ic_sol_w02d3-0
TRANSCRIPT
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a) The length of string A is given by
1A Pl y y R!= + + (1)
where R! is the arclength that the rope is in contact with the pulley. This length is
constant, and so the second derivative with respect to time is zero,
2 2 2
1
,1 ,2 2 20 A P
y y P
d l d y d ya a
dt dt dt = = + = + . (2)
Thus block 1 and the moving pulleys components of acceleration are equal in magnitude
but opposite in sign,
, ,1y P ya a=! . (3)
The length of stringB
is given by
3 2 3 2( ) ( ) 2B P P Pl y y y y R y y y R! != " + " + = + " + (4)
where R! is the arclength that the rope is in contact with the pulley. This length is also
constant so the second derivative with respect to time is zero,
2 2 2 2
2 3
, 2 ,3 ,2 2 2 20 2 2B P
y y y P
d l d y d y d ya a a
dt dt dt dt = = + ! = + ! . (5)
We can substitute Equation (3) for the pulley acceleration into Equation (5) yielding theconstraint relationbetween the components of the acceleration of the three blocks,
, 2 ,3 ,10 2
y y ya a a= + + . (6)
b) Free Body Force diagrams:
The forces acting on block 1 are: the gravitational force1m g
!
and the pulling force 1,rT!
of
the string acting on the block 1. Since the string is assumed to be massless and the pulley
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is assumed to be massless and frictionless, the tensionA
T in the string is uniform and
equal in magnitude to the pulling force of the string on the block. The free body diagramis shown below.
Newtons Second Law applied to block 1 is then
1 1 ,1 :
A ym g T m a! =j . (7)
The forces on the block 2 are the gravitational force 2m g!
and the string holding the
block, 2, rT!
, with magnitudeBT . The free body diagram for the forces acting on block 2
are shown below.
Newtons second Law applied to block 2 is
2 2 , 2 :
B ym g T m a! =j . (8)
The forces on the block 3 are the gravitational force3m g
!
and the string holding the
block,3,rT
!
, with magnitudeBT . The free body diagram for the forces acting on block 3
are shown below.
Newtons second Law applied to block 3 is
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3 3 ,3
:B y
m g T m a! =j . (9)
The forces on the moving pulley are the gravitational forceP
m =g 0!
!
(the pulley is
assumed massless); stringB
pulls down on the pulley on each side with a force, ,P BT
!
,which has magnitude
BT . String A holds the pulley up with a force
,P AT
!
with the
magnitudeA
T equal to the tension in string A . The free body diagram for the forces
acting on the moving pulley are shown below.
Newtons second Law applied to the pulley is
, : 2 0
B A P y PT T m a! = =j . (10)
Since the pulley is massless we can use this last equation to determine the condition thatthe tension in the two strings must satisfy,
2B A
T T= (11)
We are now in position to determine the accelerations of the blocks and the tension in thetwo strings. We record the relevant equations as a summary.
, 2 ,3 ,10 2
y y ya a a= + + (12)
1 1 ,1A ym g T m a! = (13)
2 2 , 2B ym g T m a! = (14)
3 3 ,3B ym g T m a! = (15)
2B A
T T= . (16)
Optional
d) Solve for the accelerations of the objects and the tensions in the ropes.
There are five equations with five unknowns, so we can solve this system. We shall first
use Equation (16) to eliminate the tensionA
T in Equation (13), yielding
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1 1 ,1
2B y
m g T m a! = . (17)
We now solve Equations (14), (15) and (17) for the accelerations,
,22
B
y
T
a g m= !
(18)
,3
3
By
Ta g
m= ! (19)
,1
1
2B
y
Ta g
m= ! . (20)
We now substitute these results for the accelerations into the constraint equation,Equation (12),
2 3 1 2 3 1
4 1 1 40 2 4
B B BB
T T Tg g g g T
m m m m m m
! "= # + # + # = # + +$ %
& '. (21)
We can now solve this last equation for the tension in string B ,
1 2 3
1 3 1 2 2 3
2 3 1
4 4
41 1 4B
g g m m mT
m m m m m m
m m m
= =
! " + ++ +# $
% &
. (22)
From Equation (16), the tension in stringAis
1 2 3
1 3 1 2 2 3
82
4A B
g m m mT T
m m m m m m= =
+ +
. (23)
We find the acceleration of block 1 from Equation (20), using Equation (22) for the
tension in stringB,
2 3 1 3 1 2 2 3
,1
1 1 3 1 2 2 3 1 3 1 2 2 3
2 8 4
4 4
By
T g m m m m m m m ma g g g
m m m m m m m m m m m m m
+ != ! = ! =
+ + + +
. (24)
We find the acceleration of block 2 from Equation (18), using Equation (22) for the
tension in stringB,
1 3 1 3 1 2 2 3
,2
2 1 3 1 2 2 3 1 3 1 2 2 3
4 3 4
4 4
By
T g m m m m m m m ma g g g
m m m m m m m m m m m m m
! + += ! = ! =
+ + + +
. (25)
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Similarly, we find the acceleration of block 3 from Equation (19), using Equation (22) forthe tension in stringB,
1 3 1 2 2 31 2
,3
3 1 3 1 2 2 3 1 3 1 2 2 3
3 44
4 4
By
m m m m m mT g m ma g g g
m m m m m m m m m m m m m
! +
= ! = ! =
+ + + +
. (26)
As a check on our algebra we note that
1, 2, 3,
1 3 1 2 2 3 1 3 1 2 2 3 1 3 1 2 2 3
1 3 1 2 2 3 1 3 1 2 2 3 1 3 1 2 2 3
2
4 3 4 3 42
4 4 4
0.
y y ya a a
m m m m m m m m m m m m m m m m m mg g g
m m m m m m m m m m m m m m m m m m
+ + =
+ ! ! + + ! ++ +
+ + + + + +
=