Transcript lecture11

Applications of Newton’s Laws (Examples)
Ropes and tension
Example 1: You tie a rope to a tree and you pull on the rope with a force of 100 N.
What is the tension in the rope?
The tension in the rope is the force that the rope “feels” across any section of it
(or that you would feel if you replaced a piece of the rope). Since you are pulling
with a force of 100 N, that is the tension in the rope.
Example 2: Two tug-of-war opponents each pull with a force of 100 N on
opposite ends of a rope. What is the tension in the rope?
This is literally the identical situation to the previous question. The tension is
not 200 N !! Whether the other end of the rope is pulled by a person, or pulled
by a tree, the tension in the rope is still 100 N !!
Example 3: You and a friend can each pull with a force of 20 N. If you want
to rip a rope in half, what is the best way?
1) You and your friend each pull on opposite ends of the rope
2) Tie the rope to a tree, and you both pull from the same end
3) It doesn’t matter -- both of the above are equivalent
Take advantage of the fact that the tree can pull with almost any force (until
it falls down, that is!). You and your friend should team up on one end, and
let the tree make the effort on the other end.
Example:
Given: m, M, f, F
T-? a-?
F-T = Ma
(1)
T-f = ma
(2)
F-f=(M+m)a
F f
a
M m
T
f
m
T=F-Ma
M
T F
(F  f )
M m
M
F
A) M=m
F f
a
;
2m
1
T  F  (F  f )  (F  f ) / 2
2
B) M>>m
F f
a
;
M
Tf
Example: Four blocks of mass m1=1kg, m2=2kg, m3=3kg, m4=4kg are on a frictionless horizontal surface as
shown on the figure below. The blocks are connected by ideal massless strings. A force F L=30N is applied to
the left block and is directed to the left. Another force F R=50N is applied to the right block, and is directed to
the right. What is the magnitude of the tension T in the string between m2 and m3.
T=?
m1=1kg
m2=2kg
m3=3kg
m4=4kg
FL=30N
FR=50N
1)
FR  Fl  (m1  m2  m3  m4 )a
a
2a)
Newton’s eq. for all masses
FR  Fl
50N  30N

 2m / s 2
(m1  m2  m3  m4 )
10kg
T  Fl  (m1  m2 )a
T  Fl  (m1  m2 )a  Fl 
Newton’s eq. for m1 and m2
(m1  m2 )
FR  Fl 
(m1  m2  m3  m4 )
3kg
(50N  30N )  36N
10kg
FR  T  (m3  m4 )a
T  30N 
2b)
T  FR  (m3  m4 )a  FR 
T  50N 
Newton’s eq. for m3 and m4
(m3  m4 )
FR  Fl 
(m1  m2  m3  m4 )
7kg
(50N  30N )  36N
10kg
Example: Sharing the weight.
In both of the cases depicted to the right, a 2-kg weight is supported by
two strings. In which case are the students holding the strings exerting a
force of greater magnitude?
1. Case A
A
2. Case B
3. Both the same
4. Need more
information
B
F1 +F2
F1
A
F2
W
F1 +F2
F2
F1
W
B
Example: Find the tension in the cables.
θ
θ
T
y
T
m
x
mg
x : T cos T cos  0
y:
2T sin  mg  0
T 
mg
2sin 
Small θ,
large T
It is impossible for a real cable (m > 0) to be completely horizontal
(it would require infinite tension, and then the cable snaps).
Pulleys
Atwood’s Machine
m1 g  T  m1 a
T  m2 g  m2 a
M=0
T
m1
m1 g
T
m1  m2 g  m1  m2 a 
m2
m2 g
2m1 m2 g
T  m2  g  a  
m1  m2
m1 g  T  m2 g

m1  m2 
a
g
m1  m2 
Example: In which case does block m experience a larger acceleration?
In (1) there is a 10 kg mass hanging from a rope and falling. In (2) a hand
is providing a constant downward force of 98 N. Assume massless ropes.
m
m
M=10kg
a
a
F = 98 N
Case (1)
Case (2)
In (2) the tension is T2 = 98 N due to the hand.
In (1) the tension T1 is less than 98 N because the block is accelerating down.
Only if the block were at rest would the tension be equal to 98 N.
Because of that in case 2 block m experience a larger acceleration.
T2  98N  Mg
Mg  T1  Ma 
T1  M ( g  a)  Mg  T2
Example: How much force does the worker have in order to support
the mass M at constant height h off the ground?
A. mg;
B. mg/2;
T2
T
T
C. 2mg;
D. 3mg;
T1  m g
T  T1 / 2  m g / 2
T2  2T  m g
T
T
T1
T1
mg
T
mg=6T
E. mg/3