Chapter 4. Force and Motion Chapter Goal: To establish a connection between force and motion.

Student Learning Objectives To recognize what does and does not constitute a force. To identify the specific forces acting on an object. To draw an accurate free-body diagram of an object. To begin the process of understanding the connection between force and motion.

A force is an interaction between two objects A force is a push or a pull on an object. If I push a book across a table, the book pushes me back (inanimate objects can exert force!) A force is a vector. It has both a magnitude and a direction. The force (interaction) has the same magnitude for both me and the book. However the direction of the force on me is opposite to the direction of the force on the book.

A force is an interaction between two objects A force on an object requires an agent. The agent is another object This is another way of confirming that a force is an interaction between objects. It takes two to tango! A force is either a contact force or a long- range force. Gravity is the only long-range force we will study this semester. It is an interaction between two objects, but they are not necessarily in contact. All other forces (this semester) only exist when the two objects are in contact.

Drawing force vectors

Combining Forces

3 N 4 N What is the net force on the object? Your answer should include both magnitude and direction (in the form of an angle from the x-axis)

Individual Forces Net Force 3 N 4 N 5 N What is the net force on the object?

A Short Catalog of Forces - Gravity Gravity is a long-range attractive force between two objects. In this class, our emphasis is on the interaction between the Earth and objects on or near its surface (the weight force). Your text uses W for the weight force.

A Short Catalog of Forces: Tension Ropes, strings, cables and hooks exert a tensional force. These agents always pull, they never push. Although it seems strange, the sled exerts a tension force on the rope.

A Short Catalog of Forces - Normal The normal force is a contact force between 2 surfaces that is perpendicular to the surfaces

A Short Catalog of Forces - friction The frictional force is a contact force between 2 surfaces that is parallel to the surfaces

A Short Catalog of Forces Other forces are usually shown as an with an identifying subscript.

You’ve just kicked a rock, and it is now sliding across the ground about 2 meters in front of you. Which of these forces act on the rock? A. Gravity, acting downward B. The normal force, acting upward C. The force of the kick, acting in the direction of motion D. Friction, acting opposite the direction of motion E. All of the above

You’ve just kicked a rock, and it is now sliding across the ground about 2 meters in front of you. Which of these forces act on the ball? You may choose more than one answer. A. Gravity, acting downward B. The normal force, acting upward C. The force of the kick, acting in the direction of motion D. Friction, acting opposite the direction of motion E. All of the above Ans: A, B, and D

Free-body diagrams Identify all forces acting on the object. Draw a coordinate system. If motion is along an incline, the coordinates system should be tilted relative to true horizontal and true vertical. Represent the object as a dot at the origin. Draw vectors representing each of the identified forces and label with the appropriate symbol.

Extension 1 – Freebody diagrams

Mass An object’s mass (m) is a measure of the amount of matter that it contains. The SI unit of mass is the kilogram (kg). The English unit of mass is the slug, which is virtually never used. The mass of an object does not change unless you add or subtract matter from it. The mass of an object is the same on any planet and in deep space. Scales do not measure mass, except for a balance scale, which compares one mass to another.

Weight The weight of an object on or above the earth is the gravitational force that the earth exerts on the object (F G or W). Since weight is a force, the SI unit of weight is the Newton (N); the English unit is the pound. The weight of an object changes on other planets since the gravitational force of other planets are different than that of the earth. Scales do not measure weight, at least not directly.

Apparent weight or “what the scale says” Scales measure a normal force or a tension force (i.e.hanging scales). In most situations, the force measured by the scale is numerically equal to the weight force. In accelerating reference frames (e.g. the ever-popular elevator with a scale), the scale reading will differ from the true weight. This is called apparent weight. Metric scales assume measurement on earth and are calibrated to read in mass units (kg). American scales read in English force units. It’s a culture thing.

4.7 The Gravitational Force Newton’s Law of Universal Gravitation Every particle in the universe exerts an attractive force on every other particle. A particle is a piece of matter, small enough in size to be regarded as a mathematical point. The force that each exerts on the other is directed along the line joining the particles.

4.7 The Gravitational Force For two particles that have masses m 1 and m 2 and are separated by a distance r, the force has a magnitude given by:

4.7 The Gravitational Force If m1 has a mass of 12 kg and m2 has a mass of 25 kg and the two objects are 1.2 meters apart, what is the gravitational force between them?

4.7 The Gravitational Force The gravitational force between 2 objects is usually pretty small, unless one of the object is a planet!

If m1 has a mass of 12 kg and m2 has a mass of 25 kg and the two objects are twice as far apart as before (i.e. 2.4 m), what is the gravitational force between them? Twice as much? Half as much? Or something else

A. one quarter as big. B. half as big. C. the same size. D. twice as big. E. four times as big. The figure shows a binary star system. The mass of star 2 is twice the mass of star 1. Compared to, the magnitude of the force is

A. one quarter as big. B. half as big. C. the same size. D. twice as big. E. four times as big. The figure shows a binary star system. The mass of star 2 is twice the mass of star 1. Compared to, the magnitude of the force is

4.7 The Gravitational Force Definition of Weight The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always acts downwards, toward the center of the earth. On or above another astronomical body, the weight is the gravitational force exerted on the object by that body. SI Unit of Weight: newton (N)

4.7 The Gravitational Force Calculate g M E = 5.98 x kg R E = 6.38 x 10 6 m (don’t forget to square!) and W = mg

4.7 The Gravitational Force Relation Between Mass and Weight g = 9.80 m/s 2 on Earth

4.7 The Gravitational Force Relation Between Mass and Weight

4.3 Newton’s Second Law of Motion Newton’s Second Law An object of mass m, subject to forces F 1, F 2, etc. will undergo an acceleration with a magnitude directly proportional to the net force and inversely proportional to the mass: The direction of the acceleration is the same as the direction of the net force.

4.3 Newton’s Second Law of Motion Newton’s First Law This is a special case of the 2 nd Law. If the net force equals 0, there is no acceleration and the velocity of the object will not change. If it is at rest, it will stay at rest. In this case the object is in equilibrium. Static equilibrium – object is at rest for a finite period Dynamic equilibrium –object is moving at constant velocity.

Problem-Solving Strategy for Newton’s Law Problems

Problem-Solving Strategy: Equilibrium Problems If the net force is equal to zero, must the object be at rest? The next example shows the answer to this:

Newton’s 1 st Law: Towing a car up a hill A car with a weight (not mass) of 15,000 N is being towed up a 20° frictionless slope at constant velocity. The tow rope is parallel to the slope surface. What is the tension on the tow rope? Is this a Newton’s 2 nd Law problem (ΣF = ma) or a Newton’s 1 st Law problem (ΣF = 0)? Draw a free body diagram for this problem.

Newton’s 1 st Law: Towing a car up a hill A car with a weight (not mass) of 15,000 N is being towed up a 20° frictionless slope at constant velocity. The tow rope is parallel to the slope surface. What is the tension on the tow rope? What part of the story indicates that ΣF = 0?

Newton’s 1 st Law: Towing a car up a hill Once the pictorial representation is completed, use Newton’s 1 st Law in component form: T and n have only one non- zero component, but F G has non-zero components in both x and y directions. How do we write the components of F G in terms of θ ?

Newton’s 1 st Law: Towing a car up a hill Solving the 2 nd equation tells us that n = F G cos θ, which is not necessary information for this problem.

Problem-Solving Strategy for Newton’s 2 nd Law Problems 1.Use the problem-solving strategy outlined for Newton’s 1 st Law problems to draw the free body diagram and determine known quantities. 2.Use Newton’s Law in component form to find the values for any individual forces and/or the acceleration. 3. If necessary, the object’s trajectory (time, velocity, position, acceleration) can be determined by using the equations of kinematics. 4.Reverse # 2 and 3 if necessary.

Newton’s 2nd Law – Speed of a Towed Car Find the acceleration using Newton’s second law in component form. Find the velocity using one of the kinematic equations. Note that in this problem, the car’s mass, not the weight, is given.

Speed of a towed car – pictorial representation of motion, motion diagram and free body diagram

Speed of a towed car

EXAMPLE 6.3 Speed of a towed car Our pictorial analysis (and common sense) indicate that there is no acceleration along the y-axis.

EXAMPLE 6.3 Speed of a towed car

Apparent Weight – What the scale says (even if there is no scale!) The value the scale reads when the object is accelerating or being acted upon by some other force (e.g the buoyant force in water) Not equal to the true weight (F G = mg) Apparent weight is either a normal force (step on scale) or tension force (hanging scale) An object inside an accelerating elevator has the same acceleration as the elevator. Acceleration is not a force so don’t include directly on free body diagram.

Typical FBD for an apparent weight problem n F G = mg, but n does not!

Apparent Weight – What the scale would say, even if there is no scale It takes the elevator in a skyscraper 4.0 s to reach a constant speed of 10.0 m/s. A 60.0 kg passenger gets on at the ground floor. What is her apparent weight during those 4 seconds? Use the problem-solving strategy (repeated in next slide).

Problem-Solving Strategy for Newton’s Law Problems

Problem-Solving Strategy for Newton’s 2 nd Law Problems 1.Use Newton’s Law in component form to find the values for any individual forces and/or the acceleration. 2. If necessary, the object’s trajectory (time, velocity, position, acceleration) can be determined by using the equations of kinematics. 3.Reverse # 1 and 2 if necessary (Hint: it is!)

Apparent Weight Problem a = +2.5 m/s 2 n = apparent weight = 738 N n FBD of elevator passenger

Graphical Interpretation of Newton’s Second Law Newton’s 2 nd Law is the equation of a line with a 0 value for the y-intercept (in this case the F- intercept!).

The following graphs plot net force vs. acceleration for different objects. Rank each situation according to mass. That is, order the situation from largest to smallest mass, using Newton’s Second Law.

Ranking Order from greatest to least mass: F C (A D) B E (Note this one is slightly different than the example in Exploration 3.

Friction

Kinetic Friction Experiments show that the kinetic friction force is nearly constant and proportional to the magnitude of the normal force. where the proportionality constant μ k is called the coefficient of kinetic friction.

Static Friction The box is in static equilibrium, so the static friction must exactly balance the pushing force:

Static friction An object remains at rest as long as f s < f s max The object slips when f s = f s max A static friction force f s > f s max is not physically possible. f s max >f k for the same surfaces where the proportionality constant μ s is called the coefficient of static friction.

A model of friction “ motion” indicates motion relative to the two surfaces the max value static friction, f s max occurs at the very instant the object begins to move (which often means 1 nanosecond before, for problem-solving purposes. for any given materials, μ s > μ k

Rank order, from largest to smallest, the size of the friction forces to in these five different situations. The box and the floor are made of the same materials in all situations.

Example Problem w tilted axis A 75-kg snowboarder starts down a 50-m high (not long!), 10 0 slope with μ k = Assume he has just started moving, but his starting velocity is essentially zero. What is his speed at the bottom? Draw the free body diagram (fbd) to determine the forces. Use Newton’s Law in component form. Determine the acceleration. Use kinematics to solve for the final speed. Start with the fbd.

Example Dynamics Problem w tilted axis A 75-kg snowboarder starts down a 50-m high, 10 0 slope, with μ k = What is his speed at the bottom? Write Newton’s Law for y-axis values. Is this 2 nd Law, or 1 st Law (a=0)? There is one force along the + y axis and one force with a negative y component. n FGFG fkfk Find: v 1 u k =.006

Example Dynamics Problem w tilted axis A 75-kg snowboarder starts down a 50-m high, 10 0 slope, with μ k = What is his speed at the bottom? Write Newton’s Law for y-axis values. Is this 2 nd Law, or 1 st Law (a=0)? ΣF y = n – mg cos θ = ma y But no acceleration along y axis so the net force is 0 there n = mg cos θ n FGFG fkfk Find: v 1 u k =.006

Example Dynamics Problem w tilted axis A 75-kg snowboarder starts down a 50-m high, 10 0 slope, with μ k = What is his speed at the bottom? ΣF x = mg sin θ – f k = ma and f k = u k n and n = mg cos θ mg sin θ – u k mg cos θ = ma You didn’t really need the mass; it cancels out. plug n chug: a = 1.12 m/s 2 now find v 1 n FGFG fkfk Find: v 1 u k =.006 n = mg cos θ

Example Dynamics Problem A 75-kg snowboarder starts down a 50-m high, 10 0 slope on a frictionless board. What is his speed at the bottom? Find v 1 a, =1.12 m/s 2

Example Dynamics Problem A 75-kg snowboarder starts down a 50-m high, 10 0 slope on a frictionless board. What is his speed at the bottom? Time is not important. Use the Indian Princess to calculate the length (hypotenuse) of the slope. That’s our x 1. v 1 = 25.4 m/s Find v 1 a, =1.12 m/s 2

Make sure the cargo doesn’t slide A box is in the back of a flatbed truck What is the maximum acceleration the truck can have without the box sliding? Coefficients of friction between truck and box are μ s = 0.40 and μ k = To solve this problem: Is the object of interest the box, truck, or both? Do the forces on the object of interest include static friction, kinetic friction, both, or neither?

Make sure the cargo doesn’t slide n FgFg fsfs Knowns μ s = 0.40 μ k = 0.20 F G = 9.8 m find a max fbd - box

Make sure the cargo doesn’t slide n FgFg f s max Knowns μ s = 0.40 μ k = 0.20 F G = mg find a max ΣF y = ma y = 0 n – F G = 0, therefore n = mg and f smax = μ s mg

Make sure the cargo doesn’t slide n FgFg f s max Knowns μ s = 0.40 μ k = 0.20 F G = mg find a max ΣF x = ma x friction is the only force! μ s mg = ma, and the m drops out a = 3.9 m/s/s

EOC #107 A person is trying to judge whether a picture of mass 1.10 kg is properly positioned by pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between picture and wall is What is the minimum amount of pressing force required? Draw a freebody diagram. In which direction is the normal force in this problem? Does it have anything to do with the weight of the picture?

EOC # 107 Freebody diagram F push FgFg fsfs n fbd - picture Knowns m = 1.10 kg μ s = Find F push Forces which are usually x are y in this problem and vice versa (with the exception of gravity). Newton’s Laws still work.

EOC # 107 Freebody diagram F push FgFg fsfs n fbd - picture Knowns m = 1.10 kg μ s = Find F push Newton’s Law in the x direction tells us that n = F push but nothing else about the value of either. Moving right along to the y- direction: ΣF y = ma y = 0 = f s – F G or f s = mg. No matter how hard you press, the picture will not levitate up. Fact. However, if you don’t push hard enough….

EOC # 107 Freebody diagram F push FGFG fsfs n fbd - picture Knowns m = 1.10 kg μ s = Find F push The minimum value of n must be the value that allows f smax to be equal to the weight of the picture: ΣF y = 0 = f smax – F G or μ s |n| = mg n = F push = 16.3 N

Push the crate (problem 1 from Exploration 4) A dockworker pushes a 50-kg crate with a downward force of N, at an angle of 30 degrees below horizontal. The coefficient of kinetic friction is A. What is the normal force? B.What is the frictional force? C.What is the acceleration? Draw a freebody diagram of the object of interest (will it have tilted axes?) Apply Newton’s Law in component form.

Push the crate - Freebody diagram n F push FgFg fkfk θ Knowns m = 50 kg F push = 300N θ = 30˚ μ k = 0.30 Find n, f k,a

Push the crate (problem 1 from Exploration 4) A. ΣF y = ma y n - F G - F p sin 30 ˚ = 0 this solves for n = 640N B. f k = μ k |n| = 192N C.ΣF x = ma x F p cos 30 ˚ - μ k |n| = ma x This will give us a x which is a = 1.36 m/s/s

Keep it up! Problem 3 from Exploration 4 A man pushes with a 100 N force on a 10- kg crate, which is sliding down the wall (despite his efforts). The pushing force is directed 53 degrees above horizontal. The coefficient of kinetic friction is Determine the magnitude of the friction force between wall and crate. Draw a freebody diagram. In which direction is the normal force in this problem? Use Newton’s law in the appropriate direction to determine the normal force.

Keep it up! Freebody diagram F push FgFg fkfk n fbd - box Knowns m = 10 kg F push = 100N θ = 53˚ μ k = 0.60 Find f k,a θ Forces which are usually x are y in this problem and vice versa (with the exception of gravity). Newton’s Laws still work.

Keep it up! Freebody diagram F push FgFg fkfk n fbd - box Knowns m = 10 kg F push = 100N θ = 53˚ μ k = 0.60 Find f k,a θ ΣF x = F p cos θ – n = 0 n = 60.2 N f k = 36.1 N ΣF y = F p sin θ + f k – F G = ma x a = m/s/s therefore box is slowing down.