PHY 171 – General Physics I
HOMEWORK #4 – Solution
An Attwood machine contains two boxes with masses m1 and m2 connected by a
light cord passing over a light pulley, as in the figure. The system is released
from rest: the mass m1 moves upward and m2 downward, each one of them a
distance h with respect to the initial position A.
A
a
a) [3] Determine an expression for the acceleration of the system using the free
body diagrams and Newton’s 2nd law applied to the two boxes. (Recall the
tutorial that we solved in class.)
The free body diagrams are represented on the figure. For each mass we consider as
positive the respective direction of motion. Hence, Newton’s 2 nd law is for each mass:
m1:
T  m1 g  m1a
T  m2 g  m2 a
m2:
a
T
m1
T
m2
m1
m1g
m2g
Then, adding these equations term by term (or substituting T), we get
m2 g  m1 g  m1a  m2 a  a  g
B
h
h
m2
m2  m1
.
m2  m1
b) [2] Use the acceleration to derive an expression for the speed v of the system in position B. (Hint: Notice that each
of the two boxes moves linearly from rest a distance h with the constant acceleration derived above. You can
consider any of these two motions, since the speeds of the boxes are the same.)
The boxes move together, so we can use any of them to calculate the speed in position B. For instance, the first mass is lifted a
distance h starting from rest, such that:
v2  v02  2ah  v  2ah  v  2 gh
m2  m1
m2  m1
c) [4] Now, as an alternative method, use the conservation of mechanical energy to derive the same expression for the
speed of the system in position B.
Since there is no non-conservative force acting on the system (and the pulley is considered very light, so the energy necessary to
rotate it is negligible), the mechanical energy of the system is conserved:
EA  EB .
Taking the ground at the floor level and noticing that the masses are released from rest, we obtain:
EB  12 m1v 2  m1 g 2h  12 m2v 2
EA  m1 gh  m2 gh
for m1
for m2
for m1
for m2
where v is the common speed of the two masses. Therefore,
1
2
m1v2  m1 g 2h  12 m2v 2  m1 gh  m2 gh ,
from where we can extract a formula for v:
1
2
v2  m1  m2   gh  m2  m1   v  2 gh
m2  m1
,
m2  m1
which is the same as the expression obtained in part (b).
d) [1] Assume h = 1.2 m, m1 = 3.7 kg, m2 = 4.1 kg, and use the expression you found out above to calculate the
numerical value for the speed v.
v  2  9.8 m s 2  1.2 m 
4.1  3.7
 1.1 m s .
4.1  3.7