Slide 1 / 139 Slide 2 / 139 Dynamics www.njctl.org Slide 3 / 139 Table of Contents Click on the topic to go to that section Introduction · Sliding Blocks · · Fixed Axis Pulley · Suspended Pulleys · Plumb bob in car · The Banked Curve Inclined Plane and a Pulley · · Falling objects with Air Resistance Non Uniform Circular Motion ·
Slide 4 / 139 Introduction Return to Table of Contents Slide 5 / 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, 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. Slide 6 / 139 Sliding Blocks Return to Table of Contents
Slide 7 / 139 Sliding Blocks A B Ground Block B, of mass 36 kg, is about to be pulled across the ground by a rope with a force of F app . The coefficient of kinetic friction between block B and the ground is 0.27 . Block A, of mass 42 kg is resting on Block B with a coefficient of static friction of 0.71 between the two blocks. Slide 8 / 139 Sliding Blocks A B Ground We're going to use dynamics concepts to find the maximum force that may be applied by the rope before block A starts sliding on block B. Slide 9 / 139 Sliding Blocks A B Ground Once someone starts pulling on the rope, what do we expect will happen?
Slide 10 / 139 Sliding Blocks A B Ground Block B will start accelerating to the right, as it now as a net 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 Sliding Blocks A B Ground The most important thing to first notice about block A is that the 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! Intuitively, we know that something will be acting in the x direction on block A. Block A will either stay right where it is, 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 both cases. If not the force due to the rope, what is this force? Slide 12 / 139 Sliding Blocks A B Ground The static friction force. 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 pulled to the right, block A would just stay where it is relative to the ground, and would slide to the left on block B. If μ s were infinitely large, then it would just stick to block B and slide with it to the right.
Slide 13 / 139 Sliding Blocks A B Ground Let's formally start the problem: Given: where μ s is between blocks A and B and μ k is between block B and the ground. Find: the maximum F app before Block A starts sliding on Block B. Slide 14 / 139 Sliding Blocks A B Ground 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 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. We will define different systems - two of them. And draw FBDs for each one. Take a few minutes with your group and propose two systems. Slide 15 / 139 Sliding Blocks A B Ground System 1 will be Block A. System 2 will be Blocks A and B. Given this starting point, try drawing free body diagrams for both systems. Remember - when working with a system, only the external forces on the system are shown. A A B System 1 System 2
Slide 16 / 139 Sliding Blocks A A B System 1 System 2 Here they are. Don't worry if you didn't get these exactly right. Each system will be now be analyzed in detail. F N(A+B) F NA a a System 2 System 1 f k F app f s (m A +m B) g m A g Slide 17 / 139 Sliding Blocks Let's look at System 1 first. A The Normal force and the force due to 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 block A. f s Note the additional subscript on the Normal force. That is to distinguish it from the Normal force acting on System 2. m A g But what of the static friction force? Why is it pointed to the right? Doesn't the static friction force oppose motion? Isn't block A moving to the right? Slide 18 / 139 Sliding Blocks F NA a A f s B Ground m A g 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 applied to an object in one direction, then the static friction force is in the opposite direction. 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 starts moving to the right, the static friction force acts to the right. This results in block A staying on the right edge of block B - their relative positions don't change.
Slide 19 / 139 Sliding Blocks F NA a A f s B Ground m A g Another way to look at this problem is using your intuition again. Again, that doesn't always work in physics, but it helps here. If you pull block B, your experience tells you that block A will move to the right. Maybe not as fast as block B is accelerating, but it will move to the right. That requires a Force (Newton's Second Law). The only force available in the x direction is the static friction force between block A and B. So, f s goes on the free body diagram pointing to the right! Slide 20 / 139 Sliding Blocks A Time for System 2 now. B Gravity is acting on the combination of blocks A and B, so it is equal to (m A + m B )g. The F N(A+B) ground exerts an upward Normal a force on the blocks, and is represented as F N(A+B) . f k F app Note the additional subscript on the Normal force. That is to distinguish it from the Normal (m A +m B) g force in System 1. Next, we'll look at the forces in the x direction. Slide 21 / 139 Sliding Blocks A B F app is acting to the right, and f k acts between the ground and block B and is directed to the left. F N(A+B) a The static friction force between blocks A and B is not drawn f k F app here. Why? (m A +m B) g
Slide 22 / 139 Sliding Blocks A B System 2 was defined as the combination of blocks A and B. Therefore, the static friction force F N(A+B) between blocks A and B is an a internal force. f k F app Internal forces are not shown on FBDs (Discuss in your groups how Newton's Third Law comes into play here). (m A +m B) g Slide 23 / 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 24 / 139
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 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 Switching to the System 1 equations, for block A not to slide, its acceleration must be equal to the acceleration of 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 backwards. So, since we're looking for the exact point where the applied force causes A to slide, we have:
Slide 28 / 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 Calculating f s from the Normal force: Putting these two equation together and canceling m A : Slide 29 / 139 Slide 30 / 139
Slide 31 / 139 Slide 32 / 139 Fixed Axis Pulley Return to Table of Contents Slide 33 / 139 Fixed Axis Pulley So far, pulleys have been used in constructions like Atwood's Machine and to allow for the horizontal and vertical motion of two objects in a system (the force is redirected) as shown below. Both of these systems used "fixed axis pulleys" as the pulley's axis of rotation does not change relative to a reference frame attached to the ground or table or ceiling.
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