[MUSIC] Okay, it's time for problems. We'll solve a couple together first, and then you'll do some for assessment. This is called the block, and that contains a pulley. We'll make the approximation that the pulley's light and the turn without friction, and that the ropes and the block are light. Meaning that their mass is much less than mine. The pulleys turn easily and the rope is light, so the tension everywhere in the rope is the same, call it t. So to our question. If my weight is 700 newtons, what force do I use to pull on the rope. The connection above the block supports me and the light ropes and block. If we assume that I'm not accelerating, this connection applies a force roughly equal to my weight, about 700 Newtons. Let's draw our free body diagram. The three ropes that support me pulled downwards on this connection. If the total force on the connection is zero, then the sum of the tensions in the three section of rope is 700 newtons. 3T approximately equal to 700 newtons, T, approximately equal to 230 newtons. 230 newtons is not a very big force, and that's why I can climb up the rope and still talk at the same time. In practice, pulleys and blocks are often used to provide extra force. Here's an example. This end of rope is one of the things that's pulling the boat along so it's under quite a bit of tension. [INAUDIBLE] move the [INAUDIBLE] aside. It's good to have some leverage. [SOUND] Wow! That was refreshing. Finally, here's a problem involving circular motion. It's a bit tricky so we'll do it together. It shows some useful techniques that you'll use in simpler problems. [LAUGH] Including some problems in the test. Sassy, the flying pig is the mascot of our second year laboratory. Let's calculate the time she takes for just one circle using just geometry. The distance from her center to the support point is 0.97m. The angle a cord makes with the vertical is 37 degrees. From that information, we want to calculate the period T. How can we do this? Well, here's a strategy. First, we have uniform circular motion, so we can relate the period T to the acceleration. Then we can use Newton's laws to obtain another equation for the acceleration and solve the two equations to get T. Okay, over to you. First, see if you can use the acceleration a to relate the period t to the given information d and theta. Yes the acceleration is centripetal, so a equals r omega squared. And we can substitute omega equals 2 pi over T. Then the right triangle gives us a equals D sine theta. And then we rearrange to get the expression for T in terms of acceleration a. Now we need the acceleration, which is where Newton's second law comes in. So the next step is to draw a diagram showing all of the external forces acting on the pig and their resultant. By the way, I've already used T for the period, so this time I'm going to use F for the tension in the string. But first, another question for you. That was easy. The acceleration is horizontal, so the total force must be horizontal. So we'll draw our free body diagram for the flying pig. The tension F in the string acts in the direction of the string. Otherwise the string would bend. Then there's the lift L generated by the wings. Is symmetrical, so this must act in the plane of symmetry. How big are F plus L? Well, we know that their horizontal components provide the only horizontal forces. So, the horizontal component of F plus L is the total force. The weight is easy. It acts downwards, and it must balance the vertical component of F plus L. Let's now use Newton's second law to involve the acceleration a. Over to you. But remember, Newton's laws are vector equations, so they often give us two or three scalar equations. So, here are the versions of Newton's second law in the two directions Here, we eliminate the unknown, F plus L, by dividing one equation by the other. Which give a in terms of the things that we know. Wow. We're nearly home. Remember our original equation, 1, the equation for the period. We substitute for a and we period equals 2 pi square root D cos theta over g. Here's a very important step. I check the units. D is in meters, g is in meter per second, per second, sign theta is a number, so the expression inside the square root is in seconds squared. So yes, my answer will be in seconds. That's good news. Next, we put in the values of the knowns, calculate the final answer, and express it with the appropriate number of significant figures. I hit the calculator and 1.8 seconds. Our camera runs at 25 frames per second, so 1.8 seconds would be 45 frames, which is what the cameras says too. Good news. By the way, Sassy's name comes from ancient Greece. And if you meet her on your Odyssey through physics, watch out for here wine. Next week we look more closely at the nature of some of the forces that we've quantified here and we'll answer that puzzle. Why does the normal force sometimes have almost exactly the same magnitude as the weight? See you then, but for now there's a quiz. [MUSIC]