Download 6. APPLICATION OF NEWTON`S LAWS Concepts: 6.1 FRICTION

6. APPLICATION OF NEWTON’S LAWS
Concepts:
6.1 FRICTION
Friction is a force of resistance that opposes the relative motion
between two surfaces. At the atomic level, It is electromagnetic in
nature and acts between atoms and molecules in the contacting
surfaces. At a larger level, friction can be seen as the jaggedness of each
surface.
Sometimes, you want to reduce friction. Other times maximize it like
when you want floor tiles that are non slip.
Two kinds of Friction:
Kinetic vs. Static
KINETIC FRICTION
Kinetic is when a moving surfaces slides over another surface. Say you
are driving down the road and of course the road slows down your
spinning wheels a bit. Or you are sanding a board of wood and your
sandpaper is slowed down or resisted by the wood surface.
Let’s write kinetic friction as Fkf= Uk * N
The force of kinetic friction is independent of the velocity that one
surface is moving past the other. Strange but true. Do you see velocity
in the equation?
The force of kinetic friction is proportional to the normal force of the
object and proportional to its coefficient of static friction.
The force of kinetic friction is independent of the area of contact.
STATIC FRICTION
Static friction tries to stop the surfaces from moving past each other. It
is a holding force.
Static friction is usually stronger than kinetic.
Example. You are about to move a heavy box which is just sitting on the
floor. Velocity =0. Kinetic friction = 0. Static friction is zero because you
haven’t started pushing on it.
As you push on it, you feel the box push back with equal force. The box
resists. Shame on it. Static friction is the force pushing back. The box is
still not moving. You push harder, and the box pushes back with equal
force.
You push a little harder and the box begins moving. You have pushed
harder than the box’s force of static friction can go.
You have given it a force larger than it maximum static force. Velocity is
now increasing as the box accelerates due to your force. Now kinetic
friction appears because the box is moving, sliding, across the floor.
Equation for static friction is
Fsf = Usf * N
Measure it with spring scale and 1kg mass.
Notice velocity, nor area is in this equation so static friction does not
depend on either.
1. What are the types of friction?
2. Which one slows and which holds?
3. Which one is dependent of the surface area that is in contact with
the sliding objects?
4. Which one increases with velocity?
5. Which is usually bigger?
6. Which has to be overcome before the object will start moving?
7. If you try to push something that is stopped, which do you
encounter first?
8. A 100 kg refrigerator has a Usf of .5? How many Newtons of force
is the normal force on it (it sits on a horizontal floor)?
9. Draw the free body diagram. Be sure to include the force of
gravity times mass, also known as weight, the normal force which
has a direction perpendicular to the supporting floor.
10.
How many Newtons of force in the horizontal direction will
it take to overcome that?
11.
Let’s say you push on it with a force double the Fsf. How
many Newtons of force remains after you subtract the Fsf?
12.
You are pushing a new bathtub (not installed) with a force of
200 N to the west. Usf=.6 Ukf=.2 The bathtub has a mass of 200
kg. What does it weigh?
13.
What is the normal force?
14.
What is the Fsf=?
15.
If you give the tub a 3000 N push, how fast, if at all, will it
accelerate?
16.
Ok, let’s say the tub is accelerating as you keep pushing with
3000 N of force. Now there is no static force because it is moving.
But now there is a weaker Fkf and its coefficient is .3 What would
the new acceleration be?
6.2 STRINGS AND SPRINGS
Mastery Objective:
By the end of this lesson, be able to find the force a spring
exerts back towards its natural length.
You will use Hooke’s formula F=-kx where F is the restorative
force that restores the spring to its natural length. K is the
spring constant which tells you how much force the spring
exerts for a given distance stretched. The direction of the
force is always towards the zero point (natural state of the
spring)
Draw a spring, a weight, a distance stretched or compressed.
Are some springs firmer than others? For the stronger spring,
would you think the K would be greater or smaller than the
weak spring? Why?
For great distances, Hooke’s law does not work. The spring
breaks down and doesn’t pull or push.
Hooke’s law works on an ideal spring that has little or no
mass.
Springs exist in the suspension of your car, in old watches.
I do one, you do one game.
Standard Problems:
1 mine
A 2.00 kg object hangs motionless from a spring with a force
constant k of 250 N/m. How far is the spring stretched from its
natural state?
F=kx
Free body diagram.
A mass is attached to a spring that sits on a table.
Force diagram…
The k is 100 n/m and x is .01 m. How much is the Frestorative?
Fr = kx = (100 N/m )(.01 m) = 1 N
Yours:
Your spring has a k of 300 N. The stretch distance x is .1 m.
What is Fs?
Mine: If the stretch is twice as much , what happens to the
force?
Yours: if the stretch is half as much, has happens to the Fr?
Mine: If a spring has a k of 70 N/m and is compressed by .5 m,
what is the Fr?
Fr=(70N/m)(.5m)=35 N.
Mine: You hang a 2 kg weight by a spring with a k or 150 N/m.
Where does it drop down to?
Fg=Fr (a mass hanging from a spring is pulled downward by
gravity) but is also pulled upward by the spring. When those
two forces are equal, the mass will stop and be at it’slowest
point.
Fg=mg
Fr=kx so mg=kx or mg/k = x
3*9.81/150 = 1.96 m
Yours: same situation:
K=280 N/m x=?
Standard Students HW
1. A spring with k=309 n/m Compressed .7m. What is the
Fr now? If a 2 kg mass is attached, what is the
acceleration at .7 of compression?
2. A mass of 104 kg is hung from a vertical spring with a k of
2000 N/m and feels two equal and opposite forces.
Where will it hang down to?
Fgravity vs Fsping pulling up. They will be equal.
Fg=Fs or mg=kx
End of Lesson
STRINGS AND TENSION
We use strings and springs often to exert forces on different objects.
The tension in string is equal at all points. (at least in an ideal string that
weighs nothing).
If a box is hanging by a rope, then the tension at the point where it
holds the box is equal to the weight of the box.
Honor students must consider the weight of the rope. If a rope weighs
10 N. then at the point of the box, the tension is 10 more. Half up the
rope the tension is 5 more. At the top of the rope, tension is zero more
or just the weight of the box.
An ideal pulley of no weight will just change the direction of the ropes
tension.
An ideal pulley change the direction of the tension of the string but not
its magnitude.