 Hi, I'm Zor. Welcome to Unisor Education. I would like to solve a few problems related to composition of different forces acting on the object. Now, this lecture is part of the course called Physics 14 presented on Unisor.com. I suggest you to watch this and all other lectures of this course and the prerequisite for this course, which is called Mass 14s from the Unisor.com because every lecture has very detailed explanation, notes, whatever you want, and also there is certain educational functionality built into the site, which basically allows supervision and then you can actually take exam, if you want to, obviously. And the site is completely free and there are no advertisements. Alright, so, few problems. I have four problems. Let's do it one by one. First of all, we have a river, obviously a straight river with the flow, with uniformly flows. Now, there is a boat in the middle of the river, which is tied to the bank by rope of the length L. Now, there are two forces acting on this particular boat. Number one is the flow of the river and number two, the wind. So, these two forces, both flow is everywhere, right? So, including this point. Well, obviously, assume that we are talking about the point object. So, this is not really a big boat, but a point object with some kind of a mass in this particular object. So, what's given is the flow of the river is pushing the boat with the force F downstream. Now, the wind is perpendicular to the flow of the river and its force is 0.75 F, slightly less than the river flow. Now, as a result of these two forces, obviously the resultant force is something here, right? The vector sum of this and this. Well, on my picture, this looks bigger than this, but basically this is F and this is 3 quarters of the F, right? So, let's just disregard this slight picture inconsistency. So, anyway, this is the resultant force. Now, at the same time, since my boat is tied to the bank, obviously this rope has certain tension and goes to this direction. And this force, the resultant of the wind and the flow of the river and the tension are all balanced each other. So, basically this is the same as this by magnitude and opposite in direction. And that's why the boat stands still on the river. So, this is the condition of whatever we have here right now. What's necessary to determine is what is the force of tension, let's call it P, which keeps the whole thing in balance. And what is this distance from the bank? Alright, so this is really kind of an easy problem because if you know these two components, what is their resultant? Well, that's basic geometry. You have this is F and this is 0.75 F or 3 quarters, if you wish, of the F. So, what is the diagonal of this right triangle? Well, obviously it's 3 quarters, square plus 1 square F, which is what? 9, 16 plus 16, 16, 25, 16, so it's 5 quarters of the F, which is 1.25 F. So, this is 1.25 F, which must be equal to the tension of the rope because otherwise our boat will not be standing still in the middle of the river. Now, since this is one line, this is exactly opposite vectors. These triangles, this one and this one, obviously are two right triangles with the same angles. So, they are similar to each other. So, to determine this, I can use this similarity. Basically, L divided by D, this is the D, is the same thing as 1.25 divided by 3 quarters, right? F divided by 0.75 F. So, it's 5 quarters divided by 3 quarters, this is 5 quarters divided by 3 quarters, so it's 5 over 3, right? So, D is equal to 3 fifths of the L or 0.6 L. So, we have determined the tension and we have determined how far from the bank the boat actually stands. And obviously, if the wind is stronger, the boat will be a little bit further and my rope will be turned a little bit this way. If my flow is stronger, then the boat will be a little bit down there and my rope will be in this position. And if the wind is zero, then the flow will eventually bring this boat down to the bank. Obviously, it will be somewhere here and this will be the rope along the bank. Okay, that's my first problem. Good, next. Alright, now next problem. You have a bridge which is in the form of an arc of a known radius. Now, there is a car of the mass m on the top of the bridge. Now, what I need to know is how my speed of the car, well, considering the speed is constant, right? So, V is constant. Now, how this speed actually depends on the pressure which I have at this particular point. Pressure onto the bridge. Now, obviously, the speed is greater, the pressure is greater. Sorry, the speed is greater, the pressure is smaller because the car would be like flying, right, a little bit. And if the speed is very, very big, then the car will just fly off the bridge. Now, if the speed is less, the pressure will be greater at this point, right? So, basically, all I need to know is how my speed on that top depends on the pressure. That's the function which I would like to know. If pressure is given, what's my speed, basically? That's the main question. Here is another question. At what speed the passengers in the car will feel weightlessness? Alright, so, first of all, let's just talk about this. Now, this is a circular trajectory. Now, what kind of forces are acting on the car? Well, first of all, obviously, since we are on a circular trajectory, there must be some force which keeps it on this particular trajectory, right? Now, what is this force? Well, if the speed is v, then this force is mv squared divided by r, right? This is my centripetal force. This is studied from the previous lectures. v squared over r is the centripetal acceleration times mass gives me, according to the second law of Newton, gives me the force. Now, at the same time, so I know that this must be the force which keeps me on the trajectory. At the same time, I know from which components this particular force actually is constructed. Well, number one, we obviously have the weight. So, the weight also goes this direction and the weight, considering my conditions of this experiment are very close to the ground, my weight is equal to m times the acceleration of the free fall. So, that's my weight and it's given because m is given, right? At the same time, there is something which is the pressure. The pressure goes from the car to the bridge. So, the pressure is not acting against the car, it's acting against the bridge. However, there is a reaction of the bridge which basically goes this way, which is equal to the pressure but opposite in sign, right? So, it goes that way. And what can I say right now about this? That the combination of the weight and the pressure, and I will put pressure with a minus sign because it goes opposite to, I mean, sorry, not the pressure, the equal to the pressure, but it's a reaction of the bridge. But that's why I put it with a minus sign because it's directed upwards. Pressure goes down, reaction goes up. And this is supposed to be equal to my force which keeps on the trajectory, keeps the car on the trajectory. Now, if I will put it a little bit more detail, it would be mv squared divided by r, from which we can derive how v, my speed, depends on the p, right? So, what that would be? v is equal to, well, let's just divide it by m. And I will have rg minus p over m and square root. That's my v, right? So, that's the dependency between p, which is the pressure, and the v, which is the speed in this particular position. And my second question was when will the passenger in the car feel weightless? Well, obviously, that's when this particular p is equal to zero because weight is actually, how do we feel the weight? Well, we press down on the floor and the floor presses on us and that's how we feel our weight. We feel it as the pressure from the floor to us, right? So, obviously, if we don't press on the floor, the floor doesn't press on us so we don't really feel anything, right? There is no feeling of the pressure anywhere. So, when p is equal to zero, we will have v0, which is the speed of the weightlessness, is rg. Where g is about 9.8 m per second square, r is whatever the r is, and that's my speed. If I will drive on this arc with this speed, at the top I will feel weightless. Next. Next is a simple problem about pendulum. So, if you have a pendulum, this is trajectory. Now, the initial position is this. The angle is phi. My question is, if given the mass, and obviously the thread is weightless, no tension, I mean, there is tension, but there is no stretching. Sorry, there is no stretching. So, unstretchable weightless thread. And this is the initial position. My question is, what's the tension of the thread in this particular position? Well, let's just think about it. What kind of forces again are acting against this particular mass? Well, obviously there is a weight. On another hand, there is a tension which basically goes this way. It's called T tension. Now, why the object goes this way? Well, because the direction of these two forces is tangential to my trajectory, right? So, the resultant of this plus this is equal to this force, which basically pushes it along the trajectory. And regardless of where exactly my pendulum is located, anywhere, wherever you are, force goes down, tension goes to the center, and the resultant will be on this side, it will be in this direction, on this side, it will be in this direction, but it's always tangential. The resultant force is always tangential to a trajectory because that's exactly what moves it along this particular trajectory, right? So, my question is, what then is my tension? Well, this is tangential, so this is the perpendicular. So, if I will put parallel to this one and parallel to this one, obviously this is the right angle and this is the right angle, right? So, I know the weight, which is basically a hypotenuse in this right triangle, from which I can determine this, which is exactly the tension. If this angle is phi, then this angle is also phi, right? So, my force, which is tangential, which is the force which pushes along the trajectory, would be w times cosine phi, and tension, which is this, would be w times sine phi, which is mg sine phi, because again, they're resuming everything on the surface of the Earth, so my weight, which is the gravity, is equal to mass times acceleration of the free-falling. So, these are two components which are the tension, which goes along this line, which connects my mass with the center, and the tension acts on the mass by pulling it up, weight pulls it down, and the resultant, which is this force, is tangential to trajectory. Well, we will actually talk about pendulum at length, it's really a complicated movement, since your angle is changing all the time, and cosine and sine of this angle are obviously changing with it, that's why the tension is also changing. Tension is the weight times sine of the angle, so we're dealing with a situation where the force is actually variable, right? So, the force of the tension is variable, and the force which pushes along the trajectory is variable, it depends on the time, which means it's not really like a simple F equals to MA, where F is a constant force, and M is a mass, and A is a constant acceleration, this is much more complex, think, it goes to differential equations, but we will spend some time on this some other time. Now, and the last problem is the following. Let's say you have the wall of the castle, which is under siege, you have some troops which are attacking this castle, now there is a cannon here on the top, and it shoots the cannon, and it shoots the projectile against the troops, right? So, the projectile obviously goes this way. Why? Well, because it has initial speed towards horizontal direction, right? On one hand, but on the other hand, there is a weight which goes down. So, my projectile moves simultaneously this way, and this way, and the result is this way, right? Now, what I'm interested in, number one, I have to find time of landing. Now, what's given? Well, given obviously the height of the wall, and there is the speed I'm shooting this projectile. Okay, so, number one, what is the time to land? Now, and let's ignore the air resistance, right? So, this projectile goes horizontally with always the same speed V. Now, vertically, it basically falls down. So, every second it falls down with certain acceleration, because the force of gravity is a constant force. It's equal to M times G, right? Where M is the mass of the projectile. So, I have a constant force. Since I have a constant force, my height H is basically taken by the familiar formula. Acceleration times square of the time divided by 2. This is how much time it takes to cover the length H with a constant acceleration G. Now, H is given. So, from here, we can find VT. T landing is equal to 2H divided by G and square root, right? From here. 2H divided by G and square root. So, I found the time. Now, next days I would like to find the distance. How far this distance, how far I will shoot this projectile? Well, obviously, since horizontally I'm always making, my motion is uniform, right? So, it's a uniform velocity time, I know. So, my distance is equal to V times square root of 2H divided by G. Now, and the last thing which I would like to know is, what's the magnitude of the speed? Now, this is my landing piece, right? And the speed is directed at this way, which is basically a continuation of horizontal movement and vertical movement. So, my question right now is, if this is the vector of speed, what is this component, horizontal component, and what is the vertical component at the point of landing? Well, I know the horizontal component is always V, because it's not changing, there is no air resistance. My vertical component is the speed which falling object gains after this time, if acceleration is G. Well, we know the formula, right? V is equal to GT, right? V landing, it's a vertical component I'm talking about, and T of landing, right? So, V landing is equal to G times this, because G goes under the square root, it would be 2HG. Now, if I know the vertical component and horizontal component of my landing speed, then obviously I can use the Pythagorean theorem and get the length of this hypotenuse, if this is my vertical and this is my horizontal components. So, my V ending, this is the speed, is equal to square of this plus square of this, V square plus 2HG. So, that's the answer. Alright, basically, that's all I wanted to present today. These are really very easy problems. I think I should come up with something a little bit more complex, and I will. But anyway, these are kind of introductory problems into basically all the dynamics of these superposition of the forces, using gravity, pendulum, and all different kinds of scenarios we can consider. Alright, what I do recommend you to do is go to the website Unisor.com to this course, Physics 14. Then you can open the section called Mechanics. Within the Mechanics, there is a Dynamics section where I'm considering superposition of the forces. And in there, there are some theoretical lectures which have been already put on the net. And there are these problems. I think it's called problems number one. Yes, problems number one within the superposition. And try to do it yourself. Whatever the problem you have over there, try to do it yourself, and then check against the answers, which also are on the web in these notes. Just to check how you're basically going through these relatively easy problems. And then next lecture will probably be more complex. Alright, thanks very much and good luck.