Chapter 5: 5.1 Vectors Remember: A Vector quantity has both _____________________ and _____________________.

A Vector quantity is represented by an _________________. o The length of the line represents the _______________.

The longer the line, the greater the ________________. o The direction of the arrow indicates the

_________________ of the quantity. o A Resultant is a vector quantity that represents the

_____________ of several vector quantities acting at once.

Vector Addition in One Dimension:

Vectors in the same direction are ________________, and the resultant is in the direction of both vectors.

+ = 20 m 10m _____ m

Vectors in the opposite direction are _______________ and the resultant is in the direction of the _____________ vector.

+ = 10 m 20m ______ m

Measuring Vectors

Directions: For each of the following examples, a) measure the vector in (cm), b) use factor label to find the quantity, c) measure the angle in reference to the x-axis, d) record the magnitude and direction of the vector.

Scale: 1 cm = 5 m/s 1) 2) 3) 4) 5) 6) 7) 8)

Vectors: Components and Resultants Vector Components Every vector can be broken down into components: 1) ___ - component East (+) or _______ (-) 2) ___ - component _______ (+) or South (-) These components represent how much of the given quantity

is directed in the ____________ (x component) and ____________ (y component) directions.

These x & y components are drawn ___________________

(90 degrees) to each other. Graphical Method: 1) Draw a vector (magnitude and direction) to a

suitable scale. 2) From the tip of the arrow, draw a vertical line

up/down. This line (magnitude and direction) represents the ____________. 3) From the tail of the arrow, draw a horizontal line (right/left), on the reference axis, to the tail of the y- component. This line (magnitude and direction) represents the ____________.

Example: Scale: 1 cm = 20 m/s x – component y- component

Drawing Vectors Directions:

1. Create a suitable scale 2. Create a horizontal reference line parallel to the bottom of the page 3. Create a starting point (position) 4. Measure the angle using the x-axis as your reference 5. Draw a vector extending from the starting point at the appropriate angle 6. Vector length should measured according to scale. 7. Label the vector: Magnitude = length, Direction = angle

Practice: 1. v = 20 m/s @ 45 degrees N of E

2. F = 400 N @ 30 degrees N of W

3. a = 5 m/s2 @ 65 degrees S of E

4. D = 450 m @ 20 degrees S of W

Vector Addition in Two Dimensions:

Tip-to-Tail Method:

1. Make sure vectors are all to scale. (Ex: 1 cm = 1km, 1in = 25m/s, 1cm = 5m/s2)

2. Draw your first vector (to scale in magnitude) with respect to the x-axis. 3. Measure the given angle (from a parallel x-axis) and draw the second vector. 4. The second vector’s tail must start at the tip of the previous vector. 5. If there are multiple vectors involved, you must continue this process of adding vectors from tip-to-tail. 6. After all given vectors are graphed, you extend the resultant vector from the tail (starting point) of the first vector graphed, to the tip (end point) of the last vector. 7. Measure the length of the resultant. 8. Using the scale you created, calculate the magnitude of the resultant vector. 9. Measure the resultant angle with respect to the first x-axis you used (in measuring the angle of the first vector graphed). 10. Write down the resultant vector (its magnitude and direction).

Adding Vectors

Directions: Use the tip-to-tail method in order to add the following vectors. Measure the resultant vector. 1) Add 40 km East to 60 km East 2) Add 40 km East to 60 km West 3) Add 40 km East to 60 km North 4) Add 5 m/s West to 10 m/s South

Sect. 5.2 Equilibrium The net force on an object is _________________of all the forces acting on it. When the net force on an object is zero, we say the object is in _________________. If the net force is zero (equilibrium): 2 different scenarios’ can happen.

1. object at rest will stay at rest or 2. object in motion will stay in motion with constant speed and direction.

__________________ says the acceleration of an object in equilibrium is zero because the net force acting on the object is zero. ____________________means neither the speed nor the direction of motion can change. The______________________ In mathematics, it means perpendicular. The ___________________ force the table exerts is perpendicular to the table’s surface. _________________________ explains why normal forces exist.

The book pushes down on the table, so the table pushes up on the book. The book’s force on the table is the ________________, and the table’s force on the book is the ____________________. The third law says that these forces are equal in strength. If the book is at rest, these forces must be ______________ but ____________ in direction. If the book were heavier, it would exert a stronger downward force on the table. The table would then exert a stronger upward force on the book.

In equilibrium, the net force in each direction must be ____________. That means the total force in the __________________ must be zero and the total force in the __________________ also must be zero. Getting the forces in each direction to cancel separately is easiest to do when all forces are expressed in x-y components. 1. Two chains are used to lift a small boat weighing 1500 newtons. As the boat moves upward at a constant speed, one chain pulls up on the boat with a force of 600 newtons. What is the force exerted by the other chain? 2. A heavy box weighing 1000 N sits on the floor. You lift upward on the box with a force of 450 N , but the box does not move. What is the normal force on the box while you are lifting? 3. A 40 N cat stands on a chair. If the normal force on each of the cat’s back feet is 12 N, what is the normal force on each front foot? (You can assume it is the same on each.)

Free Body Diagrams 1. 5 N Fnet = _____________________ 2. 3 N 20 N Fnet = ______________________ 3. 100 N 250 N Fnet = _____________________ 4. What should force x and y be equal to in order to have this object in equilibrium? 23 N x = ____________________ 7 N x y = ____________________ Or _____________________ y 5. A 25 N force acting 30 degrees S of W acts on an object. At this same time, another 75 N force is acting on the same object at 30 degrees N of E. What is the net force acting on the object? (Draw it First)

Section 5.3 Friction Friction (Ff) = any force that ______________ the motion of substances that are in contact with each other. Friction acts ______________ to the surfaces in contact and in the __________________ direction of motion. Coefficient of Friction [(u) pronounced mu] depends on the nature of the two materials and the smoothness of the surfaces in contact. Amount of friction depends on ________________ and __________ of motion. Static Friction = (Ffs) starting friction; Kinetic Friction = (Ffk) sliding friction; Coefficient of Friction (µ) = the ratio of the _______________ _____________ over the _____________ _______________ pressing the surfaces together; constant for certain materials.

µ= Ff/Fn Ff = µFn FN

Ff FA

Fw = mg Rich slides a crate over a factory floor. The force of friction acting between the crate and the floor is 10 N. a) What force is required to initiate this objects motion? b) What force is required to maintain constant velocity of the crate? c) What force must Rich apply to make this crate in equilibrium?

d) What is the acceleration of the crate when this object is in equilibrium?

Force Friction Problems 1. A 200 N force is exerted on a 25 kg object.

a) If there is a constant 100 N frictional force, what is the acceleration of the object? b) If the object is initially at rest, and accelerates for 5 s, what is the final velocity of the object? c) What displacement does the object cover in this time?

2. A 650 N force acts on a 50 kg object. If there is a 150 N frictional force acting on the car, what is the car’s acceleration? b) What displacement does it cover if it starts from rest and accelerates for 7.5 s? 3. A 1500 kg car starts from rest and reaches a speed of 75 m/s in 6.5 s. a) Calculate the force acting on the car. b) What resistive force would the car have to encounter in order to travel at a constant velocity?

Do you think you have it? Try this!! A 50 kg crate rests on a wooden floor.

a. How much does the object weigh?

b. What is the normal force acting on the object and the floor?

c. If the coefficient of friction is 0.5, what is the force of friction between the crate and the floor?

d. What force is needed to initiate motion of the crate?

e. If you apply a force of 250 N, what is the net force acting on the object?

f. What is the acceleration of the box?

g. If the box is initially at rest, and travels for 4 s, what will be the final velocity?

h. What displacement will the box cover in this time period?

5.4 Torque ____________ is a new action created by forces that are applied off-center to an object. Torque is what causes objects to ____________________. Torque is the rotational equivalent of force. If force is a push or pull, you should think of torque as a _______________. The line about which an object turns is its ________________________. Some objects have a fixed axis: a door’s axis is fixed at the hinges. A wheel on a bicycle is fixed at the axle in the center. Other objects do not have a fixed axis. The _____________________of a tumbling gymnast depends on her body position. Torque is created whenever the __________________________ of a force does not pass through the axis of rotation. The line of action is an imaginary line in the direction of the force and passing through the point where the force is applied. If the line of action passes through the axis the torque is___________, no matter how strong a force is used! A force creates more torque when its line of action is far from an object’s axis of rotation. ____________________are positioned far from the hinges to provide the greatest amount of torque. A force applied to the knob will easily open a door because the line of action of the force is the width of the door away from the hinges. The same force applied to the hinge side of the door does nothing because the line of action passes through the axis of rotation. The first force creates torque while the second does not.

Torque Problems: 1. A force of 50 N is applied to a wrench that is 0.30 m long. Calculate the torque if the force is applied perpendicular to the wrench. 2. You apply a force of 10 N to a doorknob that is 0.80 m away from the edge of the door on the hinges. If the direction of your force is straight into the door, what torque do you create? 3. Calculate the net torque in diagram A (at right). 4. Calculate the net force and the net torque in diagram B (at right).