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

Find (a) the tension in the cable.

Video transcript

Hi everyone today, we are going to find the tension of the court holding uniform aluminum bar where the bar itself is actually holding an advertising sign on the end of it. So right now we are given the mass of each things here. So first we are given the mass of the advertising sign, which is, I'm just gonna write it down as m a d fee, which is two kg. And then we are also given the mass of the uniform element um bar, which is a bar, it is going to be eight kg right here. And then we are given the length of the bar which is, I'm just going to be using the letter D distance, we're just going to be two m and this overall system right here is actually helped in equilibrium with one another, which means that the actual sigma torque of the overall system is going to be equals two zero. So in this case, I'm just using Z access as a reference access right here. So now we want to try to indicate what kind of torques are actually acting upon this uniform aluminum bar. So first, obviously we have what is actually being asked, which is the tension of the court right here. So we know that to calculate the torque or to actually find it, we have to Have a perpendicular force against the radius or against the length or distance itself. So in this case, since we have the tension at a 30° angle here, which is known, we want to try to do a projection where the tension itself is actually going to be perpendicular to this beam or to this bar. So this component of projection will then be the multiplied by the correction angle here, multiplied by sine of data, which is going to be 30°. So next we want to look into the weight of the advertising sign, which is going to be just the mass of the advertising sign here, which is known multiplied by gravity right here. And then lastly we have the last force which is just the actual mass or the actual weight of the uniform aluminum itself. And Because we know that the aluminum bar is actually going to be a uniform bar, we can determine that the center of mass is going to be right in the middle of the bar which has a length of two m. So in this case we have This whole thing to be two m. Well, the um The weight itself is going to be coming from one m away from the wall or in the middle of the actual bar itself. So now that we have indicated the three different forces or the three different talks, that's actually going to be acting upon our bar, we can write them down into this actual formula. Right? So, so not because we know that the actual system is at equilibrium, It doesn't really matter that whether you want to start with the torque going clockwise or the torque going counterclockwise, but in this case we know that the tension is going to be going counterclockwise like, so and then the weight of the advertising sign is going to be going clockwise along with the weight of the beam itself like. So so we can actually start with just um the torques going clockwise, we can start with and bar multiplied by gravity and then this is going to be the force. So to calculate the torque we have to use or multiply the force with the distance, which in this case is just going to be the I'm just going to indicate this with you bar minus d center of mass right here like, so which is just going to be D. Bar is two m and the center of mass is one m. So this value right here, all of it is going to be one m and then we know that the weight right here is already perpendicular to the distance or to the bar itself, so we don't have a correction angle and also this advertising sign weight as well as also already perpendicular. So we won't have a correction angle as well. So now we want to continue with the advertising sign, force or torque and this is going to be multiplied by the d. Of the bar itself or the overall length because it is acting upon at the end of the whole bar and then we are going to subtract this all of this thing with the torque going onto the other direction which is going counterclockwise and in this case we know that it is going to be the tension multiplied by sine data for the correction angle and then this is also going to be the bar because it is acting on the end of the bar itself. And we're going to equal this to zero because it is again at equilibrium. So now we want to actually pluck every single values that we know into this equation right here. So we start with that bar, we're just going to be eight kg multiplied by the G. We just want to use 9.81 m per second squared which is just the gravitational acceleration and then this is going to be two m minus one m. And then plus the M of advertising sign which is two kg multiplied by the G 9.81 m per second squared. And to clarify this, our multiplication science right here and then we wanna use D. Bar which is known to be two m right here. And then you want to subtract this whole thing with the Which is asked multiplied by sine of 30° And then the bar again which is just two m equals to zero. Now we can actually start solving for each of these terms. So first what I want to do is to actually remove or put the T sign 30° times two m to the other side. And we can actually do so by just like this And this will then be equals two. So all of these terms right here can actually be combined into eight kilograms times 9.81 m per second squared times one m plus. This is going to be two kg multiplied by 9.81 m per second squared, multiplied by two m. And then we know that sine of 30 degrees is going to be half, so half multiplied by two. Then this will just easily be detention. And then solving for this right side. Right here, we can get the value to be Newton, which is just going to be option d. So 118 Newton is going to be the answer or the magnitude of the tension in the court and option D. is going to be representing that and there you go.