MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
IC_W02D3-3 Blocks and Pulleys on a Table Solution
Two blocks 1 and 2 rest on a frictionless horizontal surface. They are connected by three
massless strings and two frictionless, massless pulleys as shown above. A force F is
applied to block 1. What is the resulting acceleration of block 1?
Solution:
We shall apply Newton’s Second Law to the blocks and pulleys. In addition we shall see
that we need to find the constraint condition that relates the acceleration of block 1 to the
acceleration of block 2. We begin by choosing a coordinate system. Choose an origin at
the left end of the support structure with positive x-direction to the right with unit vector
φ
i and coordinate functions x1 (t) and x2 (t) for each of the blocks. We denote the length
of the string connecting block 1 to the pulley A by s1 and the length of the string
connecting block 2 to the pulley B by s2 . We denote the distance from the origin to the
fixed block in the center by d .
Free Body Diagrams: We will only depict the horizontal forces on any object.
Two horizontal forces act on block 1, the pulling force F and the string force T A due to
the string connecting the pulley A to block 1.
Newton’s Second Law on block 1 is then
F  TA  m1
d 2 x1
dt 2
.
(1)
Two horizontal forces act on block 2, the tension T due to string that is wrapped around
the two pulleys and the tension TB due to the string connecting the pulley B to block 2.
Newton’s Second Law on block 2 is then
T  TB  m2
d 2 x2
dt 2
.
(2)
The force diagram on pulley A is shown below
Since pulley A is massless, Newton’s Second Law becomes
TA  2T  0 .
Similarly the force diagram on pulley B is
(3)
Since pulley B is also massless, Newton’s Second Law becomes
2T  TB  0 .
(4)
2T  TB  TA
(5)
Thus we have that
and the force equations for the blocks become
F  2T  m1
3T  m2
d 2 x1
(6)
dt 2
d 2 x2
(7)
dt 2
We have two equations and three unknowns, the tension T , and the two accelerations
d 2 x1 / dt 2 , and d 2 x2 / dt 2 . We need to find an additional condition relating the
accelerations. The string that wraps around the pulleys is constant in length, hence the
first and second derivatives of the length are zero
d 2l
 0 (constraint condition).
dt 2
(8)
We need to express the length l of the string in terms of x1 (t) and x2 (t) apply the
constraint condition.
d 2l
0.
dt 2
(9)
We shall examine each leg of the string and express the length of that leg in terms of the
quantities shown in our coordinate system. We first note that the distance between the
blocks is given by (x1  x2 ) . The upper leg has length
lupper  (x1  x2 )  s1 ,
(10)
lmiddle  (x1  x2 )  (s1  s2 ) ,
(11)
the middle leg has length
and the lower leg has length
llower  (d  x2 )  s2 .
(12)
Let the R denote the radius of the pulleys, then we must also include the wrap around
length
lwrap  2 R .
(13)
So the length of the string is
l  lupper  lmiddle  llower  lwrap
 (x1  x2 )  s1  (x1  x2 )  (s1  s2 )  (d  x2 )  s2  2 R
(14)
Now when we take two derivatives all constant terms vanish because the derivative of a
constant is zero. Hence
d 2 x1
d 2 x2
d 2l
0 2  2 2 3 2 .
dt
dt
dt
(15)
Therefore the accelerations satisfy
d 2 x1
2
3 d x2
.

2 dt 2
dt 2
(16)
We now solve Eqs. (6), (7), and (16) for the accelerations. We begin by substituting
d 2 x2
d 2 x1 3 d 2 x2
into Eq. (6) yielding
T  m2 2 / 3 and

2 dt 2
dt
dt 2
2
2
2 d x2
3 d x2
.
F  m2 2  m1
3
2 dt 2
dt
(17)
Thus
d 2 x2
dt
2

2F / 3
.
4
(m1  m2 )
9

F
.
4
(m1  m2 )
9
(18)
Finally
d 2 x1
dt
2
(19)