we want to draw the hinges forces. I'm gonna go with the blue here, like, so um we are not sure whether the hinge is going to be pointing or the vertical direction is pointing up or down because that is actually being asked. So essentially I'm just gonna draw it as pointing to the up words direction. And then from there we can check it later on. Okay, but we know that the X direction is going to go to the right side because there's only one other force on the horizontal direction which is our X. Which is pointing to the left. So when the system is at equilibrium, we know that we wanted to go to the right side. Okay, there, we have all the forces necessary for this particular problem. And now, what we want to do next is to actually found where this point of uh intersection or the meeting point is from the rod and the board. Okay. The way we want to do this is to actually use the category interim. So that is going to just be the casual A squared plus b squared equals c squared where c squared is the hypothalamus. And we know that this angle right here is going to be perpendicular and the board is actually a 90° angle with the wall. So therefore we can use this category in theorem with I am just going to uh draw this here to be an X. So we know the X squared plus 6.4 square will equals to 7.5 square. So X will equals to the square root of 7.5 square minus 6.4 square. And their ex will then be 3. m. Alright, so we know that the X is 3.91 m and the weight is essentially going to be Hold on this 8.4. The weight is essentially going to be exactly in the middle of the board itself, which is going to be at um Which is going to be at 8.45 x two, which is 4.2 m. So we know that the X is 3.91 and the weight is 4. m from the hinge. So this representation of the free body diagram is going to be correct and we know that the weight is going to be after the point where the road meets the board. Okay, So this is pretty much all the dimensions that we needed. And the next, the last thing that we have here is the degree angle that we haven't found yet. So the way we want to found this is to actually use the properties of the triangle that we have here. So we know that we have 6.4, we have 7.5 and we just found that this side right here is going to be 3.91. However, uh just for simplicity purposes and to make sure that we got the correct angle, I want to use 6.4 and 7.5 and that is going to be the front length over the slope or the Hippopotamus. So that's gonna be the sign. So the sine of the angle Sine Theta is 6.4 over 7.5. And recall that to find this angle here, we have to find the inverse of the sin. And in first of the sin is going to be arcs in. So the angle is going to equal to the arc sine of six point 4/ 6.0.5, which is then going to be 58.6 degrees. There we go. Okay, now that we find all the dimensions necessary for this particular problem, what we want to do next is to actually um start solving. So the way we want to solve an equilibrium condition is to utilize the 1st and 2nd equilibrium conditions. So when the system is at equilibrium the first condition is going to be utilizing all the sigma force is in a certain angle or certain access, X, Y or Z to be equal to zero. And then the second is going to be utilizing all torque at a certain axis. Usually See it's just it's just access of rotation two equals to zero. So we want to apply this to force or this to equilibrium condition into our example here. Okay first, what we want to solve is probably the easiest which is the H X. Here. The way we want to solve this is to use the first equilibrium condition because we know that there's only one other force along the same access S H X which is the X axis. Okay, so using the first equilibrium condition we can use sigma F at X. Access equals to zero. And usually it's gonna be whatever is pointing to the right, which is H X minus whatever is pointing to the left which is our multiplied by cosine of theta like so And therefore this will equals to zero. Right? So HX is going to be our multiplied by cosine of data which HX will equals to R is 1500 Newton, which is known. Sure. Um 1500 multiply by co sign of 58.6 which is the theater that we just found. And this will come out to be newton like so Okay, That is going to be a checks. And therefore we can actually just start eliminating all the answers where the H. X. is not 782. Okay, now after finding H. X. We want to next find what H. Y. Is. So the way we want to do this is we actually have a lot of unknowns in this particular problem. First we have the weight that is unknown and then we have the H. Y. Itself. That is also unknown. The only thing that we know is this our signed data here amongst all of these forces along the Y. Direction or vertical direction. So what we want to do because we want to find a church Y. And we do not know what W. Is. Using. The second equilibrium condition actually allows us to eliminate one particular or any forces that is located at that exact perfect point. So what we want to do is to use the second including condition and use our pivot point or locate our pivot point at the W. Here. I'm going to label that as our pivot point here. I'm gonna use red. This is going to be our pivot point and therefore we can actually eliminate W. Because recall that to calculate torque, we have to actually use Torque is going to equals two. R multiplied by the force, multiplied by a correction correction angle here and recall that the torque itself actually has to be calculated by multiplying distance times force at a perpendicular angle. So where when the distance from the pivot point to the force itself is zero, which is the W. Here, then this W can be neglected. Okay awesome. So this arson tater is going to be particular and okay now we can actually start by writing down our program conditions. So we are going to utilize the second equilibrium which is going to be sigma tau Z equals to zero at a pivot point of this W. Here. Okay, so we know that usually the um convention that we are using is that whatever is going clockwise is going to be negative and whatever is going counterclockwise is going to actually be positive. So that's what we want to use here. Looking at this pivot point, we know that um the our signed data here is actually going to go clockwise and H. Y. Here is also going to be going clockwise. So we wanted to be, so it's gonna be zero of the force going counterclockwise minus H Y minus our signed data. And then don't forget that this is a torque. So this is gonna be torque H. Y. And then torque our signed data, okay equals to zero. So next we can actually do minus T. H. Y. Equals torque. Are multiplied by science data. And we can plug this into our torque formula. So the distance between H. Y. And our pivot point is going to be 4.2 m here or half of the length of the board. So I'm just gonna write H. Y. Which is the force multiplied by length of board To fight it by two. And we know that H. Y. Is already perpendicular so there's no correction angle needed equals two. R. Multiplied by science data. And our science data is also already perpendicular. So that's good, multiplied by the distance between This pivot point here to this point here. So we know that this distance here distance here is 2.91. So this difference here. So I'm going to write it down as L. Of R. C. R. Sine theta. This is going to actually be 4.2 minus 3.91. Which is going to be 0.29 m like so Okay, 0.29 m. So this is going to be a L. R. Sine Theta here. And remember the minus sign on this side here. Just like so okay, I am going to start with plugging in all the values that we have. This is going to be still minus H. Y. Multiplied by L. Board over two is going to be 4.2 m. Multiplied by our signed data is going to be 1500 Newton multiplied by sine of 58. degrees multiplied by the length. Which is just what we just found here. 0.29 m. Just like so okay, there we go. And then now we can actually start solving for H. Y. So solving for this H. Y. Is then going to actually be 88. newton, that is going to be minus H. Y. So H. Y. Is going to be minus 88.4. Newton just like So okay, now that we found that H. Y is minus 88.4. Newton. We know that our representation here is actually incorrect because therefore we know that H. Y. Is supposed to actually be pointing downwards. So I'm just gonna cross this and revamp So re correct. This H. Y. is going to be pointing downwards like so with a magnitude of 88.4 Newton. Remember that in this? In this vertical direction forces. The Upward section is convention is having a convention of positive, while the downward forces will have the convention of a negative. Okay, so therefore H. Y. is going to be -88.4 Newton. So that will actually be answer option C Where we have the H. X. Of 782 Newton and H. Y of minus 88.4 Newton. If you guys are still confused on this particular example, then there are a lot of other lessons plans on our website that you guys should check out. And yeah, thank you.