Download Physics 8 — Wednesday, October 19, 2011

Physics 8 — Wednesday, October 19, 2011
Troublesome questions for HW4 (5 or more people got 0 or 1
points on them): 1, 14, 15, 16, 17, 18, 19. Yikes!
Troublesome HW4 questions
1. Two objects of inertias m1 and m2 start from rest and then
interact only with each other in some way.
(a) What is the ratio of their relative velocities at any time?
(b) What is the ratio of their kinetic energies at any time?
Troublesome HW4 questions
Problem 1
(a) Since the two objects interact only with each other, they form
an isolated system. So momentum is conserved. Since the two
objects start from rest (so momentum is initially zero), the
system’s momentum is zero at all times. Thus, at any time,
m1 v1x + m2 v2x = 0. So
m2
v1x
=−
v2x
m1
(b) The ratio of kinetic energies is
1
m1 v12
K1
m1
= 21
=
2
K2
m2
2 m2 v2
v1
v2
2
m1
=
m2
m2 2 m2
=
−
m1
m1
Troublesome HW4 questions
14. A worker pushes boxes in a factory. In each case decide which
force has the greater magnitude — the force exerted by the worker
on the box or the force exerted by the box on the worker.
(14a) The box is heavy and does not move no matter how hard she
pushes.
(A) Force exerted by worker on box is larger
(B) Force exerted by box on worker is larger
(C) The two forces have the same magnitude
Troublesome HW4 questions
14. A worker pushes boxes in a factory. In each case decide which
force has the greater magnitude — the force exerted by the worker
on the box or the force exerted by the box on the worker.
(14b) Some contents are removed, and now when pushed the box
slides across the floor at constant speed.
(A) Force exerted by worker on box is larger
(B) Force exerted by box on worker is larger
(C) The two forces have the same magnitude
Troublesome HW4 questions
14. A worker pushes boxes in a factory. In each case decide which
force has the greater magnitude — the force exerted by the worker
on the box or the force exerted by the box on the worker.
(14c) The worker pushes harder, and the box accelerates.
(A) Force exerted by worker on box is larger
(B) Force exerted by box on worker is larger
(C) The two forces have the same magnitude
Troublesome HW4 questions
Problem 14
In all three cases, the force exerted by the worker on the box is
equal in magnitude to the force exerted by the box on the worker.
Troublesome HW4 questions
15. A pitcher has thrown a fastball toward home plate.
(15a) When it is halfway to the plate, does the ball still feel the
pitcher’s push? Explain your answer.
(15b) What forces does the ball feel, if any?
(A) The ball still feels a force from the pitcher when it is halfway
to home plate. So the relevant forces are the pitch (forward),
gravity (downward), and air resistance (backward).
(B) Once the ball leaves the pitcher’s hand, it no longer feels a
force from the pitcher. So the relevant forces are gravity
(downward) and air resistance (backward).
(C) Neither answer is correct.
Troublesome HW4 questions
Problem 15
(a) The force between the ball and the pitcher is a contact force.
The ball no longer feels the pitcher’s push once the ball has left
the pitcher’s hand.
(b) The ball feels the downward force of the earth’s gravity and the
backward force of air resistance. It’s OK if you forgot to mention
air resistance, since we’ve neglected it so far in the course.
Troublesome HW4 questions
16. I push on a refrigerator (horizontally), but it does not move.
Explain how this can be.
(A) the downward force of gravity on the fridge is larger than the
force that I can exert on the fridge.
(B) the frictional force of the floor on the fridge exactly cancels
the force of my push on the fridge.
(C) the force I exert on the fridge is smaller than the force of
friction that the fridge exerts on me.
(D) the combined force of the fridge’s weight and the friction
between fridge and floor is larger than the force of my push.
(E) the force exerted by the fridge on me equals the force exerted
by me on the fridge, so the net force is zero.
(F) the fridge pushes back on me harder than I push on it.
(G) the fridge’s very large inertia causes it not to accelerate.
Troublesome HW4 questions
Problem 16
The refrigerator is pushed forward by me and is pushed backward
by its contact with the floor. If these two forces cancel (as they
will until I push hard enough to start the refrigerator sliding), then
the refrigerator will remain at zero velocity. (There is an
interesting fact about friction that it always takes a larger force to
start an object sliding initially than to keep it sliding once it is
moving. We’ll study this in some detail soon.)
17. You are in a stationary elevator, so that the contact force
exerted by the floor on you is equal in magnitude to the force of
gravity acting on you. When the elevator starts downward, which
force changes? What happens to its magnitude?
(A) The contact force exerted by the floor on me increases in
magnitude when the elevator starts downward.
(B) Both the contact force and the gravitational force must be
reduced in magnitude, so that they can remain equal and
opposite.
(C) Since the elevator is accelerating downward, the force I exert
on the elevator becomes larger.
(D) When the elevator starts downward, the force of gravity on me
does not change, but the upward contact force exerted by the
floor on me decreases in magnitude.
(E) When the elevator starts downward, the force of gravity
increases in magnitude.
Troublesome HW4 questions
Problem 17
The downward force of gravity on me does not change when the
elevator starts downward. The upward contact force between the
floor and my feet is reduced when the elevator starts downward.
Since the two forces no longer balance, my body accelerates
downward.
Troublesome HW4 questions
18. When you are standing motionless on the ground, your feet are
exerting a force on Earth. Why doesn’t Earth move away from you?
(A) Earth is so massive that even though I am pushing on it, its
acceleration is imperceptible.
(B) The ground is so firm that it pushes back up on me harder
than I push down on it.
(C) Earth pushes back on me just as hard as I push on Earth, so
there is no acceleration.
(D) Earth doesn’t move away from me because gravity is pulling
me down toward Earth.
(E) Gravity is pulling Earth up toward me with a force that exactly
cancels the downward force of my feet on the ground.
(F) Earth doesn’t move because Earth’s gravitational pull on me is
far greater than the push of my feet on the ground.
Troublesome HW4 questions
Problem 18
Since we want to know whether Earth is accelerating, let’s try to
figure out the forces acting on Earth.
Earth exerts a downward gravitational force on me. So I must
exert an upward gravitational force (of equal magnitude) on Earth.
My feet exert a downward force on Earth. That downward force on
Earth is balanced by the upward gravitational force that I exert on
Earth. So my standing on Earth does not cause Earth to
accelerate. (Whew!)
Troublesome HW4 questions
19. You push on a crate, and it starts to move but you don’t.
Draw a free-body diagram for you and one for the crate. Then use
the diagrams and Newton’s third law of motion to explain why the
crate moves but you don’t.
(A) The force I exert on the crate is larger than the force the crate
exerts on me.
(B) The crate’s force on me is equal and opposite to my force on
the crate. The frictional force between my shoes and the floor
is equal in magnitude to the crate’s push on me, while the
frictional force between the crate and the floor is smaller than
my push on the crate.
(C) The crate and I exert equal and opposite forces on each other,
but I don’t move because I am much more massive than the
crate.
Troublesome HW4 questions
Problem 19
If the crate and I were both standing on an ice rink, then it seems
clear that we would both start to move. If the crate and I were
both bolted to the floor, then it seems clear that neither one of us
would start to move. So the grip of the floor’s friction on my feet
must be greater in magnitude than the grip of the floor’s friction
on the crate.
Let’s say that I push to the right on the crate with a force
~ me,crate , so the crate pushes to the left on me with a force
F
~ crate,me = −F
~ me,crate . Meanwhile, the floor pushes to the right on
F
~ floor,me , and the floor pushes (by a smaller
me with a force F
~ floor,crate .
amount) to the left on the crate with a force F
~ floor,crate | < |F
~ floor,me |, because the bottom
It is reasonable that |F
of the crate is wood, while the soles of my shoes are rubber.
Troublesome HW4 questions
Problem 19
Chapter 10 (second half) — motion in a plane
Reading Question #1:
Can an object be accelerated without changing its kinetic energy?
Related question (part 1)
Can an object moving in only one dimension be accelerated
without changing its speed?
(A) Yes.
(B) No.
(C) Not instantaneously, no. But over a finite time interval you
could do something really tricky, like changing +v into −v .
Related question (part 2)
Can an object moving in two dimensions be accelerated without
changing its speed?
(A) Not instantaneously, no. (Like the last question.)
(B) Yes. Over a finite time interval, you can do something tricky
like +~v → − ~v . And even instantaneously, yes, if the
acceleration vector is perpendicular to the velocity vector.
(C) No, never. Period.
Chapter 10 (second half) — motion in a plane
Reading Question #1:
Can an object be accelerated without changing its kinetic energy?
Answer: Yes. You can change an object’s direction without
changing its speed. So its velocity can change without changing its
kinetic energy.
Over a finite time interval, this is easy to arrange.
Over an infinitessimal time interval, if the acceleration vector is
perpendicular to the velocity vector, then direction changes, but
speed does not. This will be important next week!
Reading Question #2
Suppose that I walk 3 miles east
and then 4 miles north. How far
am I from where I started? If you
wanted to tell a pilot to fly directly
from your starting point to your
ending point, what compass
direction would you tell her to fly?
(Tell her something like, “Your
heading should be such-and-such
degrees east of north.”)
Answer: I am 5 miles from where I
started. I ask the pilot to fly 37◦
east of north (or 53◦ north of east).
Summary of Chapter 10 results (vector math)
Various ways to write a vector:
~ = (Ax , Ay ) = Ax (1, 0) + Ay (0, 1) = Ax î + Ay ĵ
A
Can separate into two ⊥ vectors that add up to original, e.g.
~ x = Ax î,
A
~ y = Ay ĵ
A
~ =A
~x + A
~y
A
Scalar product (“dot product”) is a kind of multiplication that
accounts for how well the two vectors are aligned with each other:
~ ·B
~ = Ax Bx + Ay By = |A|
~ |B|
~ cos(θAB )
A
Summary of Chapter 10 results (friction)
Static friction and kinetic (sometimes called “sliding”) friction:
F Static ≤ µS F Normal
F Kinetic = µK F Normal
“normal” & “tangential” components are ⊥ to and k to surface
Summary of Chapter 10 results (work)
In one dimension, we learned
Z
W = Fx ∆x
→
Fx (x) dx
Sometimes the force is not parallel to the displacement. For
instance, what is the work done by gravity if you slide down a hill?
~ · ∆~r = Fx ∆x + Fy ∆y
W =F
which in the limit of many, many infinitessimal steps becomes
Z
Z
~
W = F (~r ) · d~r = (Fx (x, y ) dx + Fy (x, y ) dy )
Suppose mass = 1 kg.
∆x = +19 m
∆y = −13 m
∆~r = (+19 m, −13 m)
~ = (0, −9.8 N)
F
Sometimes the force is not parallel to the displacement. For
instance, what is the work done by gravity if you slide down a hill?
~ · ∆~r = Fx ∆x + Fy ∆y
W =F
W = (0 N) (+19 m) + (−9.8 N) (−13 m) ≈ 130 J
Same answer as W = mgh.
Let’s work through some examples using friction
How hard do you have to push a 1000 kg car (with brakes on, all
wheels, on level ground) to get it to start to slide? Let’s take
µS ≈ 1.5 for rubber on dry pavement.
Let’s work through some examples using friction
How hard do you have to push a 1000 kg car (with brakes on, all
wheels, on level ground) to get it to start to slide? Let’s take
µS ≈ 1.5 for rubber on dry pavement.
F Normal = mg = 9800 N
F Static ≤ µS F N = (1.5)(9800 N) ≈ 15000 N
Let’s work through some examples using friction
How hard do you have to push a 1000 kg car (with brakes on, all
wheels, on level ground) to get it to start to slide? Let’s take
µS ≈ 1.5 for rubber on dry pavement.
F Normal = mg = 9800 N
F Static ≤ µS F N = (1.5)(9800 N) ≈ 15000 N
So the static friction gives out (hence car starts to slide) when
your push exceeds 15000 N.
Let’s work through some examples using friction
How hard do you have to push a 1000 kg car (with brakes on, all
wheels, on level ground) to get it to start to slide? Let’s take
µS ≈ 1.5 for rubber on dry pavement.
F Normal = mg = 9800 N
F Static ≤ µS F N = (1.5)(9800 N) ≈ 15000 N
So the static friction gives out (hence car starts to slide) when
your push exceeds 15000 N.
How hard do you then have to push to keep the car sliding at
constant speed? Let’s take µK ≈ 0.8 for rubber on dry pavement.
Let’s work through some examples using friction
How hard do you have to push a 1000 kg car (with brakes on, all
wheels, on level ground) to get it to start to slide? Let’s take
µS ≈ 1.5 for rubber on dry pavement.
F Normal = mg = 9800 N
F Static ≤ µS F N = (1.5)(9800 N) ≈ 15000 N
So the static friction gives out (hence car starts to slide) when
your push exceeds 15000 N.
How hard do you then have to push to keep the car sliding at
constant speed? Let’s take µK ≈ 0.8 for rubber on dry pavement.
F Kinetic = µK F N = (0.8)(9800 N) ≈ 8000 N
How far does your car slide on dry, level pavement if you jam on
the brakes, from 60 mph (27 m/s)?
How far does your car slide on dry, level pavement if you jam on
the brakes, from 60 mph (27 m/s)?
F N = mg ,
F K = µK mg
a = −F K /m = −µK g = −(0.8)(9.8 m/s2 ) ≈ −8 m/s2
How far does your car slide on dry, level pavement if you jam on
the brakes, from 60 mph (27 m/s)?
F N = mg ,
F K = µK mg
a = −F K /m = −µK g = −(0.8)(9.8 m/s2 ) ≈ −8 m/s2
Constant force → constant acceleration from 27 m/s down to zero:
vf2 = vi2 + 2ax
x=
vi2
(27 m/s)2
=
≈ 45 m
−2a
2 × (8 m/s2 )
How far does your car slide on dry, level pavement if you jam on
the brakes, from 60 mph (27 m/s)?
F N = mg ,
F K = µK mg
a = −F K /m = −µK g = −(0.8)(9.8 m/s2 ) ≈ −8 m/s2
Constant force → constant acceleration from 27 m/s down to zero:
vf2 = vi2 + 2ax
x=
vi2
(27 m/s)2
=
≈ 45 m
−2a
2 × (8 m/s2 )
How much time elapses before you stop?
How far does your car slide on dry, level pavement if you jam on
the brakes, from 60 mph (27 m/s)?
F N = mg ,
F K = µK mg
a = −F K /m = −µK g = −(0.8)(9.8 m/s2 ) ≈ −8 m/s2
Constant force → constant acceleration from 27 m/s down to zero:
vf2 = vi2 + 2ax
x=
vi2
(27 m/s)2
=
≈ 45 m
−2a
2 × (8 m/s2 )
How much time elapses before you stop?
vf = vi + at
⇒ t=
27 m/s
= 3.4 s
8 m/s2
How does this change if you have anti-lock brakes (or good
reflexes) so that the tires never skid?
How does this change if you have anti-lock brakes (or good
reflexes) so that the tires never skid? Remember µS > µK . For
rubber on dry pavement, µS ≈ 1.5 (though there’s a wide range)
and µK ≈ 0.8. The best you can do is maximum static friction:
F S ≤ µS mg
a = −F S /m = −µS g = −(1.5)(9.8 m/s2 ) ≈ −15 m/s2
How does this change if you have anti-lock brakes (or good
reflexes) so that the tires never skid? Remember µS > µK . For
rubber on dry pavement, µS ≈ 1.5 (though there’s a wide range)
and µK ≈ 0.8. The best you can do is maximum static friction:
F S ≤ µS mg
a = −F S /m = −µS g = −(1.5)(9.8 m/s2 ) ≈ −15 m/s2
Constant force → constant acceleration from 27 m/s down to zero:
vf2 = vi2 + 2ax
x=
vi2
(27 m/s)2
=
≈ 24 m
−2a
2 × (15 m/s2 )
How much time elapses before you stop?
vf = vi + at
⇒ t=
27 m/s
= 1.8 s
15 m/s2
So you can stop in about half the time (and half the distance) if
you don’t let your tires skid. Or whatever µK /µS ratio is.
I
Steel on steel µK is about half that of rubber on concrete,
and much less than that of µS for rubber on concrete.
I
So a train can take a while to skid to a stop!
I
Even more so if the tracks are wet: µK ≈ 0.1
I
At µ = 0.1 on level ground: 360 m to stop from 60 mph.
I
At µ = 0.1 on 6◦ slope: not possible to stop.
Friction on inclined plane
Take x axis to be downhill, y axis to be upward ⊥ from surface.
~ gravity = −mg cos θ ĵ,
F
⊥
~ contact = +mg cos θ ĵ
F
⊥
~ gravity = +mg sin θ î
F
k
If block is not sliding then friction balances downhill gravity:
~ static = −mg sin θ î
F
k
Magnitude of “normal” force (“normal” is a synonym for
“perpendicular”) between surfaces is
F N = mg cos θ
Magnitude of static friction must be less than maximum:
F S ≤ µS F N = µS mg cos θ
Block begins sliding when downhill component of gravity equals
maximum magnitude of static friction . . .
Block begins sliding when downhill component of gravity equals
maximum magnitude of static friction:
µS mg cos θ = mg sin θ
µS =
mg sin θ
mg cos θ
µS = tan θ
A heavy crate has plastic skid
plates beneath it and a tilted
handle attached to one side.
Which requires a smaller force
to move the box?
(A) Pushing the crate is easier
than pulling.
(B) Pulling the crate is easier
than pushing.
(C) There is no difference.
Which path best represents the trajectory of a cantaloupe thrown
horizontally off a bridge? (What’s wrong with the other two?)