Bell Ringer: What is Newton s 3 rd Law? Which force acts downward? Which force acts upward when two bodies are in contact?

Transcription

1 Bell Ringer: What is Newton s 3 rd Law? Which force acts downward? Which force acts upward when two bodies are in contact? Does the moon attract the Earth with the same force that the Earth attracts the moon? What is friction?

2 4.3 Newtonian Mechanics: Tension, Friction, Restoring Forces Physics Honors I

3 Objectives: Explain the tension in ropes and strings in terms of Newton s 3 rd Laws. Define the friction force. Distinguish between static and kinetic friction. Describe the force of an Elastic Spring.

4 Vocabulary: Kinetic Static Tension Restoring Force

5 Active Physics Reference: Chapter 2 - Section 7: p. 212 p. 216 (Friction) Chapter 4 - Section 5: p. 396 p. 400 (Hooke s Law)

6 Further Learning: Physics: Principles and Problems - the red book we have in class Chapter Tension Page 105 Page 106 Chapter Friction Page 126 Page 130 (3) Physics Classroom: Types of Forces (4) Khan Academy Tension Force Frictional Forces Restoring Forces

7 What are Forces?

8 Types of Forces: Tension Force (T) Tension is simply a specific name for the force exerted by a string or rope. The tension in the cord is the magnitude T of the force on the body. NOTE: The cord is normally assumed to be massless and unstretchable. Pulleys are usually assumed massless and frictionless.

9 Types of Forces: Tension Force (T)

10 Types of Forces: Tension Force (T) A bucket hangs from a rope attached to the ceiling. The rope is about to break in the middle. If the rope breaks, the bucket will fall; thus before it breaks, there must be forces holding the rope together. The force that the top part of the rope exerts on the bottom part is F top on bottom. Newton s third law states that this force must be part of an interaction pair. The other member of the pair is the force that the bottom part exerts on the top, F bottom on top. These forces equal in magnitude but opposite in direction.

11 Example 1: Tension Force (T) A 50.0 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. If the acceleration is constant, is the rope in danger of breaking? Step 1: Draw a free-body diagram.

12 Example 1: Tension Force A 50.0 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. If the acceleration is constant, is the rope in danger of breaking? Step 2: Write your knowns vs unknowns Known: Unknown: m = 50.0 kg F T =? v i = 0 m/s v f = 3.0 m/s h = 3.0 m

13 Example 1: Tension Force A 50.0 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. If the acceleration is constant, is the rope in danger of breaking? Step 3: Solve for unknown force using Newton s 2 nd Law. F net = ma F T F g = ma F T = ma + F g F T = ma + mg F T = m(a +g)

14 Example 1: Tension Force A 50.0 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. If the acceleration is constant, is the rope in danger of breaking? Step 4: Before we can solve for the tension force, we must find the acceleration. v i = 0 m s Now, we don t have to account for gravitational acceleration because it is accounted for in F g. We choose our equation based on an equation not having time but having acceleration in it. v f = 3.0 m s a =? h = 3.0 m t = x a = v f 2 2h v f 2 = v i 2 + 2ah v f 2 = 0 + 2ah v f 2 = 2ah (3.0 m/s)2 = (2)(3.0 m) = 1.5 m s 2

15 Example 1: Tension Force A 50.0 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. If the acceleration is constant, is the rope in danger of breaking? Step 5: Solve for the tension force due to the acceleration. F T = m(a +g) F T = (50 kg)(1.5 F T = 565 N mτ s mτ s 2) So yes, the rope will exceed 525 N and break when accelerated by 1.5 mτ s 2

16 Comprehension Check 1: Tension Force (T) 1. Which of Newton s laws govern the tensional force? 2. How do we treat ropes, strings, and pulleys? 3. Draw the free body diagram for a weight hanging from the ceiling by a string.

17 Types of Forces: Frictional Force (f) Frictional Force is a resistance to motion. This force is directed along the surface, opposite the direction of the intended motion.

18 Types of Forces: Frictional Force (f) PHET - Friction and Temperature What happens as you bring the two books together? What happens when you start sliding the two books across one another? What happens when you slide the two books across one another quickly?

19 Types of Forces: Frictional Force (f) Frictional Forces can change in magnitude so that the two forces still balance. e.g. Push on a crate, it does not move. Push a little harder on the crate, it does not move. The frictional force opposing this movement will adjust in magnitude to counter balance the force you are exerting on it. However, there is a maximum magnitude of the frictional force as can be witnessed when you push hard enough on the crate to begin moving it.

20 Types of Forces: Frictional Force (f) There is a maximum magnitude of the frictional force that has to be exceeded before an object will move. After the object starts moving, there is a frictional force that continues to oppose the objects motion. Therefore, there are two types of frictional force: 1. Static frictional force, f s the force causing an object to not move. 2. Kinetic frictional force, f k opposes the movement of an object already in motion.

21 Types of Forces: Frictional Force (f) PHET - Friction Lab - Forces and Motion Watch how the static friction force balances the applied force. Notice what happens to the force of friction as the applied force exceeds it. Observe how the sum of forces (F net ) behaves before and after the object that is being pushed is set into motion.

22 Types of Forces: Frictional Force (f) Properties of Friction: Property 1: If a body does not move, then the static frictional force f S and the component of the force ԦF that is parallel to the surface balance each other. They are equal in magnitude, and the static frictional force f S is directed opposite that component of force ԦF.

23 Types of Forces: Frictional Force (f) Property 2: The magnitude of the static frictional force f S has a maximum value Ԧf s,max that is given by: Ԧf s,max = μ s F N μ S = the coefficient of static friction. F N = the magnitude of the normal force on the body from the surface. If the magnitude of the component of force ԦF that is parallel to the surface exceeds f S,max then the body begins to slide along the surface.

24 Types of Forces: Frictional Force (f) Property 3: If the body begins to slide along the surface, the magnitude of the frictional force rapidly decreases to a value f k given by: μ k = the coefficient of kinetic friction. Ԧf k = μ k F N F N = the magnitude of the normal force on the body from the surface. Thereafter, during the slide, a kinetic frictional force f k opposes the motion.

25 Types of Forces: Frictional Force (f) Note: The magnitude of the normal force F N appears in the frictional equations as a measure of how firmly the body presses against the surface. Remember that the normal force ԦF N is always perpendicular to the surface. Note: The coefficients of the static frictional force μ S and the kinetic frictional force μ k are dimensionless and must be determined experimentally.

26 Types of Forces: Frictional Force (f) EMPHASIS ON THE COEFFICIENT OF SLIDING FRICTION: µ (the Greek letter mu) is the coefficient of friction; static or kinetic. The coefficient of friction (µ) is a dimensionless quantity. The coefficient of friction (µ) values depend on the properties of the two surface in contact and is used to calculate the force of friction

27 Types of Forces: Frictional Force (f) EMPHASIS ON THE COEFFICIENT OF SLIDING FRICTION: µ is valid only for the pair of surfaces in contact when the value is measured; any significant change in either of the surfaces (such as the kind of material, surface texture, moisture, or lubrication on a surface, etc.) may cause the value of µ to change. µ usually is expressed in decimal form, such as 0.85 for rubber on dry concrete (0.60 on wet concrete).

28 Types of Forces: Frictional Force (f) PROBLEM-SOLVING TECHNIQUES: Most of the time when static friction is involved then the equation becomes: θ = tan 1 μ S This is related to finding the maximum of a function but is a good formula to remember if short of time or what you are coming up with doesn t seem right. PROBLEM SOLVING TECHNIQUES: If a problem says that an object (say a block) is on the verge of moving than that means that the static frictional force must be at its maximum possible value which is f S = f S,max.

29 Example 2: Frictional Force (f) You push a 25.0 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. If the coefficient of friction between the wooden box and the wooden floor is 0.20, how much force do you exert on the box? Step 1: Analyze and sketch the Problem Identify the forces and establish a coordinate system. Draw a free-body diagram including the direction of motion and the acceleration if there is any.

30 Example 2: Frictional Force (f) You push a 25.0 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. If the coefficient of friction between the wooden box and the wooden floor is 0.20, how much force do you exert on the box? Step 2: Identify your knowns and unknowns: Knowns: Unknowns: m = 25.0 kg F P =? v = 1.0 m/s a = 0.0 μ k = 0.20 mτs 2

31 Example 2: Frictional Force (f) You push a 25.0 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. If the coefficient of friction between the wooden box and the wooden floor is 0.20, how much force do you exert on the box? Step 3: Start the equation by analyzing the axis that the direction of motion is on. In this case, the x-axis. F net,x = ma x (Start with the x-axis since that is the direction of motion) F p f k = ma x (Substitute in for net force) F p f k = m(0) F p f k = 0 (Realize that since it is at a constant speed than there is not zero which means you can set the pushing force equal to the kinetic frictional force) F p = f k

32 Example 2: Frictional Force (f) You push a 25.0 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. If the coefficient of friction between the wooden box and the wooden floor is 0.20, how much force do you exert on the box? Step 4: Substitute the kinetic frictional force equation into the formula. F p = f k (Since f k = μ k F N we can substitute it into the above equation.) Therefore, we end up with the equation: F p = μ k F N However, we have a problem. We don t know what the normal force is. So how will we find it?

33 Example 2: Frictional Force (f) You push a 25.0 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. If the coefficient of friction between the wooden box and the wooden floor is 0.20, how much force do you exert on the box? Step 5: Switch to the y-axis to analyze the forces acting on that axis and determine the normal force. F net,y = ma y F N F g = ma y F N F g = 0 (Acceleration is zero because there is no movement on that axis. No movement on that axis means no acceleration. There is however, a gravitational acceleration still but that is accounted for in the F g = mg equation.) F N = F g F N = mg

34 Example 2: Frictional Force (f) You push a 25.0 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. If the coefficient of friction between the wooden box and the wooden floor is 0.20, how much force do you exert on the box? Step 6: On the last slide, we found that the normal force is equal to the gravitational force in this instance or mg. F p = μ k F N (Substitute in F N = mg) F p = μ k mg (Finally we are ready to solve) F p = (0.20)(25.0 kg)(9.8 mτ s 2) F p = 49 N (Note: The only reason we care about the velocity was because it informed us there was no acceleration.)

35 Comprehension Check 2: Friction 1. What is the opposing force of an object in motion called? 2. What is the opposing force of an object not yet in motion? 3. What does the symbol μ signify? 4. What is the SI Units of μ?

36 Types of Forces: Restoring Force / Spring Force There are many objects in nature which take the form of a spring. Stretching a rubber band or a spring requires a force. Another example is an atom within the lattice of a molecule. Technically the forces that hold the atoms together in a molecule can be modelled as springs connecting each one another. Many springs have the property that the stretch of a spring is directly proportional to the force applied to it. - This means that if you double the force, the stretch of the spring doubles. If you triple the force, the stretch of the spring triples. - Once you let go of the object after stretching it or compressing it, the spring returns to its relaxed state which is why it is called a restoring force.

37 Types of Forces: Restoring Force / Spring Force PHET - Hanging Masses Lab Hang a mass on the spring. Notice how it wants to bounce back to its original spot. That is the restoring force pulling the mass back up. Ultimately gravity wins out and the mass stretches the spring. What happens when you take the mass off of the spring?

38 Types of Forces: Hooke s Law Hooke s Law explains the restoring force on a spring after it has been stretched or compressed. Basically, the more you stretch a spring, the larger the restoring force of the spring.

39 Types of Forces: Hooke s Law A spring force is given to us by Hooke s Law which states that the force needed to extend or compress a spring by some distance is proportional to that distance. F s = kx [Hooke s Law] - F s is the force exerted by the spring. - x is the stretch (or compression) of the spring. - k is the spring constant.

40 Types of Forces: Hooke s Law F s = kx [Hooke s Law] - F s is the force exerted by the spring. - x is the stretch (or compression) of the spring. - k is the spring constant. Note: The negative sign in the equation indicates that the pull by the spring is opposite to the direction it is stretched or compressed. Note: The spring constant k is a measure of the stiffness of the spring.

41 Types of Forces: Hooke s Law

42 Example 3: Hooke s Law A 3.0 N weight is suspended from a spring. The spring stretches 2.0 cm. Calculate the spring constant. Answer: k = 150 N/m

43 Exit Ticket 1. Tension is a result of which of Newton s Laws? 2. Name the two types of frictional forces. 3. Describe why springs are associated with restoring forces?