MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
IC_W02D3-2 Group Problem Two Block and Two Pulleys
Block 1 moves on the surface of a horizontal table. The static and kinetic coffecients of
friction between the block and table are equal and have the value µ . A massless string
of fixed length is attached at one end to block 1, extends horizontally to a pulley, drops
vertically to a second pulley, and finally rises vertically to the ceiling where the second
end is attached. The two pulleys are massless and have no friction. Block 2 is suspended
a fixed distance below the movable pulley. The system is released from rest. Find an
expression for the downward acceleration of block 2 in terms of some or all of the
parameters m1 , m2 , µ and the acceleration of gravity g . Explain any situation in which
your expression may not represent the true motion of the system.
Solution:
We will first define a convenient coordinate system. We will then make a free body
diagram for each of the two bodies. From these, we will find two equations linking three
unknowns: the accelerations of each block and the tension in the string. We will find a
third, constitutive (constraint) equation linking the two accelerations. We will then solve
the three equations in three unknowns for the acceleration of block 2. Finally we will
study the behavior of this result as the masses of the blocks change.
Note that if y increases (block 2 falls), x also increases but not necessarily at the same
rate.
Block 1:
There is no vertical acceleration so N = m1 g . In the horizontal direction
T ! f = m1ax1
T ! µ N = m1ax1
T ! µ m1 g = m1ax1 (1)
T = µ m1 g + m1ax1
Block 2 plus the attached pulley:
m2 g ! 2T = m2 a y 2 (2)
Constraint: The length, call it L, of the string is fixed. Let R be the pulley radius.
L = (D1 ! x) + " R / 2 + y + " R + (D2 + y)
After two derivatives with respect to time and simple rearrangement this gives
0 = !ax1 + 2a y 2
Thus
ax1 = 2a y 2 (3)
Use(1) for T and substitute the result into (2), then use (3).
m2 g ! 2( µ m1 g + m1 2a y 2 ) = m2 a y 2
ay 2 =
(m2 ! 2 µ m1 )
g
(4m1 + m2 )
When m1 = 0 block 2 is in free fall with an acceleration g , which makes sense. This result
is positive as long as m2 > 2 µ m1 . When m2 = 2 µ m1 the tension in the string just balances
the friction force. The displayed answer is incorrect for m2 < 2 µ m1 in that it has block 1
pulling block 2 upward, which is unphysical.