Transcript Feb12-Forces

Feb 12 Discussion
•Force and Newton’s Second Law
problems
Designing an elevator
• Because of your physics background, you are asked
to choose the cable strength necessary to safely
support an elevator. The cab of the elevator has a
mass of 1200 kg and its passenger capacity is 800
kg. The maximum acceleration (up or down) of the
elevator is to be 2.0 m/s2.
• Hint: when two objects accelerate together, they can
be treated as a single combined mass from the point
of view of external forces and Newton’s Second Law.
Designing an elevator
t
a
w
• The cab of the elevator has a mass of 1200 kg and
its passenger capacity is 800 kg. The maximum
acceleration (up or down) of the elevator is to be 2.0
m/s2.
• The greatest tension occurs when the elevator is
accelerating upward. The cable must support the
total weight is the weight of the elevator plus the
passengers plus additional force to accelerate the
mtotala  t  w
total mass upward.
w  mtotalg
t  mtotala  mtotalg
mtotal  (1200kg)  (800kg) = 2000kg
 m
m 
t = (2000kg)2.0 2  9.8 2  23600N
 s
s 
Towing
• A 5000 kg truck can produce a maximum force on a
dry road so that the road pushes back hard enough
to accelerate the truck at 2.4 m/s2. The truck is to
tow a trailer with a mass of 10000 kg. What is the
maximum acceleration for the truck/trailer
combination?
• Hint: when two objects accelerate together, they can
be treated as a single combined mass from the point
of view of external forces and Newton’s Second Law.
Towing
• A 5000 kg truck can produce a maximum force on a
dry road so that the road pushes back hard enough
to accelerate the truck at 2.4 m/s2. The truck is to
tow a trailer with a mass of 10000 kg. What is the
maximum acceleration for the truck/trailer
combination?
 m 
Fmax  m amax  (5000kg)2.4 2  12000N
 s 
When combined,t he total mass wit h trailer: is
mtotal  (5000kg)  (10000kg)  15000kg
T he maximum acceleration with the trailer is
F
12000N
m
amax  max 
 0.8 2
mtotal 15000kg
s
Pulling a car out of the mud
• Your car is stuck in the mud. You have
a rope and a place to tie it securely to
the car. You try pulling on it to free the
car, but cannot pull hard enough to get
the car’s drive wheels out of the muddy
rut. You decide to see if taking physics
allows you to solve real-world
problems…
The set up
• First, of course, you get out some paper
and draw the free body diagram:
consider the balance of forces on the
midpoint of the rope.
The set up
• First, of course, you get out some paper
and draw the free body diagram:
consider the balance of forces on the
midpoint of the rope.
f
t 
t
How much force?
You decide to estimate the tension you will get. You realize
that it also depends on how hard you can pull against the
rope. Draw the free-body diagram for yourself.
What limits the force you can exert on the rope?
A) your arm strength
B) maximum static friction force from the
ground
C) the weight of the car
D) the strength of the tree
How much force?
You decide to estimate the tension you will get. You realize
that it also depends on how hard you can push against the
rope. Draw the Free-body diagram for yourself.
fs
What limits the force you can exert on the rope?
A)
B) maximum static friction force-that’s the 
only
other horizontal force acting on you
C)
D)

f
How much tension?
• Assuming that the maximum static friction force
is on the order of your weight (500-1000 N, for
most people), and the angle of the rope is
similar to the diagram (5-10 degrees), how
much tension can you produce?
A) 300 N
B) 3000 N
C) 30000 N
D) 300000 N
The weight of one ton is about 20000 N.
How much tension?
• Assuming that the maximum static friction force
is on the order of you weight, and the angle of
the rope is similar to the diagram (5-10
degrees), how much tension can you produce?
y
A)
f
B) 3000
x
N
t

C)
The forces balance in the x-direction due to the D)
symmetry. For the forces to balance in the y-direction,
 to satisfy: f  2t sin( )  0
the tension needs
 t 
f
500N

 3000N
2sin( ) 2sin(5)
Truck stopping a plane
• In a recent television ad for a pickup
truck, a cargo plane carrying the truck
lands, the truck rolls down a ramp out
the back of the plane onto the ground,
and we see that the truck and plane are
still connected by a chain. The truck’s
brakes to bring the plane and the truck
to rest. What should we be impressed
about here?
Forces on the truck
• What is the direction of the acceleration of the
truck? Consider the forces on the truck.
Draw the free body diagram.
• Which is greater,
A) the force of the road on the truck, or
B) the force of the chain on the truck, or
C) are they the same?
Forces on the truck
• What is the direction of the acceleration of the truck?
• Consider the forces on the truck. Draw the free body
diagram.
• Which is greater,
A) the force of the road on the truck, since the
acceleration is opposite from the velocity, the force
on the car from the road must be greater!
B)
v
C)
a
fs
t


Force on the plane
• Which is greater?
A) the force of the road on the truck or
B) the force of the chain on the plane, or
C) are they the same?
Force on the plane
• Which is greater?
A) the force of the road on the truck or
B)
C)
• Note that the force of the chain on the plane is the
same magnitude as the force of the chain on the
truck
• What’s impressive?
It’s not so much the amount of force involved, it’s the
fact that the truck’s brakes didn’t burn up that was
impressive.
Braking a plane
• Consider: a 3000 kg pickup stopping a 30000 kg
plane. The plane lands at 50 m/s and the braking
force from friction is no greater than 80% of the
weight of the pickup. How far do they travel together
while braking? (use g=10 m/s2)
Braking a plane
• Consider: a 3000 kg pickup stopping a 30000 kg
plane. The plane lands at 50 m/s and the braking
force from friction is no greater than 80% of the
weight of the pickup. How far do they travel together
while braking?
• The magnitude of the braking force is
fs=0.80(3000 kg)(10 m/s2)=24000 N
• Taking v0 as positive, the braking acceleration is
a=-fs/mtot=-(24000 N)/(33000 kg)=-0.73 m/s2
• The braking distance is given by
vf2=v02+2a Dx or
Dx=-v02/(2a)=-(50 m/s)2/(2(-0.73 m/s2))=1700 m!