Since the block has a tendency to slide down, the frictional force points up the inclined plane. As long as the block is in equilibrium
|
|
- Emory Gardner
- 6 years ago
- Views:
Transcription
1 Friction Whatever we have studied so far, we have always taken the force applied by one surface on an object to be normal to the surface. In doing so, we have been making an approximation i.e., we have been neglecting a very important force viz., the frictional force. In this lecture we look at the frictional force in various situations. In this lecture when we talk about friction, we would mean frictional force between two dry surfaces. This is known as Coulomb friction. Frictional forces also exist when there is a thin film of liquid between two surfaces or within a liquid itself. This is known as the viscous force. We will not be talking about such forces and will focus our attention on Coulomb friction i.e., frictional forces between two dry surfaces only. Frictional force always opposes the motion or tendency of an object to move against another object or against a surface. We distinguish between two kinds of frictional forces - static and kinetic - because it is observed that kinetic frictional force is slightly less than maximum static frictional force. Let us now perform the following experiment. Put a block on a rough surface and pull it by a force F (see figure 1). Since the force F has a tendency to move the block, the frictional force acts in the opposite direction and opposes the applied force F. All the forces acting on the block are shown in figure 1. Note that I have shown the weight and the normal reaction acting at two different points on the block. I leave it for you to think why should the weight and the normal reaction not act along the same vertical line? It is observed that the block does not move until the applied force F reaches a maximum value F max. Thus from F = 0 up to F = F max, the frictional force adjusts itself so that it is just sufficient to stop the motion. It was observed by Coulombs that F max is proportional to the normal 1
2 reaction of the surface on the object. You can observe all this while trying to push a table across the room; heavier the table, larger the push required to move it. Thus we can write where µ s is known as the coefficient of static friction. It should be emphasize again that is the maximum possible value of frictional force, applicable when the object is about to stop, otherwise frictional force could be less than, just sufficient to prevent motion. We also note that frictional force is independent of the area of contact and depends only on N. As the applied force F goes beyond F, the body starts moving now experience slightly less force compound to. This force is seem to be when is known as the coefficient of kinetic friction. At low velocities it is a constant but decrease slightly at high velocities. A schematic plot of frictional force F as a function of the applied force is as shown in figure 2. Values of frictional coefficients for different materials vary from almost zero (ice on ice) to as large as 0.9 (rubber tire on cemented road) always remaining less than 1. A quick way of estimating the value of static friction is to look at the motion an object on an inclined plane. Its free-body diagram is given in figure 3. 2
3 Since the block has a tendency to slide down, the frictional force points up the inclined plane. As long as the block is in equilibrium As θ is increased, mgsinθ increases and when it goes past the maximum possible value of friction f max the block starts sliding down. Thus at the angle at which it slides down we have Let us now solve a couple of simple standard examples involving static friction/kinetic friction. Example 1: A 50kg block is on an inclined plane of 30. The coefficient of static friction between the block and the plane is 0.3. We wish to determine the range of m under the block will be in equilibrium (see figure 4). 3
4 The very first question that we address first is: why is there a range of m? It is because of the friction. If friction were absent, there is only one value of m,, that will balance the 50kg mass. On the other hand, friction can adjust itself according to the kind of motion; it can even change direction depending upon which way is the 50kg block slipping. Thus when friction is present, there is range of m, starting from when the 50kg block has a tendency to slide down the ramp to when m pulls it up the ramp. Let us calculate these values. We first take the case when the 50kg block is about to slide down the ramp. At that point, the friction will be pointing up with its magnitude at its maximum value. In that case the free body diagram of 50kg block is as follows. A reminder here that a free body diagram is the one where we isolate a body and replace all other elements in contact with it by the forces they apply on the body. Thus the ramp surface is replaced by its normal reaction N, and the frictional force (max) µn, and the string attached with 4
5 m is replaced by the corresponding tension T in it. We reiterate that we have taken the direction of up the ramp because we have assumed the block to be sliding down and we have taken the friction at its maximum possible value. This gives us the smallest possible mass m. Taking directions along the ramp to be the x-direction and that perpendicular to it to be the y- direction gives 50g sin30 = T + µ N Similarly gives N = 50g sin30 To get T, we apply the force balance equation to mass m to get Solving these equations with g = 10ms -2 gives T = mg m = 7.68kg. Now as we start increasing m, the frictional force would become smaller and smaller than its maximum value µn, eventually changing direction and increasing up to in the µn in the opposite direction. The free body diagram of 50 kg block will then look like (note that the direction of friction is opposite to that in figure 5) figure 6. Then (taking the x-axis along the ramp) implies that 5
6 T = 50g sin30 + µn The other two equations remain the same as in the previous case. Solution of these equations gives m = 42.32kg Thus we see that due to friction, there is a range of mass m from 7.68 kg to kg that can balance the 50 kg weight on the ramp; for all the values of m between the values determined above, the frictional force will be less than its maximum value. I leave this example by asking you: at what value of m will the frictional force be zero? We now take certain specific example of friction viz. rotation of a solid cylinder against a dry surface; this is known as dry thrust waving. We then discuss the case of belt friction and finally the square screw thread and the screw jack. In these discussions we closely follow the book by Shames on Engineering Mechanics. Example 2: Let us first take the case of a cylinder of radiation R and mass m kept vertically on a rough surface. It is to be rotated about the vertical by applying a torque T. We wish to calculate T when the cylinder is about to rotate. Or in other words what is value of maximum T so that the cylinder does not rotate see figure 7). The coefficient of friction between the cylinder and the surface is µ. 6
7 In this example, we will have to consider the torque generated by friction. To do this, let us consider a ring of radius r and thickness 'dr' and see how much frictional force does this experience (see figure 8)? If we assume the normal reaction to be evenly distributed, then normal reaction on the ring is The frictional force on it at radius r is therefore And the torque due to this friction is therefore That is the maximum torque that can be tolerated against friction. Of course we have assumed in this derivation that the weight of the cylinder is evenly distributed. If the weight is concentrated more towards the centre, T would be less and it would be more if the weight is more towards the periphery. The next example that we take is that of a block on a ramp again. Example 3: A block of mass 100 kg is on a ramp of angle 30. We wish to determine the magnitude and direction of the frictional force for the applied force F = 600N, F = 500N and F = 100N. This is a problem where we do not know a priori whether the 7
8 block will be moving up or down the plane or whether it will have a tendency to move up or down the plane. So while solving we have to keep it mind : 1. Whether the block will remain stationary or move. 2. Which way does the block have tendency to move or which way does it move? That will determine the direction of friction. To get the answer, we see that if the maximum possible static friction is not able to stop the block, it will move and in that case the friction will be kinetic. We will check that as we solve the problem. To understand which way will be the friction act, let us first assume that there is no friction and calculate the corresponding F o for equilibrium. If applied F is greater F o than the block will have a tendency to move up, otherwise it will have a tendency to move down the plane. Free-body diagram of the block looks (friction = 0) as follows. Taking x-axis along the plane and y-axis perpendicular to it (see figure) we get from the equilibrium conditions 8
9 Let us see what happens if we increase F beyond 558.6N. In that case the component of F up the slope will increase and the block will have a tendency to move up. Let us now answer the second question whether at 600N we get ( Fcos30-100gsin 30 )> Max. Friction OR ( Fcos30-100gsin θ ) < Max Friction From N=100gcos30 +Fsin30 N=981 =490 = = N Thus maximum value of frictional force is On the other hand, Thus only 29.6N of frictional force is required to keep the block in equilibrium. This is well below the maximum possible frictional force. So under 600N, the block will be in equilibrium and the direction of friction will be down the plane. The free body diagram will look as follows in this case. 9
10 Now we consider the case when the horizontal push is changed to 500N. At F = 500N ( Fcos30-100gsin30 ) will be negative so the block will have a tendency to slide down. Let us again calculate the maximum possible frictional force µ s N and ( 100gsin30º - Fcos30º ) for F = 500N. and N= 100gcos30 - Fsin30 = = Further 100gsin30 -Fcos30 = =57N which is again well below the maximum frictional force of 219.7N. So the block will remain in equilibrium with its free-body diagram a follows. 10
11 Next we consider the case of the force pushing the block to be 100N. At F = 100N obviously the block as a tendency to slide down since it does so for F = 500N. For 100N case and N = 100gcos30 +Fsin30 = = N Further This force is larger than the maximum frictional (static) force. Thus the block will start sliding down, However when the block slides down, the friction will no more be given by µ s N but by µ k N. Thus the frictional force is =.17x N = 152.8N. Thus the free body diagram of the block will look as given below. 11
12 Another way of doing this problem would be to apply the external force F and calculate the frictional force required to keep the body in equilibrium by assuming a direction for the frictional force. This would give both the direction and magnitude of the frictional force required for equilibrium of the block. If this force is below µ s N, the body will remain in equilibrium; if it exceeds µ s N, it will move. Next we consider friction on a belt or a rope going around an object. For simplicity, we consider the rope to be going around a pulley and making contact with it over an angle.let T 1 >T 2 so that the rope has a tendency to slide to the right and therefore experiences a frictional force to the left. 12
13 Let us now consider a small section of angular width normal to this section be shown by the dotted line. so and see how it gets balanced. Let the Normal reaction in this direction balances the components of T and direction so that. in the opposite The frictional force is balanced by so that Solving this equation, we get 13
14 Thus tension in the rope increases exponentially with the contact angle. Notice that the relationship T 1 =T 2 does not depend on the radius of the pulley but only on the contact angle. Thus, even if the shape of the contact is not perfectly circular, the same derivation applies and the relationship between T 1 and T 2 remains the same, depending only on the contact angle. Because of the relationship, if we take a cylindrical object - say a pencil -and let a string pass over it, then. If we wrap the string once more, then, and so on. This is nicely demonstrated by balancing a heavy object like a bottle filled with water by a light object like a bunch of keys tied on two ends of a string and the string itself going many times over a rough stick. More you wrap the string, a bigger filled bottle would you be able to balance with the same bunch of keys. If you measure the weights of the objects carefully, you should be able to check the relationship derived above and also get the value of the coefficient of friction (see figure 15). We unknowingly use the effect discussed above when we wrap a clothes-string over the nail on which it is tied many times to make it tight. Similarly boats are tied with the ropes holding them going over a pole many times. We now solve an example where the surface over which the rope passes is not cylindrical. Example: A 70kg load is being lifted by tying a rope to it and passing the rope over a tree-trunk. The persons lifting it have to apply a force of 1800N just when the load starts moving up. What 14
15 is the coefficient of static friction between the tree-trunk and the rope if the contact angle between the trunk and the rope is 120? (see figure 16). Since the rope is in contact with the rope over an angle of, the relationship between T on the slide of the persons pulling the rope and T o on the side of the load is (friction is towards the load since rope is about to move towards the persons pulling it) Notice that we are not worrying about the shape of the tree-trunk but rather only about the contact angle. It is given that T = 1800N and T 0 =70 x 9.8 = 688N. Therefore we have Finally, we take the example of a square screw thread and a screw jack. In this case a screw with square threads passes through a nut and we wish to consider the action of the nut on the screw. You have seen this in a jack where a load (say a car) is lifted by rotating a screw. So a load W is lifted by applying a torque T on the screw (figure 17). 15
16 The threads on the screw make an angle α with the horizontal so that where p is the pitch and r is the mean radius of the screw. We wish to find the torque T (minimum) that is needed to lift the load W. When the screw is being lifted up the nut applies a normal reaction and a frictional force on the screw shown in figure 16. Keep in mind that the normal reaction and the friction act on the periphery of the screw. Thus they also apply a couple about the vertical axis on the screw. The figure also shows the load W. Balancing the force in the vertical direction and the torque about the vertical axis gives Solving these equations gives 16
17 This is the torque needed to lift the load W. Let us now say that we raise the load and then release the torque. Will the jack self-lock i.e., would it hold the load where it is and would not unwind. We want to find the condition for this. When the screw self-locks, it will have a tendency to move down. Thus the free body diagram of the screw in this condition looks as follows. Balancing the vertical force now gives and if the screw is not to unwind, the torque due to the friction should be large than that due N. Thus for self-locking we require that Thus is the condition for self-locking. Under this condition the screw will not unwind by itself as torque due to the frictional force would be sufficient to prevent the unwinding due to the torque arising from N. 17
18 Example: Let us now solve a household example of this. Sometimes you see that experimental tables or refrigerators at home have screw-like contraption at their legs to adjust their heights. Let the radius of such a screw be 2mm. We wish to know what should be minimum number of thread per cm so that the screw self-locks for µ S =0.15. For self locking we know that This gives on substitution If the number threads per cm is N then This gives Thus there should be a minimum of 6 threads per centimeter in order that the screw self-locks itself. We conclude this lecture with this example. In this lecture we have learnt about the frictional forces and how they act under different situations. 18
STATICS. Friction VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.
Eighth E 8 Friction CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University Contents Introduction Laws of Dry Friction.
More informationChapter 10: Friction A gem cannot be polished without friction, nor an individual perfected without
Chapter 10: Friction 10-1 Chapter 10 Friction A gem cannot be polished without friction, nor an individual perfected without trials. Lucius Annaeus Seneca (4 BC - 65 AD) 10.1 Overview When two bodies are
More informationEngineering Mechanics. Friction in Action
Engineering Mechanics Friction in Action What is friction? Friction is a retarding force that opposes motion. Friction types: Static friction Kinetic friction Fluid friction Sources of dry friction Dry
More informationThe Laws of Motion. Newton s first law Force Mass Newton s second law Gravitational Force Newton s third law Examples
The Laws of Motion Newton s first law Force Mass Newton s second law Gravitational Force Newton s third law Examples Gravitational Force Gravitational force is a vector Expressed by Newton s Law of Universal
More informationPhysics 101 Lecture 5 Newton`s Laws
Physics 101 Lecture 5 Newton`s Laws Dr. Ali ÖVGÜN EMU Physics Department The Laws of Motion q Newton s first law q Force q Mass q Newton s second law q Newton s third law qfrictional forces q Examples
More informationPhys101 Second Major-152 Zero Version Coordinator: Dr. W. Basheer Monday, March 07, 2016 Page: 1
Phys101 Second Major-15 Zero Version Coordinator: Dr. W. Basheer Monday, March 07, 016 Page: 1 Q1. Figure 1 shows two masses; m 1 = 4.0 and m = 6.0 which are connected by a massless rope passing over a
More informationPart A Atwood Machines Please try this link:
LAST NAME FIRST NAME DATE Assignment 2 Inclined Planes, Pulleys and Accelerating Fluids Problems 83, 108 & 109 (and some handouts) Part A Atwood Machines Please try this link: http://www.wiley.com/college/halliday/0470469080/simulations/sim20/sim20.html
More informationName Date Period PROBLEM SET: ROTATIONAL DYNAMICS
Accelerated Physics Rotational Dynamics Problem Set Page 1 of 5 Name Date Period PROBLEM SET: ROTATIONAL DYNAMICS Directions: Show all work on a separate piece of paper. Box your final answer. Don t forget
More informationFRICTIONAL FORCES. Direction of frictional forces... (not always obvious)... CHAPTER 5 APPLICATIONS OF NEWTON S LAWS
RICTIONAL ORCES CHAPTER 5 APPLICATIONS O NEWTON S LAWS rictional forces Static friction Kinetic friction Centripetal force Centripetal acceleration Loop-the-loop Drag force Terminal velocity Direction
More informationOnline homework #6 due on Tue March 24
Online homework #6 due on Tue March 24 Problem 5.22 Part A: give your answer with only 2 significant digits (i.e. round answer and drop less significant digits) 51 Equilibrium Question 52 1 Using Newton
More informationEngineering Mechanics: Statics
Engineering Mechanics: Statics Chapter 6B: Applications of Friction in Machines Wedges Used to produce small position adjustments of a body or to apply large forces When sliding is impending, the resultant
More informationReview: Advanced Applications of Newton's Laws
Review: Advanced Applications of Newton's Laws 1. The free-body diagram of a wagon being pulled along a horizontal surface is best represented by a. A d. D b. B e. E c. C 2. The free-body diagram of a
More informationA. B. C. D. E. v x. ΣF x
Q4.3 The graph to the right shows the velocity of an object as a function of time. Which of the graphs below best shows the net force versus time for this object? 0 v x t ΣF x ΣF x ΣF x ΣF x ΣF x 0 t 0
More informationEngineering Mechanics: Statics
Engineering Mechanics: Statics Chapter 6B: Applications of Friction in Machines Wedges Used to produce small position adjustments of a body or to apply large forces When sliding is impending, the resultant
More informationAssignment 9. to roll without slipping, how large must F be? Ans: F = R d mgsinθ.
Assignment 9 1. A heavy cylindrical container is being rolled up an incline as shown, by applying a force parallel to the incline. The static friction coefficient is µ s. The cylinder has radius R, mass
More informationPhysics B Newton s Laws AP Review Packet
Force A force is a push or pull on an object. Forces cause an object to accelerate To speed up To slow down To change direction Unit: Newton (SI system) Newton s First Law The Law of Inertia. A body in
More informationChapter 4. Forces and Newton s Laws of Motion. continued
Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting
More informationPhysics 8 Monday, October 12, 2015
Physics 8 Monday, October 12, 2015 HW5 will be due Friday. (HW5 is just Ch9 and Ch10 problems.) You re reading Chapter 12 ( torque ) this week, even though in class we re just finishing Ch10 / starting
More informationGeneral Physics I Spring Applying Newton s Laws
General Physics I Spring 2011 pplying Newton s Laws 1 Friction When you push horizontally on a heavy box at rest on a horizontal floor with a steadily increasing force, the box will remain at rest initially,
More informationMarch 10, P12 Inclined Planes.notebook. Physics 12. Inclined Planes. Push it Up Song
Physics 12 Inclined Planes Push it Up Song 1 Bell Work A box is pushed up a ramp at constant velocity. Draw a neatly labeled FBD showing all of the forces acting on the box. direction of motion θ F p F
More informationd. Determine the power output of the boy required to sustain this velocity.
AP Physics C Dynamics Free Response Problems 1. A 45 kg boy stands on 30 kg platform suspended by a rope passing over a stationary pulley that is free to rotate. The other end of the rope is held by the
More informationh p://edugen.wileyplus.com/edugen/courses/crs1404/pc/b02/c2hlch...
If you a empt to slide one... 1 of 1 16-Sep-12 19:29 APPENDIX B If you attempt to slide one solid object across another, the sliding is resisted by interactions between the surfaces of the two objects.
More informationReading Quiz. Chapter 5. Physics 111, Concordia College
Reading Quiz Chapter 5 1. The coefficient of static friction is A. smaller than the coefficient of kinetic friction. B. equal to the coefficient of kinetic friction. C. larger than the coefficient of kinetic
More informationPH 2213 : Chapter 05 Homework Solutions
PH 2213 : Chapter 05 Homework Solutions Problem 5.4 : The coefficient of static friction between hard rubber and normal street pavement is about 0.90. On how steep a hill (maximum angle) can you leave
More informationHSC PHYSICS ONLINE B F BA. repulsion between two negatively charged objects. attraction between a negative charge and a positive charge
HSC PHYSICS ONLINE DYNAMICS TYPES O ORCES Electrostatic force (force mediated by a field - long range: action at a distance) the attractive or repulsion between two stationary charged objects. AB A B BA
More informationAnnouncements 23 Sep 2014
Announcements 23 Sep 2014 1. After today, just one more lecture of new material before Exam 1!! a. Exam 1: Oct 2 Oct 7 (2 pm) in the Testing Center, late fee after Oct 6 2 pm b. Exam review sessions by
More informationPhysics. TOPIC : Friction. 1. To avoid slipping while walking on ice, one should take smaller steps because of the
TOPIC : Friction Date : Marks : 0 mks Time : ½ hr. To avoid slipping while walking on ice, one should take smaller steps because of the Friction of ice is large (b Larger normal reaction (c Friction of
More informationOverview. Dry Friction Wedges Flatbelts Screws Bearings Rolling Resistance
Friction Chapter 8 Overview Dry Friction Wedges Flatbelts Screws Bearings Rolling Resistance Dry Friction Friction is defined as a force of resistance acting on a body which prevents slipping of the body
More informationSolution of HW4. and m 2
Solution of HW4 9. REASONING AND SOLUION he magnitude of the gravitational force between any two of the particles is given by Newton's law of universal gravitation: F = Gm 1 m / r where m 1 and m are the
More informationConsider the case of a 100 N. mass on a horizontal surface as shown below:
1.9.1 Introduction The study of friction is called: The force of friction is defined as: The force of friction acting between two surfaces has three properties: i) ii) iii) Consider the case of a 100 N.
More informationActually, no perfectly frictionless surface exists. For two surfaces in contact, tangential forces, called friction forces, will develop if one
Chapter 8 Friction Actually, no perfectly frictionless surface exists. For two surfaces in contact, tangential forces, called friction forces, will develop if one attempts to move one relative to the other.
More information4.1.1 Extra Practice 4.1 Analyze the effects of a uniform force (magnitude and direction.)
4.1.1 Extra Practice 4.1 Analyze the effects of a uniform force (magnitude and direction.) Frictional Forces LEVEL 2 1. (HRW 6-3) A bedroom bureau with a mass of 45 kg, including drawers and clothing,
More informationLecture 8: Friction Part 1 Friction Phenomena
Lecture 8: Friction Part 1 Friction Phenomena Friction is a force which is generated between. Its direction is always to that of the motion, or tendency for a motion. F mg P N Figure 1 Figure 1 shows a
More informationCourse Material Engineering Mechanics. Topic: Friction
Course Material Engineering Mechanics Topic: Friction by Dr.M.Madhavi, Professor, Department of Mechanical Engineering, M.V.S.R.Engineering College, Hyderabad. Contents PART I : Introduction to Friction
More informationSo now that we ve practiced with Newton s Laws, we can go back to combine kinematics with Newton s Laws in this example.
Lecture 7 Force and Motion Practice with Free-body Diagrams and ewton s Laws So now that we ve practiced with ewton s Laws, we can go back to combine kinematics with ewton s Laws in this example. Example
More informationAnnouncements 24 Sep 2013
Announcements 24 Sep 2013 1. If you have questions on exam 1 2. Newton s 2 nd Law Problems: F m a. Inclined planes b. Pulleys c. Ropes d. Friction e. Etc Remember N2 is a blueprint for obtaining a useful
More informationIsaac Newton ( ) 1687 Published Principia Invented Calculus 3 Laws of Motion Universal Law of Gravity
Isaac Newton (1642-1727) 1687 Published Principia Invented Calculus 3 Laws of Motion Universal Law of Gravity Newton s First Law (Law of Inertia) An object will remain at rest or in a constant state of
More informationResolving Forces. This idea can be applied to forces:
Page 1 Statics esolving Forces... 2 Example 1... 3 Example 2... 5 esolving Forces into Components... 6 esolving Several Forces into Components... 6 Example 3... 7 Equilibrium of Coplanar Forces...8 Example
More informationE X P E R I M E N T 6
E X P E R I M E N T 6 Static & Kinetic Friction Produced by the Physics Staff at Collin College Copyright Collin College Physics Department. All Rights Reserved. University Physics, Exp 6: Static and Kinetic
More information7.1 Interacting Systems p Action/reaction pairs p Newton s Third Law p Ropes and Pulleys p.
7.1 Interacting Systems p. 183-185 7.2 Action/reaction pairs p. 185-189 7.3 Newton s Third Law p. 189-194 7.4 Ropes and Pulleys p. 194-198 7.5 Interacting-system Problems p. 198-202 1 7.1 Interacting Systems
More informationFriction is always opposite to the direction of motion.
6. Forces and Motion-II Friction: The resistance between two surfaces when attempting to slide one object across the other. Friction is due to interactions at molecular level where rough edges bond together:
More informationFriction Can Be Rough
8.1 Observe and Find a Pattern Friction Can Be Rough Perform the following experiment: Rest a brick on a rough surface. Tie a string around the brick and attach a large spring scale to it. Pull the scale
More informationFigure 5.1a, b IDENTIFY: Apply to the car. EXECUTE: gives.. EVALUATE: The force required is less than the weight of the car by the factor.
51 IDENTIFY: for each object Apply to each weight and to the pulley SET UP: Take upward The pulley has negligible mass Let be the tension in the rope and let be the tension in the chain EXECUTE: (a) The
More informationLECTURE 11 FRICTION AND DRAG
LECTURE 11 FRICTION AND DRAG 5.5 Friction Static friction Kinetic friction 5.6 Drag Terminal speed Penguins travel on ice for miles by sliding on ice, made possible by small frictional force between their
More information1. A sphere with a radius of 1.7 cm has a volume of: A) m 3 B) m 3 C) m 3 D) 0.11 m 3 E) 21 m 3
1. A sphere with a radius of 1.7 cm has a volume of: A) 2.1 10 5 m 3 B) 9.1 10 4 m 3 C) 3.6 10 3 m 3 D) 0.11 m 3 E) 21 m 3 2. A 25-N crate slides down a frictionless incline that is 25 above the horizontal.
More informationForces Part 1: Newton s Laws
Forces Part 1: Newton s Laws Last modified: 13/12/2017 Forces Introduction Inertia & Newton s First Law Mass & Momentum Change in Momentum & Force Newton s Second Law Example 1 Newton s Third Law Common
More informationGet Solution of These Packages & Learn by Video Tutorials on FRICTION
1. FRICTION : When two bodies are kept in contact, electromagnetic forces act between the charged particles (molecules) at the surfaces of the bodies. Thus, each body exerts a contact force of the other.
More information3. A bicycle tire of radius 0.33 m and a mass 1.5 kg is rotating at 98.7 rad/s. What torque is necessary to stop the tire in 2.0 s?
Practice 8A Torque 1. Find the torque produced by a 3.0 N force applied at an angle of 60.0 to a door 0.25 m from the hinge. What is the maximum torque this force could exert? 2. If the torque required
More informationEnd-of-Chapter Exercises
End-of-Chapter Exercises For all these exercises, assume that all strings are massless and all pulleys are both massless and frictionless. We will improve our model and learn how to account for the mass
More informationPHYSICS 149: Lecture 21
PHYSICS 149: Lecture 21 Chapter 8: Torque and Angular Momentum 8.2 Torque 8.4 Equilibrium Revisited 8.8 Angular Momentum Lecture 21 Purdue University, Physics 149 1 Midterm Exam 2 Wednesday, April 6, 6:30
More informationChapter 6. Force and Motion II
Chapter 6 Force and Motion II 6 Force and Motion II 2 Announcement: Sample Answer Key 3 4 6-2 Friction Force Question: If the friction were absent, what would happen? Answer: You could not stop without
More informationPractice. Newton s 3 Laws of Motion. Recall. Forces a push or pull acting on an object; a vector quantity measured in Newtons (kg m/s²)
Practice A car starts from rest and travels upwards along a straight road inclined at an angle of 5 from the horizontal. The length of the road is 450 m and the mass of the car is 800 kg. The speed of
More informationLecture 2 - Force Analysis
Lecture 2 - orce Analysis A Puzzle... Triangle or quadrilateral? 4 distinct points in a plane can either be arrange as a triangle with a point inside or as a quadrilateral. Extra Brownie Points: Use the
More informationWebreview practice test. Forces (again)
Please do not write on test. ID A Webreview 4.3 - practice test. Forces (again) Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A 5.0-kg mass is suspended
More informationAP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force).
AP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force). 1981M1. A block of mass m, acted on by a force of magnitude F directed horizontally to the
More informationOutline: Types of Friction Dry Friction Static vs. Kinetic Angles Applications of Friction. ENGR 1205 Appendix B
Outline: Types of Friction Dry Friction Static vs. Kinetic Angles Applications of Friction ENGR 1205 Appendix B 1 Contacting surfaces typically support normal and tangential forces Friction is a tangential
More informationPulling force $ % 6 Least
B3-RT2: ROPES PULLING BOXES ACCELERATION Boxes are pulled by ropes along frictionless surfaces, accelerating toward the left. of the boxes are identical. The pulling force applied to the left-most rope
More informationForce a push or a pull exerted on some object the cause of an acceleration, or the change in an objects velocity
Chapter 4 Physics Notes Changes in Motion Force a push or a pull exerted on some object the cause of an acceleration, or the change in an objects velocity Forces cause changes in velocity Causes a stationary
More informationCoefficient of Friction
HOUSTON COMMUNITY COLLEGE SYSTEMS SOUTHWEST COLLEGE COLLEGE PHYSICS I PHYS 1401 PRE LAB QUESTIONS Due before lab begins. Coefficient of Friction 1) Explain briefly the different types of frictional forces.
More information第 1 頁, 共 7 頁 Chap10 1. Test Bank, Question 3 One revolution per minute is about: 0.0524 rad/s 0.105 rad/s 0.95 rad/s 1.57 rad/s 6.28 rad/s 2. *Chapter 10, Problem 8 The angular acceleration of a wheel
More informationInclined Planes Worksheet Answers
Physics 12 Name: Inclined Planes Worksheet Answers 1. An 18.0 kg box is released on a 33.0 o incline and accelerates at 0.300 m/s 2. What is the coefficient of friction? m 18.0kg 33.0? a y 0 a 0.300m /
More informationEngineering Mechanics
Engineering Mechanics Continued (5) Mohammed Ameen, Ph.D Professor of Civil Engineering B Section Forces in Beams Beams are thin prismatic members that are loaded transversely. Shear Force, Aial Force
More information(35+70) 35 g (m 1+m 2)a=m1g a = 35 a= =3.27 g 105
Coordinator: Dr. W. L-Basheer Monday, March 16, 2015 Page: 1 Q1. 70 N block and a 35 N block are connected by a massless inextendable string which is wrapped over a frictionless pulley as shown in Figure
More informationPhysics 207 Lecture 7. Lecture 7
Lecture 7 "Professor Goddard does not know the relation between action and reaction and the need to have something better than a vacuum against which to react. He seems to lack the basic knowledge ladled
More informationBroughton High School
1 Physical Science Vocabulary Vocabulary for Chapter 5 - Work and Machines No.# Term Page # Definition 2 1. Compound Machine 2. Efficiency 3. Inclined Plane 4. Input force 5. Lever 6. Machine 7. Mechanical
More informationAngular Speed and Angular Acceleration Relations between Angular and Linear Quantities
Angular Speed and Angular Acceleration Relations between Angular and Linear Quantities 1. The tires on a new compact car have a diameter of 2.0 ft and are warranted for 60 000 miles. (a) Determine the
More informationCHAPTER 8: ROTATIONAL OF RIGID BODY PHYSICS. 1. Define Torque
7 1. Define Torque 2. State the conditions for equilibrium of rigid body (Hint: 2 conditions) 3. Define angular displacement 4. Define average angular velocity 5. Define instantaneous angular velocity
More informationIsaac Newton ( )
Isaac Newton (1642-1727) In the beginning of 1665 I found the rule for reducing any degree of binomial to a series. The same year in May I found the method of tangents and in November the method of fluxions
More informationPhys 1401: General Physics I
1. (0 Points) What course is this? a. PHYS 1401 b. PHYS 1402 c. PHYS 2425 d. PHYS 2426 2. (0 Points) Which exam is this? a. Exam 1 b. Exam 2 c. Final Exam 3. (0 Points) What version of the exam is this?
More informationLecture PowerPoints. Chapter 5 Physics for Scientists & Engineers, with Modern Physics, 4 th edition. Giancoli
Lecture PowerPoints Chapter 5 Physics for Scientists & Engineers, with Modern Physics, 4 th edition 2009 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely
More informationTotal 0/15. 0/1 points POE MC.17. [ ]
Sample Problems to KSEA (2383954) Current Score: 0/15 Question Points 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 Total 0/15 1. 0/1 points POE1 2000.MC.17.
More informationThere are two main types of friction:
Section 4.15: Friction Friction is needed to move. Without friction, a car would sit in one spot spinning its tires, and a person would not be able to step forward. However, the motion of an object along
More informationWebreview Torque and Rotation Practice Test
Please do not write on test. ID A Webreview - 8.2 Torque and Rotation Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A 0.30-m-radius automobile
More informationProblem Set 5: Universal Law of Gravitation; Circular Planetary Orbits.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01T Fall Term 2004 Problem Set 5: Universal Law of Gravitation; Circular Planetary Orbits. Available on-line October 1; Due: October
More informationYou may use g = 10 m/s 2, sin 60 = 0.87, and cos 60 = 0.50.
1. A child pulls a 15kg sled containing a 5kg dog along a straight path on a horizontal surface. He exerts a force of a 55N on the sled at an angle of 20º above the horizontal. The coefficient of friction
More informationacceleration weight load
Instructions for Vocabulary Cards: Please photocopy the following pages onto heavy card stock (back to back, so the word is printed on the back side of the matching definition). Then, laminate each page.
More information2. A 10 kg box is being pushed by a 100 N force 30 above the horizontal. The acceleration of the box is 5 m/s 2. What is the value of µ k?
Physics Whiteboard Forces with Friction 1. A 70 kg block is being pushed across a tabletop with a constant force of 350 N exerted in the direction of travel. If the coefficient of kinetic friction (µ k
More informationUnit 1: Equilibrium and Center of Mass
Unit 1: Equilibrium and Center of Mass FORCES What is a force? Forces are a result of the interaction between two objects. They push things, pull things, keep things together, pull things apart. It s really
More informationb) Fluid friction: occurs when adjacent layers in a fluid are moving at different velocities.
Ch.6 Friction Types of friction a) Dry friction: occurs when non smooth (non ideal) surfaces of two solids are in contact under a condition of sliding or a tendency to slide. (also called Coulomb friction)
More informationPhys 1401: General Physics I
1. (0 Points) What course is this? a. PHYS 1401 b. PHYS 1402 c. PHYS 2425 d. PHYS 2426 2. (0 Points) Which exam is this? a. Exam 1 b. Exam 2 c. Final Exam 3. (0 Points) What version of the exam is this?
More informationRandom sample problems
UNIVERSITY OF ALABAMA Department of Physics and Astronomy PH 125 / LeClair Spring 2009 Random sample problems 1. The position of a particle in meters can be described by x = 10t 2.5t 2, where t is in seconds.
More informationHowever, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied.
FRICTION 1 Introduction In preceding chapters, it was assumed that surfaces in contact were either frictionless (surfaces could move freely with respect to each other) or rough (tangential forces prevent
More informationDry Friction Static vs. Kinetic Angles
Outline: Types of Friction Dry Friction Static vs. Kinetic Angles Applications of Friction 1 Contacting surfaces typically support normal and tangential forces Friction is a tangential force Friction occurs
More informationPhys101 Second Major-131 Zero Version Coordinator: Dr. A. A. Naqvi Sunday, November 03, 2013 Page: 1
Coordinator: Dr. A. A. Naqvi Sunday, November 03, 2013 Page: 1 Q1. Two forces are acting on a 2.00 kg box. In the overhead view of Figure 1 only one force F 1 and the acceleration of the box are shown.
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Common Quiz Mistakes / Practice for Final Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A ball is thrown directly upward and experiences
More informationQuestions from all units
Questions from all units S.NO 1. 1 UNT NO QUESTON Explain the concept of force and its characteristics. BLOOMS LEVEL LEVEL 2. 2 Explain different types of force systems with examples. Determine the magnitude
More informationIn the last lecture the concept of kinetic energy was introduced. Kinetic energy (KE) is the energy that an object has by virtue of its motion
1 PHYS:100 LETUE 9 MEHANIS (8) I. onservation of Energy In the last lecture the concept of kinetic energy was introduced. Kinetic energy (KE) is the energy that an object has by virtue of its motion KINETI
More informationFORCE. Definition: Combining Forces (Resultant Force)
1 FORCE Definition: A force is either push or pull. A Force is a vector quantity that means it has magnitude and direction. Force is measured in a unit called Newtons (N). Some examples of forces are:
More informationLecture 11 - Advanced Rotational Dynamics
Lecture 11 - Advanced Rotational Dynamics A Puzzle... A moldable blob of matter of mass M and uniform density is to be situated between the planes z = 0 and z = 1 so that the moment of inertia around the
More information2. FORCE AND MOTION. In the above, the objects are being moved by a push or pull. A push or pull acting on objects is called a force.
2. FORCE AND MOTION Force We do many jobs in our daily life like lifting things, moving things from one place to another, cutting objects, etc. To do these jobs, we have to move. We are surrounded by a
More informationPHY218 SPRING 2016 Review for Final Exam: Week 14 Final Review: Chapters 1-11, 13-14
Final Review: Chapters 1-11, 13-14 These are selected problems that you are to solve independently or in a team of 2-3 in order to better prepare for your Final Exam 1 Problem 1: Chasing a motorist This
More informationEnd-of-Chapter Exercises
End-of-Chapter Exercises Exercises 1 12 are conceptual questions that are designed to see if you have understood the main concepts of the chapter. 1. Figure 11.21 shows four different cases involving a
More informationReview PHYS114 Chapters 4-7
Review PHYS114 Chapters 4-7 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A 27 kg object is accelerated at a rate of 1.7 m/s 2. What force does
More informationPHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011
PHYSICS 1, FALL 011 EXAM SOLUTIONS WEDNESDAY, NOVEMBER, 011 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively. In this
More informationApplying Newton s Laws
Chapter 5 Applying Newton s Laws PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman Lectures by James Pazun Goals for Chapter 5 To use and apply Newton s Laws
More informationSTEP Support Programme. Mechanics STEP Questions
STEP Support Programme Mechanics STEP Questions This is a selection of mainly STEP I questions with a couple of STEP II questions at the end. STEP I and STEP II papers follow the same specification, the
More informationIn this lecture we will discuss three topics: conservation of energy, friction, and uniform circular motion.
1 PHYS:100 LECTURE 9 MECHANICS (8) In this lecture we will discuss three topics: conservation of energy, friction, and uniform circular motion. 9 1. Conservation of Energy. Energy is one of the most fundamental
More informationPhysics 2A Chapter 4: Forces and Newton s Laws of Motion
Physics 2A Chapter 4: Forces and Newton s Laws of Motion There is nothing either good or bad, but thinking makes it so. William Shakespeare It s not what happens to you that determines how far you will
More informationLecture 6. Applying Newton s Laws Free body diagrams Friction
Lecture 6 Applying Newton s Laws Free body diagrams Friction ACT: Bowling on the Moon An astronaut on Earth kicks a bowling ball horizontally and hurts his foot. A year later, the same astronaut kicks
More informationPhysics 207: Lecture 24. Announcements. No labs next week, May 2 5 Exam 3 review session: Wed, May 4 from 8:00 9:30 pm; here.
Physics 07: Lecture 4 Announcements No labs next week, May 5 Exam 3 review session: Wed, May 4 from 8:00 9:30 pm; here Today s Agenda ecap: otational dynamics and torque Work and energy with example Many
More information