Strings, Pulleys, and Gravity
There are many situations where the gravitational force
acting on one object indirectly produces motion in another
object. In some cases, a failing object is connected to another
object by a light string or rope passing over a nearly
frictionless pulley. To understand this type of motion, we
make the following assumptions about strings and pulleys:
 Strings are considered to have negligible mass and are
capable of exerting only "pulling" forces on objects to
which they are attached (tension forces).
 Strings transmit forces undiminished: the tension force in
a string is the same throughout its length.
 A frictionless pulley changes the direction of a string
without diminishing its tension.
 Strings are assumed not to stretch.
Sample problems:
1.
A 2.0 kg mass, placed on a smooth, level table, is attached by a light string
passing over a frictionless pulley to a 5.0 kg mass hanging freely over the edge of the
table, as illustrated. Calculate the tension in the string, and the acceleration of the
2.0 kg mass.
FBD for the mass on the table (m2 = 2 kg)
FNET  T  FG  FN
FNET  T
FNET  T
m2  a  T
(1)
FBD for the suspended mass (m1 = 5 kg)
FNET  FG  T
FNET  FG  T
m1  a  m1  g  T
(2)
two equations combined :
m2  a  T
(1)
m1  a  m1  g  T
(2)
m2  a  m1  a  T  m1  g  T
 m1  m2   a  m1  g
m
5 kg  9.8 2
s
a 

 5 kg  2 kg 
form (1)
check (2)
49 N
7 kg

m
7 2
s
m
T  m2  a  2kg  7 2  14 N
s
m1  a  m1  g  T
m
m
5kg  7 2  5kg  9.8 2  14 N
s
s
2. Two spheres of masses 1.5 kg and 3.0 kg are tied together by a light string looped
over a frictionless pulley. They are allowed to hang freely. What will be the
acceleration of each mass and tension in the string?
FBD for the small suspended mass (m1 = 1.5 kg)
FNET  T  FG
FNET  T  FG
m1  a  T  m1  g
(1)
FBD for the big suspended mass (m2 = 3 kg)
FNET  FG  T
FNET  FG  T
m2  a  m2  g  T
(2)
two equations combined :
m1  a  T  m1  g
(1)
m2  a  m2  g  T
(2)
m2  a  m1  a  m2  g  m1  g
 m1  m2   a  m2  g  m1  g
a 
m
m
3 kg  9.8 2  1.5 kg  9.8 2
s
s
1.5 kg  3 kg 
a 
29.4 N  14.7 N
4.5 kg
form (1)

m
3.27 2
s
m1  a  T  m1  g
m
m
1.5kg  3.27 2  T  1.5kg  9.8 2
s
s
T  14.7 N  4.905 N  19.61 N
form (2)
m2  a  m2  g  T
m
m
3kg  3.27 2  3kg  9.8 2  T
s
s
T  29.4 N  9.81 N  19.59 N
3. For the following system of three interconnected masses, draw a free-body
diagram for each mass. Determine the rate at which the masses accelerate and
the tension(s) in both strings.
FBD for the big suspended mass (m1 = 300 g)
FNET  FG  T1
FNET  FG  T1
m1  a  m1  g  T1
(1)
FBD for the big suspended mass (m2 = 200 g)
FNET  T1  T2  FG  FN
FNET  T1  T2
FNET  T1  T2
m2  a  T1  T2
(2)
FBD for the big suspended mass (m3 = 100 g)
FNET  T2  FG
FNET  T2  FG
m3  a  T2  m3  g
(3)
three equations combined :
m1  a  m1  g  T1
(1)
m2  a  T1  T2
(2)
m3  a  T2  m3  g
(3)
m1  a  m1  g  T1
(1)
m2  a  T1  T2
(2)
m3  a  T2  m3  g
(3)
m1  a  m2  a  m3  a  m1  g  m3  g
 m1  m2  m3   a  m1  g  m3  g
a 
m
m
0.3 kg  9.8 2  0.1 kg  9.8 2
s
s
 0.3 kg  0.2 kg  0.1 kg 
a 
2.94 N  0.98 N
0.6 kg
form (1)

m
3.27 2
s
m1  a  m1  g  T1
m
m
0.3kg  3.27 2  0.3kg  9.8 2  T1
s
s
T1  2.94 N  0.981 N  1.961 N
form (3)
m3  a  T2  m3  g
m
m
0.1kg  3.27 2  T2  0.1kg  9.8 2
s
s
T2  0.327 N  0.98 N  1.307 N
check with (2)
m2  a  T1  T2
m
0.2kg  3.27 2  1.961 N  1.307 N
s
0.654 N  0.654 N
Practice problems # 4 - 6:
For each of the following systems of interconnected masses,
determine the common acceleration and tension(s) in the
string(s).
#4
two equations combined :
m2  a  T
(1)
m1  a  m1  g  T
(2)
m2  a  m1  a  T  m1  g  T
 m1  m2   a  m1  g
m
3.2kg  9.8 2
31.36 N
s
a 

7.2 kg
 3.2 kg  4 kg 
form (1)
check (2)

m
4.36 2
s
m
T  m2  a  4kg  4.36 2  17.44 N
s
m1  a  m1  g  T
m
m
3.2kg  4.36 2  3.2kg  9.8 2  17.44 N
s
s
13.95 N  31.36 N  17.44 N
#5
two equations combined :
m1  a  T  m1  g
(1)
m2  a  m2  g  T
(2)
m2  a  m1  a  m2  g  m1  g
 m1  m2   a  m2  g  m1  g
a 
m
m
4 kg  9.8 2  2 kg  9.8 2
s
s
 4 kg  2 kg 
39.2 N  19.6 N
a 
6 kg
form (1)

m
3.27 2
s
m1  a  T  m1  g
m
m
2kg  3.27 2  T  2kg  9.8 2
s
s
T  6.54 N  19.6 N 
26.14 N
#6
m1  a  m1  g  T1
(1)
m2  a  T1  T2
(2)
m3  a  T2  m3  g
(3)
m1  a  m2  a  m3  a  m1  g  m3  g
 m1  m2  m3   a  m1  g  m3  g
a 
m
m
3.2 kg  9.8 2  2 kg  9.8 2
s
s
 3.2 kg  4.2 kg  2 kg 
a 
31.36 N  19.6 N
9.4 kg
form (1)

m
1.25 2
s
m1  a  m1  g  T1
m
m
3.2kg  1.25 2  3.2kg  9.8 2  T1
s
s
T1  31.36 N  4.0 N  27.36 N
form (3)
m3  a  T2  m3  g
m
m
2kg  1.25 2  T2  2kg  9.8 2
s
s
T2  2.5 N  19.6 N  22.10 N