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Lecture 5

PHYS 1110 Lecture Notes - Lecture 5: Acceleration, Centripetal Force, Centrifugal Force


Department
Physics
Course Code
PHYS 1110
Professor
Marino
Lecture
5

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9/28/2013 © University of Colorado at Boulder
Some Applications of Newton’s Laws.
In this chapter, we get some practice applying Newton’s laws to various physical problems. We
do not introduce any new laws of physics.
Solving Fnet = ma problems with multiple bodies
Problem: A mass m1 is pulled up a frictionless incline by a string over a pulley and a hanging
mass m2.
We know m1, m2, and the angle .
We seek :
T = tension in the cord,
a = acceleration of the mass,
N = normal force on m1
Step 1
Draw a free-body diagram for each moving object.
Label the force arrows with
the magnitudes of the forces.
Notice that T is the same for
both objects (by NIII).
Use m1, m2, in the diagrams,
not m.
Step 2
For each object, choose xy axes so that the acceleration vector a is in the (+) direction.
Notice that we can choose
different axes for the different
objects. And we can tilt the
axes if necessary.
m1
m2
no friction
T
N
m1g
T
m2g
T
N
T
m2g
+x
+y
a
+y
a

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9/28/2013 © University of Colorado at Boulder
Step 3 For each object, write the equations
x x y y
F m a , F m a

m1 :
11
1
x-eqn (1) T m gsin m a
y-eqn (2) N m gcos 0
 
 
m2 :
22
(3) m g T m a  
Notice that
aa
is the same for both m1 and m2 since they are connected by a string that
doesn't stretch.
Now we have a messy algebra problem with 3 equations in 3 unknowns:
11
1
22
(1) T m gsin m a
(2) N m gcos 0
(3) m g T m a
 
 
 
The unknowns are T , N , and a.
We can solve for N right away. Eqn (2)
1
N m gcos
Now we have 2 equations [(1) & (3)] in 2 unknowns ( a & T ).
Solve (1) for T and plug into (3) to get an equation without an T :
(1)
1 1 1
T m gsin m a m (gsin a)    
(3)
2 1 2
m g [m (gsin a)] m a  
T
N
T
m2g
x
y
+y
a
m1g
m1g cos
m1g sin
2
1
a

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9/28/2013 © University of Colorado at Boulder
Now solve this last equation for a:
Finally, if you have any strength left, we can solve for tension T by plugging our expression for
acceleration a back into either (1) or (3) and solving for T. From (3), we have
2 2 2
T m g m a m (g a)  
.
Plugging in our big expression for a, we get
21
22
12
m g m gsin
T m (g a) m g (m m )


 


We can simplify:
1 2 2 1 1 2 2 1
22
1 2 1 2 1 2
1 1 2 1
2
1 2 1 2
g(m m ) m g m gsin m g m g m g m gsin
T m m
(m m ) (m m ) (m m )
m g m gsin m m g
T m T (1 sin )
(m m ) (m m )
 
  
 
 
 
 


 



Do these expressions make sense? Let’s check some limits.
If m1 = 0, then m2 should be freely falling with a = g and the tension T should be zero. Check
that this is so.
If m2 = 0, then m1 should slide down the incline with acceleration a = g sin (since it would be
accelerating in the negative direction). Check that this is so.
2 1 1 2 1 2
21
12
m g m gsin m a m a a(m m )
m g m gsin
a(m m )
  

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