A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it (Fig. E11.20). The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam. Find the tension T in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

Video transcript

Hi everyone. Today, we are going to first determine the tension in the court as well as the force exerted by the pivot on the bar. So a free body diagram will be very useful on this particular problem because we do have a lot of different forces exerted and a lot of different dimensions. So first what we want to do is to make a list of what is actually given. I'm gonna start with the traffic light here. So it is given the mass of the traffic light is three kg. So that will actually be this weight of the traffic light here. This is just going to be m traffic multiplied by G. Like so and it is actually hanging from the right end of a non uniform bar. As shown in this image, the length of the bar, It's actually seven m and it has a mass of 7, 10 kg like. So because it is a non uniform bar, the center of gravity or center of mass is not exactly in the middle and it is listed here that the center of gravity of the bar is four m from its left. And so I'm just gonna write that down as L. C. M. Which is four m here. It is given that the center of gravity is going to be here. I'm gonna just extend that and this is going to be M. Bar multiplied by G. Like. So okay, Next we have that the bar is actually held by a frictionless pivot at its left hand and a light court at a distance of 2m away from the pivot. So I'm gonna write that down as I'll cord is equals to two m. And then after that it is also given that the light cord itself is perpendicular to the bar. So this is going to be at a perpendicular angle At the same time in its equilibrium position. The bar makes an angle of 30° below the horizontal. So that is going to be this angle right here. This is going to be this angle which is going to be theta. So Tina is going to be 30 degrees like so okay, we now have all of what is actually given in our problem. So what we want to do next is to create the actual free body diagram. Okay, so obviously we have the weight of the traffic light, the weight of the bar itself and then we will also have the attention of the court which is being asked. So this is gonna be T and T is actually perpendicular to the bar but as usual we want to actually do the projection along the y and the X axis like so okay, just like so okay, so what is missing here is this angle Which is, I'm just gonna call it with a five. We know that this data right here is 30° and for two parallel lines along the same line, we know that it will have the same Angle. So this angle here is also going to be data therefore because tension is going to be perpendicular to the bar, we can calculate what five is, which is actually going to be 90° because its perpendicular minus data, which is going to be 60°. Okay, so this will actually be the multiplied by sine of five, which is uh 60° and the x component is going to be the multiplied by cosine of five, which is going to be 60° as well. Okay, next we want to actually do the projection for M bar and M traf times G I'm gonna switch color here and just start the projection. So we want anything in essentially the perpendicular to the actual bar itself to calculate for our talks in the end, just like so, okay, next we want to actually calculate what this, essentially what this projection is based on the embargo times G. Right, we know that this angle right here is also going to be data, which is going to be using the same reasoning and we know that this right here is essentially an x and y axis and they're actually going to be perpendicular to one another along with this side here And this side here. Okay, so this angle right here is essentially going to be 90°C data, which is also going to be five. So this is essentially going to be M bar times G multiply by sine of phi because it is opposite to this angle right here, this is going to be exactly the same, which is just going to be M Treff times G multiplied by sine of phi like so okay, the last force that we have here is what is being asked, which is the exerted force by the pivot on the bar. So we actually, I'm not going to be sure on whether the pivot here is going to be going downwards or upwards on the vertical direction but we can in the end just pretty much um determine whether it actually is facing onto the right direction or not. Okay and then on the horizontal direction we kind of see here that most of the forces is pointing to the right on the horizontal position. So we can just start with the first pivot force going to the left side and we can also determine India and whether we are having the correct direction or not. Through our calculations. Okay, now that we have all the free body diagram here, we can start by solving each components. I am gonna first start to find what the tension is here so to find what the tension is here right now we do not know what PX and py is and we only know the M bar and the em drive. So in order to solve for the tension, we want to use the second equilibrium condition, the second equilibrium condition is essentially sigma torque at a certain rotation access equals to zero. And using this at a certain pivot point will actually allow us to eliminate the forces acting directly onto the pivot point itself. So what we want to do here is to actually use the pivot point to be on the hinge here. And therefore we can actually eliminate PX and py. Because if you guys recall torque itself is going to be calculated with multiplying distance times, force times a correction angle if the two is actually not perpendicular to one ***. So um using this form well right here, the forces exactly at the pivot point will have a torque of zero because the our or the distance is zero. Okay, so I am just going to start with the second equilibrium condition here and recall that in the torque formula, we have the convention where everything going clockwise is going to be negative and everything going counterclockwise is going to be positive. So in this particular problem with the pivot point being in this uh place right here we have the tension to be going counterclockwise while the ember G sci fi and M Treff is going to be going clockwise. Okay, so I'm just gonna start off with the attention because it is going counterclockwise. So attention this distance, right distance right here is going to be L. Court. O. L. Yeah, L court and then we have two minus that With first, it's going to be the weight and bar G sine phi and this will be multiplied by the distance which is L. C. M. And then we want to also minus this with M. Treff G sci fi multiply this by the length of the overall bar which is a bar and this really goes to zero. So essentially the convention does not really matter here because it's at equilibrium but just for consistency purposes, we want to keep it this way and then we will have the multiplied by two m subtracted with um Bar G. Which is going to be 10 kg, multiplied by 9.81 m per second squared multiplied by four m. Which is L C. M. Multiplied by the sine of five sine of 60° and then subtract this by traffic. It's just three kg, multiplied by 9.81 m per second squared, multiplied by seven m. Which is the overall L bar multiplied by sine of degrees. Like so And we will equals this to zero. And then solving for all of this, we can find that the tea is essentially going to be 259.1 Newton like. So okay, now that we have find the t we can actually start with solving for PX and py so the way we want to do this is to essentially use the first equilibrium condition and the first equilibrium condition is essentially just sigma force at a certain X or Y. Or even Z. Access when it goes to zero. And let's just start with the X axis because that will be easier for us since we have last forces on the X axis. Okay. Along the X axis we have sigma F X equals to zero. And then we will subtract everything going to the right minus to the left. So in this case we will actually have uh T multiplied by co sign of five minus P X equals to zero. So P X is going to be T multiplied by cosine of five. And that will actually be 259. newton, which is the one that we just found, multiplied by cosine of 60 degrees like so and this will actually come up to be 129.6 newton. And that will be the answer for our P. X. Okay, so we can actually start eliminating now we know that the T. Is going to be 259.1 and we also know that the P is going to be 1 29.6 newton. And now we can start with solving for Y. So you solving for Y. We can look into the Y axis or yeah, solving for P. Y. Look into the Y axis. See my F Y equals to zero. We want to still use the same equilibrium condition and we are just going to actually um subtract everything going upwards minus downwards. Just like our normal convention. So the thing going upwards essentially just be multiplied by sci fi. And we want to subtract this with py subtract this with M bar G and then subtract this with Amtrak G as well, it goes to zero, these are all the forces going in the vertical direction. Now we can actually just start plugging things in and rearranging. So py is going to be T multiplied by sine of five plus M Bar G oh minus amber G. Sorry, minus M bar G. And then minus M. Two F G. As well. And there, we can actually just start plugging everything in. Remember that the T that we just found is 259.1 newton sign of five sine of 60 degrees minus M. Bar times G is essentially 10 kg multiplied by 9.81 m per second squared minus and G is three kg multiplied by 9. m per second squared as well. And therefore py is going to be 96.8 newton like, so, okay, so we know that P X is going to the left direction, which is correct because it is positive after we find everything here and we also know that P Y. Is going to the downward direction which is positive from the one that we found here. So based on our usual convention where everything going um to the left is essentially negative and everything going downwards is also negative. Then we can actually find the answer to this particular problem which is then going to actually be option C. The negative signs here essentially is just ah re emphasizing that P. X. Is going to the left and Py is going downwards and this positive values here is essentially just saying the magnitude of both PX and PY, therefore the correct answer for this particular problem is option. See if you can still have any questions or any confusion. There are a lot of other lessons on this particular example or example similar to this on our website. Thank you.