Chapter 5 - Applying Newton s Laws w./ QuickCheck Questions

Chapter 5 - Applying Newton s Laws w./ QuickCheck Questions 2015 Pearson Education, Inc. Anastasia Ierides Department of Physics and Astronomy University of New Mexico September 17, 2015

Review of Last Time Force - push/pull; the cause of acceleration; vector; units N or lb; net force Newton s Laws of Motion (3); interactions; equilibrium; dynamics Free body diagrams - identifying forces Different types of forces: weight, tension, normal, friction (static/kinetic), drag, thrust Mass versus weight; apparent weight/weightlessness

QuickCheck Question 4.4 A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows to a halt. What forces act on the sled just after she s jumped in? A. Gravity and kinetic friction B. Gravity and a normal force C. Gravity and the force of the push D. Gravity, a normal force, and kinetic friction E. Gravity, a normal force, kinetic friction, and the force of the push

QuickCheck Question 4.5 A cart is pulled to the right with a constant, steady force. How will its acceleration graph look?

QuickCheck Question 4.5 A cart is pulled to the right with a constant, steady force. How will its acceleration graph look? Remember: A constant force means a constant acceleration!

Free-Body Diagrams

Look at the Skier n T fk w

QuickCheck Question 4.11 An elevator, lifted by a cable, is moving upward and slowing. Which is the correct free-body diagram? A. B. C. D. E.

QuickCheck Question 4.11 An elevator, lifted by a cable, is moving upward and slowing. Which is the correct free-body diagram? T T T T w w w w A. B. C. D. E.

QuickCheck Question 4.12 A ball has been tossed straight up. Which is the correct freebody diagram just after the ball has left the hand? Ignore air resistance. A. B. C. D.

QuickCheck Question 4.12 A ball has been tossed straight up. Which is the correct freebody diagram just after the ball has left the hand? Ignore air resistance. HINT: This problem is an example of free fall! The only force is that due to gravity! w w w w A. B. C. D.

QuickCheck Question 4.13 A ball, hanging from the ceiling by a string, is pulled back and released. Which is the correct free-body diagram just after its release? A. B. C. D. E.

QuickCheck Question 4.13 A ball, hanging from the ceiling by a string, is pulled back and released. Which is the correct free-body diagram just after its release? Let s look at the options again - which one has tension and weight? T T T T w w w A. B. C. D. E.

QuickCheck Question 4.14 A car is parked on a hill. Which is the correct free-body diagram?

QuickCheck Question 4.14 A car is parked on a hill. Which is the correct free-body diagram? Let s look at those options again which one gives you all three? n n n n f f f w w w C.

QuickCheck Question 4.14 A car is towed to the right at constant speed. Which is the correct free-body diagram?

QuickCheck Question 4.14 A car is towed to the right at constant speed. Which is the correct free-body diagram? D.

QuickCheck Question 4.16 10-year-old Sarah stands on a skateboard. Her older brother Jack starts pushing her backward and she starts speeding up. The force of Jack on Sarah is A. Greater than the force of Sarah on Jack. B. Equal to the force of Sarah on Jack. C. Less than the force of Sarah on Jack.

QuickCheck Question 4.16 10-year-old Sarah stands on a skateboard. Her older brother Jack starts pushing her backward and she starts speeding up. The force of Jack on Sarah is A. Greater than the force of Sarah on Jack. B. Equal to the force of Sarah on Jack. C. Less than the force of Sarah on Jack. Remember: Action/reaction pair forces come in pairs, are equal in magnitude and opposite in direction (Newton s third law)

QuickCheck Question 4.17 A mosquito runs head-on into a truck. Splat! Which is true during the collision? A. The mosquito exerts more force on the truck than the truck exerts on the mosquito. B. The truck exerts more force on the mosquito than the mosquito exerts on the truck. C. The mosquito exerts the same force on the truck as the truck exerts on the mosquito. D. The truck exerts a force on the mosquito but the mosquito does not exert a force on the truck. E. The mosquito exerts a force on the truck but the truck does not exert a force on the mosquito.

Equilibrium in Motion

QuickCheck Question 5.1 A ring, seen from above, is pulled on by three forces. The ring is not moving. How big is the force F? A. 20 N B. 10 cosθ N C. 10 sinθ N D. 20 cosθ N E. 20 sinθ N

QuickCheck Question 5.1 A ring, seen from above, is pulled on by three forces. The ring is not moving. How big is the force F? A. 20 N B. 10 cosθ N C. 10 sinθ N x-components cancel out leaving only y-components Fy θ θ Fy D. 20 cosθ N E. 20 sinθ N Fy = 2Fy - F = 0 F = 2Fy = 2 (10 N) sin θ = 20 sin θ N

Dynamics - non-equilibrium Forces acting on an object determine its acceleration This acceleration can be used in the equations of motion to describe the motion of the object Nonzero acceleration results in dynamic behavior

Mass versus Weight Weight is a force that always points straight down, w = mg, where m is the mass It is easy to convert weight to mass and viceversa

Apparent Weight Weight is the force you feel due to the Earth s gravitational pull Sensation of weight is due to contact forces supporting you (action/reaction pair) Your apparent weight, w app, in terms of the force you feel is

Apparent Weight in an Elevator Forces acting on the man are the upward normal force due to the elevator floor, and his weight n = w + ma = wapp Since w app > w, the man feels heavier

Weightlessness An object in free fall, with no contact forces, has zero apparent weight - weightlessness Weightlessness no weight

QuickCheck Question 5.4 What are the components of w in the coordinate system shown?

QuickCheck Question 5.4 What are the components of w in the coordinate system shown? y x w wy θ wx

QuickCheck Question 5.5 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the scale reads A. > 490 N B. 490 N C. < 490 N but not 0 N D. 0 N

QuickCheck Question 5.5 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the scale reads A. > 490 N B. 490 N C. < 490 N but not 0 N D. 0 N Remember the apparent weight of the man in the elevator accelerating upward!

QuickCheck Question 5.6 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. Sadly, the elevator cable breaks. What is the reading on the scale during the few seconds it takes the student to plunge to his doom? A. > 490 N B. 490 N C. < 490 N but not 0 N D. 0 N

QuickCheck Question 5.6 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. Sadly, the elevator cable breaks. What is the reading on the scale during the few seconds it takes the student to plunge to his doom? A. > 490 N The student feels weightless! B. 490 N C. < 490 N but not 0 N D. 0 N

Surface Forces An object on a desk or a person sitting on a chair is subject to an upward (perpendicular or normal to the surface) force due to the desk or chair

Surface Forces An object on a desk or a person sitting on a chair is subject to an upward (perpendicular or normal to the surface) force due to the desk or chair This type of force is known as the normal force, denoted by n, and adjusts itself so the object stays on the surface

Remember: Towing Car Example A car with a mass of 1500 kg is being towed at a steady speed by a rope held at a 20 angle from the horizontal. A friction force of 320 N opposes the car s motion. What is the tension in the rope?

Example 5.4: Tension in towing a car Here we found that the normal force adjusted itself due to the y-component of the tension in the rope n = w - T sinθ

Example 5.9: Normal force on a pressed book A 1.2 kg book lies on a table. The book is pressed down from above with a force of 15 N. What is the normal force acting on the book from the table below?

Example 5.9: Normal force on a pressed book A 1.2 kg book lies on a table. The book is pressed down from above with a force of 15 N. What is the normal force acting on the book from the table below? PREPARE The book is not moving and is thus in static equilibrium. We need to identify the forces acting on the book and prepare a free-body diagram showing these forces. These steps are illustrated in the figure.

Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y

Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F =

Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F = 0

Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F = 0 n = w + F

Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F = 0 n = w + F We learned during last lecture that the weight force is w = mg, and we know what the force, F, is.

Example 5.9: Normal force on a pressed book The weight of the book is thus w = mg = (1.2 kg)(9.8 m/s 2 ) = 12 N

Example 5.9: Normal force on a pressed book The weight of the book is thus w = mg = (1.2 kg)(9.8 m/s 2 ) = 12 N With this information, we see that the normal force exerted by the table is n = w + F

Example 5.9: Normal force on a pressed book The weight of the book is thus w = mg = (1.2 kg)(9.8 m/s 2 ) = 12 N With this information, we see that the normal force exerted by the table is n = w + F = 12 N + 15 N = 27 N

Example 5.9: Normal force on a pressed book ASSESS The magnitude of the normal force is larger than the weight of the book. From the table s perspective, the extra force from the hand pushes the book further into the atomic springs of the table. These springs then push back harder, giving a normal force that is greater than the weight of the book.

Normal Forces

QuickCheck Question 5.2 The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is n A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to tell

QuickCheck Question 5.2 The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is n A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to tell Remember: apparent weight n = w + ma = wapp

QuickCheck Question 5.2 The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is upward acceleration requires a net upward force n A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to tell Remember: apparent weight n = w + ma = wapp

QuickCheck Question 5.3 A box is being pulled to the right at steady speed by a rope that angles upward. In this situation: A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to judge the size of the normal force

QuickCheck Question 5.3 A box is being pulled to the right at steady speed by a rope that angles upward. In this situation: Remember: the towed car example A. n > mg n = w - T sin θ B. n = mg C. n < mg D. n = 0 E. Not enough information to judge the size of the normal force

Example 5.10: Acceleration of a downhill skier A skier slides down a steep 27 slope. On a slope this steep, friction is much smaller than the other forces at work and can be ignored. What is the skier s acceleration?

Example 5.10: Acceleration of a downhill skier A skier slides down a steep 27 slope. On a slope this steep, friction is much smaller than the other forces at work and can be ignored. What is the skier s acceleration? PREPARE We choose a coordinate system tilted so that the x-axis points down the slope. This greatly simplifies the analysis because with this choice a y = 0 (the skier doesn t move in the y-direction).

Example 5.10: Acceleration of a downhill skier SOLVE We can now use Newton s second law in component form to find the skier s acceleration: ΣF x = w x + n x = ma x ΣF y = w y + n y = ma y

Example 5.10: Acceleration of a downhill skier SOLVE Because the normal force, n, points directly in the positive y- direction, n y = n and n x = 0.

Example 5.10: Acceleration of a downhill skier SOLVE Because the normal force, n, points directly in the positive y- direction, n y = n and n x = 0. The angle between the weight, w, and the negative y-axis is the same as the slope angle θ. With this information, the components of w are w x = m g sin θ & w y = m g cos θ

Example 5.10: Acceleration of a downhill skier SOLVE With these components in hand, Newton s second law becomes ΣF x = mg sin θ = ma x ΣF y = mg cos θ + n = ma y = 0

Example 5.10: Acceleration of a downhill skier SOLVE With these components in hand, Newton s second law becomes ΣF x = mg sin θ = ma x ΣF y = mg cos θ + n = ma y = 0 The m cancels out in the first equation, leaving us with a x = g sinθ

Example 5.10: Acceleration of a downhill skier SOLVE With these components in hand, Newton s second law becomes ΣF x = mg sin θ = ma x ΣF y = mg cos θ + n = ma y = 0 The m cancels out in the first equation, leaving us with a x = g sinθ This is the expression for acceleration on a frictionless surface that we presented, without proof, in Chapter 3.

Example 5.10: Acceleration of a downhill skier SOLVE With these components in hand, Newton s second law becomes ΣF x = mg sin θ = ma x ΣF y = mg cos θ + n = ma y = 0 The m cancels out in the first equation, leaving us with a x = g sinθ This is the expression for acceleration on a frictionless surface that we presented, without proof, in Chapter 3. We can use this to calculate the skier s acceleration: a x = g sinθ = (9.8 m/s 2 ) sin 27 = 4.4 m/s 2

Example 5.10: Acceleration of a downhill skier ASSESS Our result shows that when θ = 0, so that the slope is horizontal, the skier s acceleration is zero, as it should be. Further, when θ = 90 (a vertical slope), his acceleration is g, which makes sense because he s in free fall when θ = 90. Notice that the mass canceled out, so we didn t need to know the skier s mass. We first saw the formula for the acceleration in Section 3.4, but now we see the physical reasons behind it.

What about Friction? Surfaces are very rough on a microscopic scale

What about Friction? Surfaces are very rough on a microscopic scale High points of surfaces become jammed against each other

What about Friction? Surfaces are very rough on a microscopic scale High points of surfaces become jammed against each other Amount of contact depends force between objects

What Types of Friction? Static - the force a surface exerts on an object to keep it from moving (stuck)

What Types of Friction? Static - the force a surface exerts on an object to keep it from moving (stuck) Kinetic - the force a surface exerts on an object in the opposite direction of the object s motion

What Types of Friction? Static - the force a surface exerts on an object to keep it from moving (stuck) Kinetic - the force a surface exerts on an object in the opposite direction of the object s motion Rolling - the force a surface exerts on an object rolling across it

Static Friction Denoted f s, points in opposite direction of pushing force

Static Friction Denoted f s, points in opposite direction of pushing force Object is in static equilibrium, Fpush = fs

Static Friction Denoted f s, points in opposite direction of pushing force Object is in static equilibrium, Fpush = fs Maximum magnitude, f s,max = µ s n

Static Friction, fs Maximum magnitude, f s,max = µ s n where, µ s, is the coefficient of static friction

Static Friction, fs Maximum magnitude, f s,max = µ s n where, µ s, is the coefficient of static friction Adjusts to oppose motion

Static Friction, fs Maximum magnitude, f s,max = µ s n where, µ s, is the coefficient of static friction Adjusts to oppose motion Cannot exceed f s,max

QuickCheck Question 5.7 A box on a rough surface is pulled by a horizontal rope with tension T. The box is not moving. In this situation: A. f s > T B. f s = T C. f s < T D. f s = µsmg E. f s = 0

QuickCheck Question 5.7 A box on a rough surface is pulled by a horizontal rope with tension T. The box is not moving. In this situation: Newton s second law ΣF = ma = 0 A. f s > T B. f s = T C. f s < T D. f s = µsmg E. f s = 0

Kinetic Friction, fk Nearly constant (doesn t adjust) magnitude f k = µ k n where µ k is called the coefficient of kinetic friction.

Kinetic Friction, fk Allows motion in the opposite direction, either at constant speed, or even acceleration ΣF = ma

Rolling Friction, fr A wheel rolling on a surface experiences friction on the portion that touches the surface; stationary, not sliding

Rolling Friction, fr A wheel rolling on a surface experiences friction on the portion that touches the surface; stationary, not sliding Interaction between a rolling wheel and the road can be quite complicated

Rolling Friction, fr A wheel rolling on a surface experiences friction on the portion that touches the surface; stationary, not sliding Interaction between a rolling wheel and the road can be quite complicated Friction force that opposes the motion, one defined by a coefficient of rolling friction µ r : f r = µ r n

Coefficients of Friction

Working With Friction

QuickCheck Question 5.8 A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? A. Yes B. No C. Not enough information to say

QuickCheck Question 5.8 A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? A. Yes n = w = 100 N B. No C. Not enough information to say

QuickCheck Question 5.8 A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? A. Yes B. No C. Not enough information to say n = w = 100 N f s,max = µ s n = 40 N

QuickCheck Question 5.8 A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? A. Yes B. No C. Not enough information to say n = w = 100 N f s,max = µ s n = 40 N > 30 N

QuickCheck Question 5.9 A box is being pulled to the right over a rough surface. T > f k, so the box is speeding up. Suddenly the rope breaks. What happens? The box A. Stops immediately. B. Continues with the speed it had when the rope broke. C. Continues speeding up for a short while, then slows and stops. D. Keeps its speed for a short while, then slows and stops. E. Slows steadily until it stops.

What a Drag! Like friction, it is in the opposite direction of the speed, v, of the object

What a Drag! Like friction, it is in the opposite direction of the speed, v, of the object Increases as the speed increases; usually ignored when the speed, v, is low

What a Drag! Like friction, it is in the opposite direction of the speed, v, of the object Increases as the speed increases; usually ignored when the speed, v, is low Simple model

Terminal Speed As an object falls, after some time it feels the Drag force

Terminal Speed As an object falls, after some time it feels the Drag force At some point during the fall, the Drag force is equal to the weight

Terminal Speed As an object falls, after some time it feels the Drag force At some point during the fall, the Drag force is equal to the weight Speed of object at this point is the terminal speed

Catalog of Forces

Newton s 3 rd and Interactions How do describe the motion of an object in contact with another?

Newton s 3 rd and Interactions How do describe the motion of an object in contact with another? We know about free body diagrams

Newton s 3 rd and Interactions How do describe the motion of an object in contact with another? We know about free body diagrams And we know about action/reaction pairs of forces

Newton s 3 rd and Interactions

QuickCheck Question 5.11 Consider the situation in the figure. Which pair of forces is an action/reaction pair? A. The tension of the string and the friction force acting on A B. The normal force on A due to B and the weight of A C. The normal force on A due to B and the weight of B D. The friction force acting on A and the friction force acting on B

Example 5.15: Pushing 2 Blocks The figure below shows a 5.0 kg block A being pushed with a 3.0 N force. In front of this block is a 10 kg block B; the two blocks move together. What force does block A exert on block B?

Example 5.15: Pushing 2 Blocks PREPARE The visual overview of the figure lists the known information and identifies F A on B as what we re trying to find. Following the steps of Tactics Box 5.2, we draw separate force identification diagrams and separate free-body diagrams for the two blocks.

Example 5.15: Pushing 2 Blocks The force is the contact force that block A exerts on B; it forms an action/reaction pair with the force that block B exerts on A. Notice that force is drawn acting on block B; it is the force of A on B. Force vectors are always drawn on the free-body diagram of the object that experiences the force, not the object exerting the force. Because action/reaction pairs act in opposite directions, force pushes backward on block A and appears on A s free-body diagram.

Example 5.15: Pushing 2 Blocks SOLVE We begin by writing Newton s second law in component form for each block. Because the motion is only in the x-direction, we need only the x-component of the second law. For block A, ΣF x = (F H ) x + (F B on A ) x = m A a Ax The force components can be read from the free-body diagram, where we see pointing to the right and pointing to the left. Thus Σ H F B on A = m A a Ax

Example 5.15: Pushing 2 Blocks For B, we have ΣF x = (F A on B ) x = F A on B = m B a Bx We have two additional pieces of information: First, Newton s third law tells us that F B on A = F A on B. Second, the boxes are in contact and must have the same acceleration a x ; that is, a Ax = a Bx = a x. With this information, the two x-component equations become F H F A on B = m A a x F A on B = m B a x

Example 5.15: Pushing 2 Blocks F H F A on B = m A a x F A on B = m B a x Our goal is to find F A on B, so we need to eliminate the unknown acceleration a x. From the second equation, a x = F A on B /m B. Substituting this into the first equation gives This can be solved for the force of block A on block B, giving

Example 5.15: Pushing 2 Blocks ASSESS Force F H accelerates both blocks, a total mass of 15 kg, but force F A on B accelerates only block B, with a mass of 10 kg. Thus it makes sense that F A on B < F H.

Interactions With Ropes Consider a box is pulled by a rope with tension, T

Interactions With Ropes Consider a box is pulled by a rope with tension, T Make the massless string approximation that m rope = 0

Interactions With Ropes Consider a box is pulled by a rope with tension, T Make the massless string approximation that m rope = 0 Newton s second law for the rope is thus ΣF x = F box on rope = F T = m rope a x = 0

Interactions With Ropes The tension, T, is the same from one end of the rope to the other

Interactions With Ropes The tension, T, is the same from one end of the rope to the other The pulling force does not change across the rope either

Interactions With Ropes The tension, T, is the same from one end of the rope to the other The pulling force does not change across the rope either The magnitude of the force pulling the string or rope is equal to the tension in the string

QuickCheck Question 5.12 Boxes A and B are being pulled to the right on a frictionless surface; the boxes are speeding up. Box A has a larger mass than Box B. How do the two tension forces compare? A. T 1 > T 2 B. T 1 = T 2 C. T 1 < T 2 D. Not enough information to tell

QuickCheck Question 5.13 Boxes A and B are sliding to the right on a frictionless surface. Hand H is slowing them. Box A has a larger mass than Box B. Considering only the horizontal forces: A. F B on H = F H on B = F A on B = F B on A B. F B on H = F H on B > F A on B = F B on A C. F B on H = F H on B < F A on B = F B on A D. F H on B = F H on A > F A on B

Summary

Things that are due Homework #4 Due September 23, 2015 by 11:59 pm Reading Quiz #5 Due September 17, 2015 by 4:59 pm Reading Quiz #6 Due September 22, 2015 by 4:59 pm

QUESTIONS?