Download An object placed on a tilted surface will often slide

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Coriolis force wikipedia , lookup

Jerk (physics) wikipedia , lookup

Rolling resistance wikipedia , lookup

Rotating locomotion in living systems wikipedia , lookup

Fictitious force wikipedia , lookup

Frictional contact mechanics wikipedia , lookup

Centrifugal force wikipedia , lookup

Classical central-force problem wikipedia , lookup

Buoyancy wikipedia , lookup

Mass versus weight wikipedia , lookup

Newton's laws of motion wikipedia , lookup

Gravity wikipedia , lookup

Force wikipedia , lookup

G-force wikipedia , lookup

Centripetal force wikipedia , lookup

Friction wikipedia , lookup

Transcript
Ramps with and without friction
Questions for practice a link at the end
An object placed on a tilted surface will often slide down the surface. The rate at which the object slides down the surface is dependent upon how tilted the surface is; the greater the tilt of the surface, the faster the rate at which the object will slide down it. In physics, a tilted surface is called an inclined plane.
1
Here is a picture of an object on a ramp
The force of gravity (also known as weight) acts in a downward direction; yet the normal force acts in a direction perpendicular to the surface The normal force is tricky to visualize because we are so
used to seeing the normal force straight up! Now we must
extend our thinking
2
The task of determining the net force acting upon an object on an inclined plane needs components
Fn
gx
gy
Thus you can see that without any friction the only net force is the force going down the ramp which is Fgx
F = F
net
gx
ma = mgsin
a = gsin
3
Now suppose there is friction
gx
gy
How could we find the acceleration of the block now?
4
Case 1: Sliding down a ramp with friction
What is this?
click
k=0.2
m
Find the acceleration of the block
= 30
Fn
F
f
This is the Free Body Diagram
Fg
Fn
F
f
Fgx
30
F
=
net
Fgx ­ F
f
and Fn = Fgy
Fg
Fgy
5
Case 2: Sliding up a ramp with Friction
3 m
= 0.2
/s
How long does it take block to stop?
2kg
o
40
Here is the Free Body Diagram
Fn
Now put in the components
Fn
Ff
Ff
Hint
Pull
Fg
Fgx
Fgy
40
Fg
Can you write the equations
F
net =
F + F
gx
f
and
Fn = Fgy
6
With your partner
A loonie with a mass of 7.0 g is at rest on an incline of 25o. Calculate the force of
static friction acting on the loonie
loonie
25o
7
One more to try with your partner
A 55.0 kg skier is accelerating down an incline of 35o. The magnitude of the skier's acceleration is 4.41 m/s2. Calculate the coefficient of kinetic friction
that the snow exerts on the skis.
8
I lied one more
end of day 1
Calculate:
a) The acceleration of the block
b) How far up the ramp the block will travel
c) The time it takes the block to return to the bottom of the ramp
s
v =
m/
5
1
= 0.15
k
9
There are two types of friction, and friction depends upon the surfaces involved
Ice has little friction Pushing a wagon on the beach has lots of friction!
Friction
Ever tried to push a car?
Hard to get it started but then it gets easier!
because of this
μ
How do we calculate the force of friction from the coefficient of friction, (meu)
Look in your text on Page 182­183
Ff = μFn
μs μ k
The question will specify which or If we want force required just to start something from moving or to keep something from moving always use
μs
back to question
click
10
Start of Day 2
Really just a few more examples and opportunity for us to practice more FBD
A sled with waxed hickory runners rests on a horizontal, dry snowy surface. Calculate the mass
of the sled if an applied force of 46 N causes the sled to start moving (not accelerate)
11
Example #2
A 50 kg toboggan is on a snowy hill. If the hill forms an angle of 20o with the horizontal,
the tobaggan just begins to slide downhill. Calculate the coefficient of static friction for the
toboggan and the snow
20o
12
Skidding vs Non­Skidding Tires
When tires on a vehicle lock, (no longer rotate), the tires skid on the road surface
therefore the road exerts a force of kinetic friction on the car bringing it to stop (eventuall)
If the tires were alloowed to continue to rotate, ie antilock brakes, the road now exerts
a force of static friction on the car which should bring the car to a stop faster.
hmm, if wheels lock and skid I must use μk
hmm, if wheels continue to rotate I must use μs
Example #3
A 1640 lift truck with rubber tires is skidding on wet concrete with all four wheels locked.
Calculate the acceleration of the truck
13
Example #4
A person wants to drag a 40 kg box up a hill that forms a 25o. The coefficient of kinetic friction for the sled on the snow is 0.04. Calculate the force on the rope in order to accelerate the crate at 2.5 m/s2
40 kg
25o
14
Time for you to practice!
Page 185 1, 2
Page 186 1, 2
Page 188 1, 2
Page 189 1
Here is a nice review package
to go along with the textbook review
Click the pic
Page 190 1 ­ 11
15
Attachments
Physics 20 Blocks Sliding down ramps Day 1.doc
Force Reveiw Questions.doc