Physics: Handout #1 – 9, 12, 14, 15, but all are highly recommended. Forces Homework: Physics: Handout #1 – 9, 12, 14, 15, but all are highly recommended.

Forces: push pull A force is a _____________ or a ____________ on an object by another object.

Forces can be applied either through direct contact, as shown in the picture below (a), (b), and (c). Forces shown on the right are called “action at a distance”, where no physical contact is required between the objects to cause a force.

inertia A force is tied to the motion of an object through ________________ . Inertia is the ability of an object to ________________ a change in its motion. resist Inertia is measured as the__________ of an object. The standard unit is the kilogram, kg. mass The more massive an object is, the more it is able to resist a change in its motion. It is easier to push a 10 kg object into motion than it is a 3000 kg car. dynamics All of this begins the topic of _______________, or why objects move.

Newton’s Laws of Motion: The relationship between motion and force was stated clearly by Isaac Newton in his 3 laws. Each law relates a different aspect of motion. Newton’s First Law: If the total, or net, force on an object is zero, then an object will not accelerate. First variation: If an object is at rest, it will continue to remain at rest until acted upon by some external agent. Example: A book placed on a desk will remain on the desk until someone removes it. There are forces acting on the book: gravity pulls down on it and the table pushed upwards on it. These forces balance out, and the object does not move.

Newton’s First Law: continued Second variation: If an object is moving, it will continue to move with a constant velocity. Since velocity is a vector, this means the speed (how fast) and the direction remain unchanged. Example: Take a look at a hockey game. When the puck (small black projectile) is hit, the puck travels at nearly the same speed and in a straight line across the ice. All real objects are subject to friction for anything sliding on a surface, and an icy surface is as close to frictionless as we can get.

Newton’s Second Law: If the net force on an object is not zero, then an object will accelerate. The acceleration of the object is directly proportional to the total force acting upon it, and the acceleration is inversely proportional to the mass of the object. As an equation: Here, m is the mass of the object, measured in kg. a is the acceleration of the object, measured in m/s2. This is a vector quantity F refers to the amount of force. Fnet or SF is the sum of the forces acting on an object. Force is a vector, so it will have magnitude and direction. The net force and the acceleration point in the same direction.

Units: Force is measured in units called newtons, and this unit is represented by the letter N. Newtons can be written in term of fundamental units through the statement of Newton’s Second Law: Example #1: What is the acceleration of a 2.00 kg mass if a force of 4.00 N acts upon it?

Example #2: An object with a mass of 3 Example #2: An object with a mass of 3.00 kg moves from rest to a speed of 20.0 m/s in a time of 5.00 s. a. What is the acceleration of the object? Start with: and solve for a: b. What is the net force acting on the object? Both force and acceleration are vectors. The net force points in the same direction as the acceleration of the object.

Example #2: c. Let the object’s acceleration point to the right Example #2: c. Let the object’s acceleration point to the right. If there is a drag force of 30.0 N opposing the motion, what other applied force must there be on the object to give the desired acceleration? A picture helps!

For one dimension, the vector signs are dropped. It is understood all vectors point along the same line. Just use a (+) sign for the vectors pointing right and a (–) sign for the vectors pointing left. The variables are assumed to be only the magnitude of the vectors.

Newton’s Third Law: If one object (object A) exerts a force on another object (object B), then object B exerts the same force back on object A. The force is equal in magnitude, but opposite in direction. Alternate interpretation: For every action there is an equal but opposite reaction. Example: If you drive your fist against a wooden wall as hard as you can, you will possibly leave a mark on the wall with your fist. The mark is made by the force you exerted against the wall. The wall also exerts an equal force on your fist. You perceive this as the severe pain felt in your knuckles. object A object B Drawing: FB on A FA on B

Example #3: A student pushes with a 30 Example #3: A student pushes with a 30.0 lb force towards the South on a heavy box. There is a friction (drag) force between the floor and the box that prevents the movement of the box. What is the magnitude and direction of the friction force? Since the box does not move, the net force must be zero.

Example #4: A baseball is thrown at a speed of 40 Example #4: A baseball is thrown at a speed of 40.0 m/s towards the right. The ball is hit by a baseball bat, and then travels at 50.0 m/s towards the left. The ball is in contact with the bat for 2.00 ms and the ball has a mass of 200 grams. What is the net force acting on the ball? Initial information: Let right be positive and left be negative. vo = +40.0 m/s, v = – 50.0 m/s, t = 2.00 x 10-3 s, m = 0.200 kg

Example #5 A horse is hitched to a wagon. Which statement is correct? A. The force that the horse exerts on the wagon is greater than the force that the wagon exerts on the horse. B. The force that the horse exerts on the wagon is less than the force that the wagon exerts on the horse. C. The force that the horse exerts on the wagon is just as strong as the force that the wagon exerts on the horse. D. The answer depends on the velocity of horse and wagon. E. The answer depends on the acceleration of horse and wagon.

Example #5 A horse is hitched to a wagon. Which statement is correct? A. The force that the horse exerts on the wagon is greater than the force that the wagon exerts on the horse . B. The force that the horse exerts on the wagon is less than the force that the wagon exerts on the horse. C. The force that the horse exerts on the wagon is just as strong as the force that the wagon exerts on the horse. D. The answer depends on the velocity of horse and wagon. E. The answer depends on the acceleration of horse and wagon.

Example #6 You are standing at rest and begin to walk forward. What force pushes you forward? A. the force of your feet on the ground B. the force of your acceleration C. the force of your velocity D. the force of your momentum E. the force of the ground on your feet

Example #6 You are standing at rest and begin to walk forward. What force pushes you forward? A. the force of your feet on the ground B. the force of your acceleration C. the force of your velocity D. the force of your momentum E. the force of the ground on your feet

Table Cloth Pull: An example of the Law of Inertia.

Tension, Gravity, and Normal Force Homework: Physics Day 1: #1 – 7, 10 Physics Day 2: #8, 9, 11 – 14, 15 a,b,c.

w = weight of the object. This is a force measured in N (newtons). Definitions: Forces are any kind of push or pull on a system or on an object. There are some special types of forces that we will define today. The first is the weight of an object. gravity The weight of an object is the force of ___________ acting on the object by Earth. mass The amount of force is directly proportional to the ____________ of the object. The equation for weight is as follows: w = weight of the object. This is a force measured in N (newtons). m = mass of the object, in kg (kilograms). g = acceleration of gravity, 9.80 m/s2. The direction of the force is downwards, towards the center of Earth.

Example #1: What is the weight of a 55.0 kg person? Example #2: In the British system, mass is measured in slugs and force is measured in pounds. Weight is still measured as mg, with one slug mass weighing 32.2 pounds. Convert the mass of the person above into slugs. Conversion Factor: 1.0000 kg mass weighs 2.2046 lbs

normal Definition: The __________ force on an object is defined as the force exerted on an object by a surface. The term “normal” is another term used in mathematics for perpendicular. The normal force exerted by a surface is always perpendicular to that surface. The normal force is represented by the letter _________ . Do not confuse this with newtons of force! n Problem Solving: Always draw a picture, and label all the forces acting on the object. This is referred to as drawing a “Free Body Diagram”. Only draw the forces acting on the object. By Newton’s third law, the forces on an object result also in forces exerted by the object on the surroundings. We are only interested in the forces on the object by the environment.

Example #3: A 1. 4 kg book is placed on a flat surface at rest Example #3: A 1.4 kg book is placed on a flat surface at rest. What is the normal force acting on the book? Start with a labeled picture: normal = n Next apply Newton’s Second Law: m = 1.4 kg For one dimension, define up as positive and down as negative. The forces may be written as follows: weight = mg The variables show the magnitude, or size of the vector. The ± sign gives the direction. and

The net force becomes: Since the object is at rest, the acceleration of the object is zero. Combine the above equations and solve for n.

Example #4: {Classic Physics Problem Example #4: {Classic Physics Problem!!!} A person stands in an elevator on a scale. (Don’t ask why…..) When the elevator is at rest, the scale reads 588 N. What will the scale read if the elevator accelerates upwards at 2.20 m/s2? Solution: The scale is designed to measure a force. The scale does not measure the weight of the person, rather the force exerted upwards on the person to support them. In other words, the scale measures the normal force acting on the person. When the person is at rest, the normal force equals the weight of the person. When the person is accelerated in the elevator, the net force on the person is no longer zero. For an upward acceleration, the normal force must be larger than the weight of the person. Let’s find out why.

Before anything, determine the mass of the person: At rest, the person weighs 588 N. This equals mg, so the mass is:

up = positive Start by drawing a complete picture and then apply Newton’s Second Law. normal = n Draw in a coordinate axis and choose a positive direction. elevator Apply Newton’s 2nd Law: person, m weight = mg down = negative

Substitute numbers: Solve for the new normal force for the accelerated elevator. The person appears to weigh more when the elevator accelerates upwards.

Example #5: What is the motion of the elevator if the person’s weight appears to be 360 N? Just follow the same reasoning as above and show again: Solve this expression for a. Since up was chosen as the positive direction, a negative acceleration means that the elevator is accelerating downwards.

Example #6: What would the scale read if the elevator was in freefall? Just follow the same reasoning as above and show again: Solve this expression for n. Note that a = – g. For freefall, the scale would read zero. The person would feel weightless.

tension pulling Definition: A __________ force is a ____________ force, usually applied through a string or rope. Ropes are to be treated as ideal, meaning that the mass of the ropes can be ignored and the ropes do not stretch or sag when used. The force of tension is then the same throughout the rope.

Example #7: Two masses rest on a frictionless horizontal surface Example #7: Two masses rest on a frictionless horizontal surface. The mass on the left is 60.0 kg and the mass on the right is 40.0 kg. The masses are tied together with a light cord and a 25.0 N force is applied towards the right on the 40.0 kg mass. Find the acceleration of the system and the tension force in the cord. m1 = 60.0 kg applied force Fa = 25.0N m2 = 40.0 kg cord

Solution: First treat the two masses as a single object Solution: First treat the two masses as a single object. They move the same: same velocity and acceleration. applied force Fa = 25.0N mtotal = m1 +m2 Apply Newton’s 2nd Law:

Now isolate one mass to solve for the force of tension in the cord Now isolate one mass to solve for the force of tension in the cord. If the mass on the left (m1) is chosen, the solution is as follows: T = tension force m1 Apply Newton’s 2nd Law and solve for the tension force.

An alternate solution would be to choose the mass on the right (m2). applied force Fa = 25.0N T = tension force m2 Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be towards the right.

Example #8: A mass (m1 = 10.0 kg) hangs from the ceiling of an elevator from a light cord. A second mass (m2 = 20.0 kg) is suspended from the first mass by another cord. If the elevator accelerates upwards at a rate of 4.70 m/s2, determine the tension in each cord. up = positive cord with tension T1 m1 elevator cord with tension T2 m2 down = negative

Solution: First treat the two masses as a single object and solve for the tension in the top cord. Start with a simple picture. Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be upwards. T1 mtot mtotg

Now isolate one mass to solve for the force of tension in the lower cord. If the mass on the bottom (m2) is chosen, the solution is as follows: Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be upwards. T2 m2 m2g

If the mass on the top (m1) is chosen, the solution is as follows: Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be upwards. T1 m1 T2 m1g

Q4.1 Motor An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable is Cable v Elevator A. greater than the downward force of gravity. B. equal to the force of gravity. C. less than the force of gravity. D. any of the above, depending on the speed of the elevator.

A4.1 Motor An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable is Cable v Elevator A. greater than the downward force of gravity. B. equal to the force of gravity. C. less than the force of gravity. D. any of the above, depending on the speed of the elevator.

Q4.9 A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg crate by a light rope. The light rope remains taut. Compared to the 6.00–kg crate, the lighter 4.00-kg crate A. is subjected to the same net force and has the same acceleration. B. is subjected to a smaller net force and has the same acceleration. C. is subjected to the same net force and has a smaller acceleration. D. is subjected to a smaller net force and has a smaller acceleration. E. none of the above

A4.9 A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg crate by a light rope. The light rope remains taut. Compared to the 6.00–kg crate, the lighter 4.00-kg crate A. is subjected to the same net force and has the same acceleration. B. is subjected to a smaller net force and has the same acceleration. C. is subjected to the same net force and has a smaller acceleration. D. is subjected to a smaller net force and has a smaller acceleration. E. none of the above

Homework: Handout #3, all problems. 2 – Dimensional Forces Homework: Handout #3, all problems.

ConcepTest 4.17 Three Blocks Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? 1) T1 > T2 > T3 2) T1 < T2 < T3 3) T1 = T2 = T3 4) all tensions are zero 5) tensions are random T3 T2 T1 3m 2m m a

ConcepTest 4.17 Three Blocks T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass. 1) T1 > T2 > T3 2) T1 < T2 < T3 3) T1 = T2 = T3 4) all tensions are zero 5) tensions are random T3 T2 T1 3m 2m m a

vector magnitude Forces are _________ quantities: they have both a _____________ and a ____________. direction For Newton’s 2nd Law, both the net force and the acceleration are vectors. It is easier to resolve the vectors into components to solve 2 – dimensional problems. On any given problem, draw in a coordinate set and label which directions are positive . Example #1: A 40.0 kg block rests on a smooth surface. A force of 400 N is applied to the block at an angle of 36.9o above the horizontal. a. Label all forces acting on the object, and resolve these forces into horizontal and vertical components. Fa n = normal q = 36.9o m mg = weight

Fa n = normal q = 36.9o +y m mg = weight +x The normal force has only a y – component: nx = 0, ny = +n The weight has only a y – component: wx = 0, wy = –mg The x – component of the applied force, Fa, is: Fa,x = +Facosq The y – component of the applied force, Fa, is: Fa,y = +Fasinq

Ex. #1 b. Determine the size of the normal force acting on the block. Apply Newton’s 2nd Law to the y – components and solve for the normal force. The y – component of the acceleration is zero. The y – component of the applied force is not larger than the weight of the object, so the object will not be lifted from the surface. It will only slide parallel to the surface.

Ex. #1 c. What is the acceleration of the block Ex. #1 c. What is the acceleration of the block? How fast will it be moving after traveling 10.0 meters? Apply Newton’s 2nd Law to the x – components and solve for the acceleration. Reach back to chapter 2 for a motion equation:

ConcepTest 4.10 Normal Force Below you see two cases: a physics student pulling or pushing a sled with a force F which is applied at an angle q. In which case is the normal force greater? 1) case 1 2) case 2 3) it’s the same for both 4) depends on the magnitude of the force F 5) depends on the ice surface In Case 1, the force F is pushing down (in addition to mg), so the normal force needs to be larger. In Case 2, the force F is pulling up, against gravity, so the normal force is lessened. Case 1 Case 2

Example #2: An object has a two forces acting upon it Example #2: An object has a two forces acting upon it. One force is 454 N, pointing due north, and the other force is 844 N, pointing due west. What is the net force on the object and the acceleration of the object? Give the answer as magnitude and direction. What one force would counter this net force? a. solve with Pythagorean theorem. N, +y Ftotal F1 q E, +x F2

Example #3: Statics. A lady with a weight of 522 N balances at the center of a long rope, as shown. If the rope deflects 10.0o from the horizontal, determine the tension in the rope. +y +x T1 T2 mg find the components of each vector.

This is a static system, meaning there is no acceleration This is a static system, meaning there is no acceleration. Solve this with Newton’s second law. Start with the x – component. therefore

Next use the y – component.

find the components of each vector. Example #4: A mass of 125 kg is supported by two ropes. One rope extends horizontally, towards the left from the object to a support wall. The second rope extends towards the right of the object, at an angle of 53.1o above the horizontal. Find the tension in each rope. draw each force vector. find the components of each vector. tension T2 q = 53.1o tension T1 +y mg +x

This is a static system, meaning there is no acceleration This is a static system, meaning there is no acceleration. Solve this with Newton’s second law. Start with the y – component. (Random choice. You could start with the x – component.)

Next use the x – component. therefore

find the components of each vector. Example #5: A 615 kg mass is suspended as shown below. Determine the tension in each rope. q1 = 20.0o q2 = 60.0o draw each force vector. tension T2 tension T1 find the components of each vector. q2 = 60.0o q1 = 20.0o +y mg +x

This is a static system, meaning there is no acceleration This is a static system, meaning there is no acceleration. Solve this with Newton’s second law. Start with the x – component. (Random choice. You could start with the y – component.) This is one equation with two unknowns. Use the y – component to get another equation with the same two unknowns.

Next use the y – component. Substitute the equation from the x – direction.

Solve for T2. Substitute the equation from the x – direction.

2 – Dimensional Forces Day #2 Homework: Handout #4, problems #1 – 3, 12

Example #6: Two masses are connected by a string over a pulley, as shown. The bottom mass (m1) is 25.0 kg and the hanging mass (m2) is 10.0 kg. What is the normal force on the bottom mass? m2 m1

Example #6: Two masses are connected by a string over a pulley, as shown. The bottom mass (m1) is 25.0 kg and the hanging mass (m2) is 10.0 kg. What is the normal force on the bottom mass? Draw in the forces and a coordinate line. +y This system is at rest. The suspended mass is too small to make the bottom mass rise upwards on its own. T m2 Solve this problem by isolating each mass and apply Newton’s 2nd law to each mass. T m2g n m1 m1g – y

Start with the hanging mass, m2. Set the acceleration equal to zero and solve for T. m2g

Next look at the bottom mass, m1: Set the acceleration equal to zero and solve for n. n m1 m1g

Example #7: Given the two masses shown below, find the acceleration of the system and the tension in the string. Ignore friction. Mass m1 moves and accelerates towards the right. Mass m2 falls and accelerates downwards. First determine the motion of each mass: Let these directions define the positive direction for each mass. Now label the forces. Apply Newton’s 2nd law to each mass separately.

Start with mass m1. There are two unknowns: acceleration and tension Next look at mass m2. This mass also has the same two unknowns: acceleration and tension. This will give us a second equation to solve the two unknowns. By choosing the positive directions to match the directions of the accelerations, the two objects will have the same acceleration, both in magnitude and direction.

Solve the simultaneous equations: Notice T’s will subtract out

Solve for the tension by substituting into one of the initial equations:

Example #8: Determine the acceleration of the system and the tension in the string. Assume the mass on the right is the heavier mass. Ignore friction.

Solve for the tension by substituting into one of the initial equations:

problems #5 and complete chapter reading. Force of Friction Homework: Handout #4 problems #5 and complete chapter reading. Test Friday

drag Definition: Friction is a form of a ____________ force. This is a force that opposes the motion of an object. In general, this force may be very complex, but the form we will study will be simple. static kinetic The friction force comes in two types: ___________ and ___________ . Static friction ____________ ____________ refers to the force that will keep an object at rest. Static means that there is no motion between the surfaces in contact. The static friction force is a variable force: The force will take on whatever value it needs to prevent an object from moving, up to a maximum amount. The maximum amount of force is proportional to the normal force between the two surfaces. Kinetic friction ____________ ____________ refers to the force that resists an object’s motion. Kinetic friction only operates when the object is moving. For our purposes, this force is constant, regardless of the speed of the object.

Friction always tries to oppose the motion of two contacting surfaces relative to one another. Static friction balances any applied force, up to a maximum limit. limit of static friction forces balance When the applied force exceeds the maximum static friction, the object breaks free and moves with constant kinetic friction.

The size of the maximum static friction force is given in terms of the normal force the surface exerts on the object: Fs is the static friction force between two surfaces. Fs,max is the maximum static friction force between two surfaces. n is the normal force between two surfaces. The maximum static friction force can be made equal to the normal force through a proportionality constant. This constant is called the coefficient of static friction, and it is represented by the symbol ms.

The coefficient of static friction rates the “roughness” of a surface The coefficient of static friction rates the “roughness” of a surface. The higher the value of the coefficient, the stronger the static friction force can be. The coefficient does not have any units! This value is unique to a given surface, and it is usually given in some way. Example #1: A 400 lb block rests on an icy horizontal surface. The coefficient of static friction between the block and the ice is 0.100. a. What will be the static friction force if a horizontal force of 20.0 lb is applied to the block? b. What if the applied force is 40.0 lb? c. 50.0 lb? SOLUTION: Start by finding the maximum static friction force available. The coefficient has a value of ms = 0.100, so the maximum static friction force is:

For a block resting on a flat surface, the normal force equals the weight of the object. Combining the above, get: When the applied force is 20.0 lb, the static friction force will match it at 20.0 lb in the opposite direction. When the applied force is 40.0 lb, that is right at the limit that static friction can hold the object still, or move it at constant speed. When the applied force is 50.0 lb, the static friction will fail and the object will accelerate.

Definition: Kinetic friction refers to the force that opposes a moving object’s motion. The kinetic friction force is proportional to the normal force between the object and the surface. The kinetic friction force is constant in magnitude. This force is independent of the speed of the object. It is also independent of the surface area of the contact area. Fk is the kinetic friction force between two surfaces. n is the normal force between two surfaces. The kinetic friction force can be made equal to the normal force through a proportionality constant. This constant is called the coefficient of kinetic friction, and it is represented by the symbol mk.

Example #2: A 40.0 kg box is made to slide on a horizontal surface that has a coefficient of kinetic friction of 0.510. How much force must be applied to the object to make it accelerate at 3.00 m/s2 horizontally?

Since the block never leaves the surface, the vertical forces balance. Use the definition of kinetic friction: Now use Newton’s 2nd law in the horizontal direction:

Solve for the applied force:

1. (Review) A force of 90.0 N pulls horizontally towards the right on an object with a mass of 20.0 kg along a surface with a coefficient of kinetic friction of 0.300. (a) What is the acceleration of this block?

Since the block never leaves the surface, the vertical forces balance. Use the definition of kinetic friction: Now use Newton’s 2nd law in the horizontal direction:

Solve for the acceleration: (b) Now use kinematics to get the final velocity:

2. A force of 90.0 N pulls the same object towards the right at an angle of 16.7 degrees above the horizontal. If the mass is still 20.0 kg and the coefficient of kinetic friction is still 0.300, (a) determine the normal force and kinetic friction force acting on the object.

There is no upward acceleration: The normal force is smaller than the weight of the mass since the applied force is also lifting upwards. A smaller normal force corresponds to a smaller friction force.

A couple of algebraic steps later……. (b) Determine the horizontal acceleration. A couple of algebraic steps later…….

Force of Friction Day #2

3. A 1.20 kg book is pushed against a vertical wall by a horizontal applied force. The coefficient of static friction between the book and the wall is 0.425. What is the amount of applied force needed to hold the book up without sliding? The forces balance in each direction: Use the definition of friction to combine the two equations:

4. Two masses are tied together, with one suspended over a pulley as shown. Let m1 = 20.0 kg and m2 = 5.00 kg. Let the coefficient of kinetic friction between m1 and the surface be 0.150. (a) Determine the acceleration of the system, assuming m2 is initially moving downwards. Mass m1 moves and accelerates towards the right. Mass m2 moves and accelerates downwards. Let these define the positive direction for the motion of each of the two masses.

Friction Positive

Tension m2g Positive

Isolate the first mass. The vertical forces balance. Solve the acceleration in the horizontal direction: Isolate the second mass and solve the acceleration :

Solve the simultaneous equations.

(b) What is the tension in the string? Pick one equation and solve for T:

(c) What would be the acceleration if the coefficient of kinetic friction increased to 0.300? What does this mean physically? The acceleration is negative. If the first mass were initially moving towards the right, then it would slow to a stop. If the system starts at rest, it would remain at rest {static friction}.

Complete Homework Set 5 Problems.

Homework: Set 6 {take two days to finish…} Inclined Planes Homework: Set 6 {take two days to finish…}

Introduction: An inclined plane (or incline or ramp) is a flat surface set to some angle relative to the horizontal. q = angle of the incline to the horizontal. Motion of the object will be parallel to the surface of the incline. The following slides include examples of inclined planes.

Johnstown, Pennsylvania

Reconstruction of Galileo’s device for determining the law of odd numbers.

Problem Solving Strategy: Since the motion is parallel to the ramp, take coordinates that are parallel and perpendicular to the ramp: This choice puts the acceleration only along one axis. This will simplify the problems.

When the cart is released on the ramp, it accelerates down the ramp When the cart is released on the ramp, it accelerates down the ramp. Note the water level when at rest or accelerating down the ramp. http://www.physics.umd.edu/lecdem/services/demos/demosc4/c4-12.htm

Ex. #1: Determine the components of gravity for a mass on an inclined surface. The weight force pulls straight downwards. Determine the components parallel and perpendicular to the surface of the incline. ǁ= mg sin θ

Ex. #2: A mass m slides down a ramp without friction Ex. #2: A mass m slides down a ramp without friction. (a) What is the normal force acting on the mass and (b) what is the acceleration of the mass? The perpendicular components balance: The parallel components produce the acceleration:

Ex. #3: A mass m slides down a ramp with friction Ex. #3: A mass m slides down a ramp with friction. (a) What is the normal force acting on the mass and (b) what is the acceleration of the mass? The perpendicular components still balance. By definition, the friction force is: The parallel components produce the acceleration. Take to the bottom of the ramp as the positive direction.

Substitute values and solve for the acceleration: Note: The acceleration will be zero when:

Ex. #4: A mass m is pulled to the top of a frictionless ramp by an applied force f. (a) If the mass is pulled to the top at a constant acceleration of a = 4.90 m/s2, what is the required force? Let q = 36.9° and m = 80.0 kg. Since there is no drag, the perpendicular components still balance are not needed. Apply Newton’s 2nd law to the parallel components:

Solve for f:

Ex. #5: Repeat the previous problem, but this time include friction on the surface of the ramp. The coefficient of kinetic friction is 0.400. There is an additional drag force pointing to the bottom of the ramp. By definition, the friction force is:

Now compute the net force parallel to the incline: plug in values:

Ex. #6: A mass m is placed on a frictionless incline of angle q to the horizontal. What horizontal force, F, is needed to hold the mass at rest? Values are given in the diagram. Balance the vertical forces: Balance the horizontal forces:

Divide the two equations to eliminate the normal force:

Ex. #7: Solve for the acceleration of this system. Mass m1 is 30 Ex. #7: Solve for the acceleration of this system. Mass m1 is 30.0 kg, mass m2 is 25.0 kg, and the coefficient of kinetic friction mk is 0.200. Assume m2 is initially moving downwards. Tackle this beast with the same technique as used on the Atwood’s machine. The perpendicular components for mass m1 balance:

Taking to the top of the ramp as the positive direction, solve for the parallel direction of incline: Now find Fnet on mass m2.

Add the two equations together:

Substituting numbers gives:

Ex. #8: {Bonus type question} Solve for the acceleration of this system. Assume the ramp is frictionless.

Solution Hints: