lecture 7 - courses | department of physics &...
TRANSCRIPT
Lecture 7
ForcesImportant note:
First Exam is next Tuesday, Feb. 6, 8:15-9:45 pm(see link on Canvas for locations)
Today’s Topics:
• Forces– The gravitational force– The normal force– Frictional Forces
• Next lecture…. Lots of examples
Newton’s Law of GravitationEvery particle exerts an attractive force on every other particle.
A particle is a piece of matter, small enough in size to be regarded as a mathematical point.
The force that each exerts on the other is directed along the line joining the particles.
221
rmmGF = 2211 kgmN10673.6 ×´= -G
Defining weightThe weight of an object on or above the earth is the gravitational force that the earth exerts on the object.
The weight always acts downwards, toward the center of the earth.
On or above another astronomical body, the weight is the gravitational force exerted on the object by that body.
2rmM
GW E=
mgW =
2rMGg E=
At the earth’s surface:
( )( )( )
2
26
242211
2
sm 80.9m 106.38kg 1098.5kgmN1067.6
=
´
´×´=
=
-
E
E
RMGg
a) Fg is greater on the featherb) Fg is greater on the stonec) Fg is zero on both due to vacuumd) Fg is equal on both alwayse) Fg is zero on both always
What can you say about
the force of gravity Fg
acting on a stone and a
feather on the moon?
The force of gravity (weight) depends on the mass of the object!! The stone has more mass, and therefore more weight.
ACT: Stone and feather I
a) it is greater on the featherb) it is greater on the stonec) it is zero on both due to vacuumd) it is equal on both alwayse) it is zero on both always
What can you say
about the acceleration
of gravity acting on the
stone and the feather?
The acceleration is given by F/m so here the mass divides out. Because we know that the force of gravity (weight) is mg, then we end up with acceleration g for both objects.
Follow-up: Which one hits the bottom first?
ACT: Stone and feather II
Defining the Normal forceThe normal force is one component of the force that a surfaceexerts on an object with which it is in contact – namely, thecomponent that is perpendicular to the surface.
Block on two tablesA block is balanced in the space between two tables as shown below. What forces are acting on the block?
Wby Earth
Nby right tableNby left table
DEMO: Nail bed
If the block rests on 100 mini-tables, each table exerts a relatively small force:
=each table 100WN
!F
net= 0
⇓
NR= N
L=W
2(½ with enough symmetry)
The acceleration of both masses together is the same in either case. But the contact force is the only force that accelerates m1
in case A (or m2 in case B). Because m1 is the larger mass, it requires the larger contact force to achieve the same acceleration.
If you push with force F on either the heavy box (m1) or the light box (m2), in which of the two cases below is the contact force between the two boxes larger?
a) case A
b) case B
c) same in both cases
F m2
m1
A
F m2
m1
B
ACT: Contact force
In case 1, the force F is pushing down (in
addition to mg), so the normal force
needs to be larger. In case 2, the force F
is pulling up, against gravity, so the
normal force is lessened.
Case 1
Case 2
Below you see two cases: a man pulling or pushing a sled with a force F that is applied at an angle 𝜃. In which case is the normal force greater?
a) case 1
b) case 2
c) it’s the same for both
d) depends on the magnitude of the force F
5) depends on the ice surface
ACT: Normal force
Inclined plane (no friction)
θgmW !!
=
NF!
mamgmgFN
==-
sin0cos
sincos
gamgFN
==
θmg
qsinmg
qcosmg
NF
a) case Ab) case Bc) both the same (N = mg)d) both the same (0 < N < mg)e) both the same (N = 0)
Consider two identical blocks, one resting on a flat surface (case A) and the other resting on an incline (case B). For which case is the normal force greater?
N
WWy
x
y
f
𝜃
𝜃
In case A, N = W.
In case B, due to the angle of
the incline, N < W. In fact,
N = W cos 𝜃.
ACT: Incline
Frictional Forces
When an object is in contact with a surface, the component of the contact force that is parallel to the surface and opposes motion is called the frictional force.
Static FrictionWhen the two surfaces are not sliding across one anotherthe friction is called static friction.
MAXss ff £
NsMAXs Ff µ=
Kinetic FrictionOnce there is motion:
Kinetic friction opposes the relative sliding motion motions that actually does occur.
fk = µkFNµk < µs
1
2
In case 1, the force F is pushing down(in addition to mg), so the normal force is increased. In case 2, the force F is pulling up, against gravity, so the normal force is lessened. The frictional force is proportional to the normal force.
a) pushing her from behindb) pulling her from the frontc) both are equivalent
Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this?
ACT: Going sledding
Inclined plane with friction
θgmW !!
=
NF!
θmg
qsinmg
qcosmg
NF
0sin
0cos
=-
=-MAXS
N
fmgmgF
Increasing 𝜃, when will the block start to slide down the plane?
a) component of the gravity force parallel to the plane increased
b) coefficient of static friction decreased
c) normal force exerted by the board decreased
d) both a) and c)e) all of a), b), and c)
A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why?
As the angle increases, the component of weight parallel to the plane increases and the component perpendicular to the plane decreases (and so does the normal force). Because friction depends on normal force, we see that the friction force gets smallerand the force pulling the box down the plane gets bigger.
Static frictionNormal
Weight
ACT: Sliding box
ExampleAn ice skater is gliding across the ice with an initial velocity of +6.3 m/s. The coefficient of kinetic friction between the ice and the skate blades is 0.081, and air resistance is negligible. How much time elapses before his velocity is reduced to +2.8 m/s?
v0 = +6.3 m/s v = +2.8 m/s Δx = ?a = ?Δt = ?
fk = -µkFN = -µkmg
From Newton’s Second Law:
ma = −µkmga = −µkg
v = v0 + aΔtΔt = (v −v0 )/( − µkg)
=(2.8 m/s − 6.3 m/s)/(– (0.081)(9.80 m/s2))
Δt = 4.4 s