Slide 1 / 139 Slide 2 / 139 AP Physics C - Mechanics Dynamics - Applications of Newtons Laws 2015-12-03 www.njctl.org Slide 3 / 139 Slide 4 / 139 Table of Contents Click on the topic to go to that section Introduction · Sliding Blocks · · Fixed Axis Pulley Introduction · Suspended Pulleys · Plumb bob in car · The Banked Curve Inclined Plane and a Pulley · · Falling objects with Air Resistance Non Uniform Circular Motion · Return to Table of Contents Slide 5 / 139 Slide 6 / 139 Introduction This is not a typical chapter presentation. There are no multiple choice or free response questions to test student knowledge. Rather, eight comprehensive Dynamics questions are solved, Sliding Blocks step by step. Questions are built into the material - you should try to answer them before you move onto the next slide. These can be done in class, led by the teacher, or they can be done by the students outside of class. Return to Table of Contents
Slide 7 / 139 Slide 8 / 139 Sliding Blocks Sliding Blocks A A B B Ground Ground Block B, of mass 36 kg, is about to be pulled across the ground We're going to use dynamics concepts to find the maximum force by a rope with a force of F app . The coefficient of kinetic friction that may be applied by the rope before block A starts sliding on between block B and the ground is 0.27 . Block A, of mass 42 kg block B. is resting on Block B with a coefficient of static friction of 0.71 between the two blocks. Slide 9 / 139 Slide 10 / 139 Sliding Blocks Sliding Blocks A A B B Ground Ground Once someone starts pulling on the rope, what do we expect will Block B will start accelerating to the right, as it now as a net happen? applied force to the right. The friction between the ground and block B will oppose the applied force, so that will reduce the acceleration. What about block A? Slide 11 / 139 Slide 12 / 139 Sliding Blocks Sliding Blocks A A B B Ground Ground The most important thing to first notice about block A is that the The static friction force. applied force is NOT acting on block A, and when the free body diagrams are drawn, this force will not be shown on block A! Let's take two limiting cases to explain this. If μ s = 0, then that implies a frictionless surface between blocks A and B. When B is Intuitively, we know that something will be acting in the x pulled to the right, block A would just stay where it is relative to direction on block A. Block A will either stay right where it is, the ground, and would slide to the left on block B. relative to block B, and move to the right, or it will start sliding to the left. Relative to the ground, it will be moving to the right in If μ s were infinitely large, then it would just stick to block B and both cases. If not the force due to the rope, what is this force? slide with it to the right.
Slide 13 / 139 Slide 14 / 139 Sliding Blocks Sliding Blocks A A B B Ground Ground Let's formally start the problem: As in all dynamics problems, the first step is to draw the free body diagram. In this case, there are two objects of interest, blocks A and Given: B, so there should be two free body diagrams (FBD). But, they will not be the ones you'd expect! Here's a clue - there will not be a FBD for block A and a FBD for block B. where μ s is between blocks A and B and μ k is between block We will define different systems - two of them. And draw FBDs for B and the ground. each one. Take a few minutes with your group and propose two systems. Find: the maximum F app before Block A starts sliding on Block B. Slide 15 / 139 Slide 16 / 139 Sliding Blocks Sliding Blocks A A A B B Ground System 1 System 2 System 1 will be Block A. System 2 will be Blocks A and B. Here they are. Don't worry if you didn't get these exactly right. Each Given this starting point, try drawing free body diagrams for both system will be now be analyzed in detail. systems. Remember - when working with a system, only the external forces on the system are shown. F N(A+B) F NA a a System 2 System 1 A A f k F app f s B System 1 System 2 (m A +m B) g m A g Slide 17 / 139 Slide 18 / 139 Sliding Blocks Sliding Blocks F NA a Let's look at System 1 first. A A B f s The Normal force and the force due to Ground gravity are pretty straight forward. Gravity F NA pulls down on block A (m A g), and block B a exerts an upwards Normal force (F NA ) on m A g block A. f s Static friction seeks to maintain the relative position of the objects that are initially at rest. In most cases, this means if a force is Note the additional subscript on the applied to an object in one direction, then the static friction force is Normal force. That is to distinguish it from in the opposite direction. the Normal force acting on System 2. m A g But, here, the applied force is not on block A, it is on block B. And for block A to maintain its position relative to block B, as block B But what of the static friction force? Why is it pointed starts moving to the right, the static friction force acts to the right. to the right? Doesn't the static friction force oppose This results in block A staying on the right edge of block B - their motion? Isn't block A moving to the right? relative positions don't change.
Slide 19 / 139 Slide 20 / 139 Sliding Blocks Sliding Blocks F NA a A A B Time for System 2 now. B f s Ground Gravity is acting on the combination of blocks A and B, so it is equal to (m A + m B )g. The m A g F N(A+B) ground exerts an upward Normal a Another way to look at this problem is using your intuition again. force on the blocks, and is Again, that doesn't always work in physics, but it helps here. represented as F N(A+B) . f k F app If you pull block B, your experience tells you that block A will move Note the additional subscript on to the right. Maybe not as fast as block B is accelerating, but it will the Normal force. That is to move to the right. That requires a Force (Newton's Second Law). distinguish it from the Normal (m A +m B) g The only force available in the x direction is the static friction force force in System 1. between block A and B. So, f s goes on the free body diagram pointing to the right! Next, we'll look at the forces in the x direction. Slide 21 / 139 Slide 22 / 139 Sliding Blocks Sliding Blocks A A B B F app is acting to the right, and f k System 2 was defined as the acts between the ground and combination of blocks A and B. block B and is directed to the left. Therefore, the static friction force F N(A+B) F N(A+B) between blocks A and B is an a a The static friction force between internal force. blocks A and B is not drawn f k F app f k F app here. Internal forces are not shown on FBDs (Discuss in your groups Why? how Newton's Third Law comes into play here). (m A +m B) g (m A +m B) g Slide 23 / 139 Slide 24 / 139 Sliding Blocks F NA a Time to apply Newton's f s A Second Law to the FBDs. We'll start with System 1. System 1 m A g x direction y direction Took a little shortcut on the notation - since the objects are moving in the x direction, the subscript, x, will be omitted on the acceleration vector.
Slide 25 / 139 Slide 26 / 139 Sliding Blocks F NA F N(A+B) a a f s f k F app System 1 System 2 m A g (m A +m B) g System 2 equations: This gives us an expression for the acceleration of the total system (blocks A and B). If we want to find the point at which block A starts to slide, what assumption can we make about the acceleration of block A? Slide 27 / 139 Slide 28 / 139 Sliding Blocks Sliding Blocks F NA F N(A+B) F NA F N(A+B) a a a a f s f k F app f s f k F app System 1 System 2 System 1 System 2 m A g (m A +m B) g m A g (m A +m B) g Switching to the System 1 equations, for block A not to slide, its acceleration must be equal to the acceleration of Calculating f s from the Normal force: the total system, so a A = a A+B . Why? If a A is greater than a A+B , then block A would slide forward on block B. If a A is less than a A+B then it would slide Putting these two equation together and canceling m A : backwards. So, since we're looking for the exact point where the applied force causes A to slide, we have: Slide 29 / 139 Slide 30 / 139
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