Transcript Springs & Strings

Springs & Strings
Springs
Spring Force
A stretched or compressed spring exerts one of the most common
contact forces. A spring can either push (when compressed) or pull
(when stretched).
In either case, the tail of the vector force is attached to the contact
point.
There is no special symbol for the spring force, but we can use Fsp.
Hooke’s Law Springs
Hooke’s law for springs states that the force
increases linearly with the amount the spring is
stretched or compressed:
k spring constant
F  k x
units N/m
F vs x
5
F (N)
0
-0.3
0
-5
x (m)
0.3
Hooke’s Law Springs
Hooke’s law for springs states that the force increases linearly with the
amount the spring is stretched or compressed. The force is negative
because it always opposes the compression or extension of the spring.
F  k x
27•• IP The equilibrium length of a certain spring with a force constant
of k = 250 N/m is 0.18 m. (a) What force is required to stretch this
spring to twice its equilibrium length? (b) Is the force required to
compress the spring to half its length the same as in part (a)? Explain.
(a)
0.18 m
F = -k x
F = -(250 N/m)(0.18 m)
F = -45 N
x = 0.18 m
(b)
x = - 0.09 m
F = -(250 N/m)(- 0.09 m)
F = 22.5 N
Tension
The tension transmitted by a string, rope, cable, etc. is
the force exerted by the string at either end and along
its length.
Tension
The tension in a real rope will vary along its length, due
to the weight of the rope.
We will assume that all ropes,
strings, wires, etc. are
massless.
So… T1 = T2 = T3
Tension ACT
What is the reading in N on each the two spring scales shown above?
9.8 N
Pulleys & Tension
Ideal pulleys simply change
the direction of the tension.
There is no friction in the
pulley and the pulleys are
massless.
Tension – Connected Objects
When forces are exerted on connected objects, their
accelerations are the same.
In the situation below, if we know the force and the masses, we
can find the acceleration and the tension
In addition to the motion of the entire system, we can treat each box
as a separate system. Use the entire system to find the acceleration
and then use one of the boxes to find the tension.
Tension – Connected Objects
If there is a pulley, it is easiest to have the coordinate
system follow the string.
42.
••• Two buckets of sand hang from opposite ends of a rope that
passes over a pulley. One bucket is full and weighs 110 N; the other is
only partly filled and weighs 63 N. (a) Initially, you hold onto the lighter
bucket to keep it from moving. What is the tension in the rope? (b) You
release the bucket and the heavier one descends. What is the tension in
the rope now? (c) Eventually the heavier bucket lands and the two
buckets come to rest. What is the tension in the rope now?
(a) Tension will be 110 N
(b)
Examine entire system.
Fnet  110 N  63N  47 N
Fnet
a
msystem
msystem 
msystem
Fgsystem
g
110 N  63N

 17.65kg
N
9.8
kg
Fnet
47 N
a

 2.66 m 2
s
msystem 17.65kg
Now examine one of the individual masses.
110 N bucket:
110 N
m
 11.22kg
9.8 N
kg
Fnet  110 N  T

Fnet  ma  (11.22kg ) 2.66 m
T  110 N  Fnet
T  110N  29.86N  80.1N
 29.86 N

s
2
The other mass could have been analyzed
in order to reach the same value.
(c)
63 N
Translational Equilibrium
in translational equilibrium, the acceleration is 0 (the
velocity is constant
(6-5)
Tension ACT 2
A
B
In which situation is the tension on the rope larger?
a. A
b. B
c. Both tensions are the same
Equilibrium Tension
A 90 kg mountain
climber is suspended
from ropes as shown.
Rope 3 can sustain a
maximum tension of
1500 N before
breaking.
What is the smallest
that angle q can
become before the
rope breaks?
Equilibrium Tension
Climber:
 (F
)  TK on C  w  TK on C  mg  0
on M y
Knot:
 (F
 (F
)  T3  T1 cos q  0
on K x
)  T1 sin q  TC on K  0
on K y
TK on C  TC on K  mg
T1 cos q  T3
T1 sin q  mg
so tan q  mg / T3
qmax  tan 1 (mg / T3max )
 tan 1[(90 kg)(9.81 m/s 2 ) / (1500 N)]  30.5