Chapter 1 - Velocity

The following is an adaptation of the original. Source: http://www.ck12.org/physics/Velocity/lesson/Velocity/

Students will learn how analyze and solve problems in two dimensions containing velocity vectors but no acceleration.

Key Equations

v x = v cos θ The x component of velocity equals the product of the velocity times cos theta, where theta is the angle between the velocity vector and the horizontal.

v y = v sin θ The y component of velocity equals the product of velocity times sin theta, where theta is the angle between the velocity vector and the horizontal.

Δ x = v x t The change in distance x equals the product of the x component of velocity and time.

Δ y = v y t The change in distance y equals the product of the y component of velocity and time.

Watch this Explanation

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Try viewing on Youtube (requires Internet connection): Relative Velocity on Youtube.

Guidance

The only equation you need is that displacement in a certain direction equals the component of velocity in that direction multiplied by the time it takes. You'll use this once for the x-direction and once for the y-direction and solve for what is asked.

Example 1

Question: If a river is flowing north at 2 m/s and you swim straight across (i.e. east) at 1.5 m/s, how far up shore will you be from your starting point once you reach the other side? The river is 9 m wide.

Answer: First solve for the time it takes you to reach the other side. Let's let north be the y-direction and the direction across the river be the x-direction.

Δ x = v x t

The change in distance x equals the product of the x component of velocity and time.

9 m =1.5 m/s × t

The river is 9 meters wide, so the change in distance is 9 meters. This is equal to the product of the velocity of swimming straight across at 1.5 meters per second and the total time it takes to travel across the river.

t = 6 s

Solving for time t, it takes 6 seconds to swim straight across a 9 meter wide river at a velocity of 1.5 meters per second.

Time for Practice

Question 1

If a river is flowing south at 4 m/s and you swim straight across (i.e. east) at 2 m/s; admittedly, you're going to drift a bit south. That said, calculate that distance that you drifted south from your starting point. The river is 16 m wide.

Answer: You would have drifted 32 m south of the starting point.

Question 2

If a river is flowing south at 3 m/s and you swim at an angle of 30 degrees north of directly east at 1 m/s, how far did you drift up or down stream from your starting point once you reach the other side? The river is 10 m wide.

Answer: You would have drifted 28.9 m south of starting point.

Question 3

If a river is flowing north at 2 m/s and you can swim at 4 m/s what angle should you swim at such that you arrive directly across the river (i.e. no drift north or south from starting point on other side)? The river is 10 m wide.

Answer: You would swim at a 30 degree angle to arrive directly across the river.

Question 4

If a river is flowing south at 5 m/s and you can swim at 4 m/s maximum, is it possible to arrive directly across? Why or why not?

Answer: No, even if you swim directly north the river will still take you south at 1 m/s.

Chapter 2 - Types of Forces

The following is an adaptation of the original. Source: http://www.ck12.org/physics/Types-of-Forces/lesson/Types-of-Forces/
Used under CK-12 Foundation CC BY-NC license.

The various types of common forces are discussed and analyzed.

Watch this Explanation

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Key Equations

Common forces include:

F g = m g Force of gravity equals mass of the object times gravitational acceleration.

F N Normal force: acts perpendicular to surfaces.

F T Force of tension in strings and wires.

F sp = - k Δ x Force of spring stretched equals the product of the spring coefficient and the distance from equilibrium.

Guidance

Normal Force

Often, objects experience a force that pushes them into another object, but once the objects are in contact they do not any move closer together. For instance, when you stand on the surface of the earth you are obviously not accelerating toward its center. According to Newton's Laws, there must be a force opposing the earth's gravity acting on you, so that the net force on you is zero. The same also applies for your gravity acting on the earth. We call such a force the Normal Force. The normal force acts between any two surfaces in contact, balancing what ever force is pushing the objects together. It is actually electromagnetic in nature (like other contact forces), and arises due to the repulsion of atoms in the two objects. Here is an illustration of the Normal force on a block sitting on earth:

Tension

Another force that often opposes gravity is known as tension. This force is provided by wires and strings when they hold objects above the earth. Like the Normal Force, it is electromagnetic in nature and arises due to the intermolecular bonds in the wire or string:

If the object is in equilibrium, tension must be equal in magnitude and opposite in direction to gravity. This force transfers the gravity acting on the object to whatever the wire or string is attached to; in the end it is usually a Normal Force --- between the earth and whatever the wire is attached to --- that ends up balancing out the force of gravity on the object.

Friction

Friction is a force that opposes motion. Any two objects in contact have what is called a mutual coefficient of friction. To find the force of friction between them, we multiply the normal force by this coefficient. Like the forces above, it arises due to electromagnetic interactions of atoms in two objects. There are actually two coefficients of friction: static and kinetic. Static friction will oppose initial motion of two objects relative to each other. Once the objects are moving, however, kinetic friction will oppose their continuing motion. Kinetic friction is lower than static friction, so it is easier to keep an object in motion than to set it in motion.

f s ≤ µ s | F → N |

Static friction opposes motion on surfaces in contact.

f k = µ k | F → N |

Kinetic fructions opposes motion of surfaces in contact.

There are some things about friction that are not very intuitive:

The magnitude of the friction force does not depend on the surface areas in contact.

The magnitude of kinetic friction does not depend on the relative velocity or acceleration of the two objects.

Friction always points in the direction opposing motion. If the net force (not counting friction) on an object is lower than the maximum possible value of static friction, friction will be equal to the net force in magnitude and opposite in direction.

Spring Force

Any spring has some equilibrium length, and if stretched in either direction it will push or pull with a force equal to:

F → sp = - k Δ x →

Force of spring Δ x → from equilibrium.

Example 1

Questions: A woman of mass 70.0 kg weighs herself in an elevator. A drawing of a woman in an elevator standing on a weight scale.

If she wants to weigh less, should she weigh herself when accelerating upward or downward?

When the elevator is not accelerating, what does the scale read (i.e., what is the normal force that the scale exerts on the woman)?

When the elevator is accelerating upward at 2.00 m/s², what does the scale read?

Answers:

If she wants to weigh less, she has to decrease her force (her weight is the force) on the scale. We will use the equation F = m a (force of an object equals mass of times acceleration of the object) to determine in which situation she exerts less force on the scale.

If the elevator is accelerating upward then the acceleration would be greater. She would be pushed toward the floor of the elevator making her weight increase. Therefore, she should weigh herself when the elevator is going down.

When the elevator is not accelerating, the scale would read 70.0 kg

Starting with the equation force equals mass times acceleration, we know the mass to be 70 kg For total acceleration, we add the acceleration required to overcome the force of gravity (9.8 m/s²) and the accelration of the elevator (2.0 m/s²) to give 11.8 m/s² total upward acceleration. Now multiplying mass of 70 kg with 11.8 m/s² gives 826 N. Converting newtons to kilograms, the scale would read 82.6 kg.

F =ma =70 kg × ( 9.8 m/s 2 + 2 m/s 2 ) = 826 N = 82.6 kg

Example 2

Question: A spring with a spring constant of k = 400 N/m has an uncompressed length of 0.23 m and a fully compressed length of 0.15 m. What is the force required to fully compress the spring?

Solution:

We will use the equation F = k x where k is the spring constant, and x is the distance of the spring when fully extended. The spring constant we are given and it is 400 N/m. To get x, the distance covered by the spring, we subtract the compressed length 0.15 m from the fully extended length 0.23 m to get 0.08 m.

Finally we multiply 400 N/m with 0.08 m to get 32 N

F = k x = ( 400 N/m ) ( 0.23 m - 0.15 m ) = 32 N

Chapter 3 - Adding Forces

Forces can be added and subtracted to give a resulting net force. To understand the basics of calculating net force, let us consider a game of tug-of-war.

Players from two teams form groups on opposite ends of a rope and pull against each other until one team wins by pulling the other team over the centre line. When the match starts, each team exerts force in opposite directions, and the team with the greatest combined force will win because the net force is going in their direction.

Example:

Left team chooses two members capable pulling 50 N and 100 N to the left. Right team selects three members who can pull 50 N, 50 N, and 100 N to the right.

To figure out which team will win, calculate the net force by adding up the forces of each team.

For the left team, the total force is 50 N + 100 N = 150 N.

For the right team, the total force is 50 N + 50 N + 100 N = 200 N.

( 50 + 100 ) - ( 50 + 50 + 100 )

= ( 150 ) - ( 200 )

= -50 N

Net force is 150 N minus 200 N (we are subtracting 200 N because right team is pulling in the opposite direction of left team). Therefore the total force is -50 N, which means right team wins.

Forces and Motion Interactive Games

The following is a collection of interactive games exploring the concepts of forces and motion. There are 4 games to choose from: Tug-of-War, Motion, Friction, and Acceleration. Feel free to choose and experiment with these games.

The following is a detailed description of the tug-of-war game.

A team is on the left (the blue team) and another team (the red team) is on the right. Each team can consist of 4 pullers, and each member pulls with different force. The team with the greatest combined force will win the round.

Instructions

Assemble the left team by choosing from any of the four available team members: large, medium, or one of two small team members. Place people onto the rope by dragging and dropping the pullers onto the rope.

Next, assemble the right team by the same way you assembled the left team.

Once both teams are assembled, select the "Go" button to start the match! If using a keyboard or switch device, you may need to move focus a few times until focus appears on the "Go" button.

Watch the two teams pull their way to a victory or to a draw.

Select the "Return" button to reset the match.