Upload
walker-craft
View
41
Download
0
Embed Size (px)
DESCRIPTION
Newton’s Laws Continued. Question: For each question, make sure to draw an accurate free body diagram!. What is the net force on a 15 kg box that is sitting on a table?. What is the net force on a 15 kg box that is being pulled to the left with 40 N of force and 28 N to the right?. - PowerPoint PPT Presentation
Citation preview
Newton’s Laws Continued
Question:For each question, make sure to draw an accurate free body diagram!
What is the net force on a 15 kg box that is sitting on a table?
What is the net force on a 15 kg box that is being pulled to the left with 40 N of force and 28 N to the right?
Question:For each question, make sure to draw an accurate free body diagram!
What is the net force on a 15 kg box on a frictionless surface that is moving at a constant velocity of 4 m/s to the right?
What is the net force on a 15 kg box on a frictionless surface that is accelerating to the right at a rate of 3 m/s2?
Can you do this one?What is the acceleration to the right of a 15 kg box sitting on a frictionless surface if you pull it from a rope at an angle of 30 degrees from the horizontal?
Ack! We don’t know how to do this!!!!! I know! We need…
TRIGONOMETRY!
TrigonometryTrig is a set of useful rules for finding various parts of a triangle.
Vector ComponentsAny vector can be broken up into two perpendicular components that lie along the x axis and along the y axis.
y c
ompo
nent
x component
-We call these the x and y components of a vector.
-You can think of an individual component of a vector as the shadow it would cast on either the x axis or the y axis.
Components
y c
ompo
nent
x component
y c
ompo
nent
al
so
Back to the original question:What is the acceleration to the right of a 15 kg box sitting on a frictionless surface if you pull it from a rope at an angle of 30 degrees from the horizontal?
Lets break it down into components:
Naming right trianglesWhen using trigonometricratios, once you pick anangle, you give newnames to the legs.
• For instance, if you chose , you now will renamethe sides.
Hypotenuse
adjacent
opp
osite
• If you choose β, you willrename the sides asshown.
β a
djac
ent
opposite
Hypotenuse
The story of Sohcahtoa
SOHSin Opposite Hypotenuse
CAHCosine Adjacent Hypotenuse
TOATangent Opposite Adjacent
Taking the inverse
How would I find the opposite side using the sin function?
(hyp)*sin() = opp
Back to the original question:What is the acceleration to the right of a 15 kg box sitting on a frictionless surface if you pull it from a rope at an angle of 30 degrees from the horizontal?
Now we can use trig to figure out what the components of the force are, and in tern, the answer to the question!
What if I want to find an angle?
20 cm
12 c
m
Here we know the opposite side and the hypotenuse.
So we will use the sin function
Sin-1(opp/hyp) =
- this is called the inverse sin function
= 36.87 degrees
Frictional ForceFriction is a force that opposes motion.
Draw a free body diagram of a 20 kg box that is being pushed to the right with a force of 50 N and there is a 30 N frictional force.
Types of frictionStatic Friction: The frictional force that exists between two surfaces when they are stationary
Kinetic Friction: The frictional force that exists between two surfaces when the objects are moving across each other (like a book sliding on a table)
Both of these types of friction can be calculated by using the coefficient of friction (either the coefficient of static friction or the coefficient of kinetic friction.)
Coefficient of FrictionA value that is different for any two surfaces.
- It is a constant that can be used to help calculate the force due to friction if you know the normal force of an object.
Ff = FN
Mu (like the pokemon)
is usually in between zero and one
Coefficient of FrictionThe coefficient of static friction tends to be greater than that of kinetic friction
Another way to think about what is, is how “sticky” the interaction between two surfaces is.
- The greater the , the greater the friction.
QuestionsIf the coefficient of friction between your tires and the road is 0.2, and the car has a mass of 1700 kg, what is the frictional force between the tires and the road?
QuestionsA car with a mass of 700 kg is breaking hard. If the car is traveling at an initial velocity of 15 m/s, and takes 3 seconds to stop, what is the coefficient of friction between the tires and the road? (draw a free body diagram)
Mu of the shoe
Here’s a doozyIf this 15 kg box is sliding down the hill at a constant velocity, what is the frictional force between the box and the incline?
30o
Thought questionWhy is it impossible to make a piece of string completely horizontal and straight if you hang a mass from the center?