A series of blocks is connected to a
pulley in the manner depicted below.
Find the minimum mass that block C
must have in order to keep block A
from sliding on the table top.
C
ms
mA=4 Kg
mB=2 Kg
A
ms=0.2 (coefficient of static friction)
B
y
x
The pulley and rope are to be
considered massless.
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Reflection
1. Which physics principle should
we use to solve this problem.
A) Work-Energy Theorem
B) Newton’s 2nd Law
C) Law of conservation of energy
Choice A
Incorrect
Since the blocks are not in motion, there
is no work being done. The work-energy
theorem will not help us solve this
problem.
Choice B
Correct
Since we are considering a situation where the blocks are not
slipping on the table and the hanging block is not moving up or
down, the system is in static equilibrium.
Newton’s Laws tell us that the sum of the forces on an object or the
entire system is zero for mechanical equilibrium. In component
form in two dimensions, the conditions are:
F

x
0
and
F
y
0
Choice C
Incorrect
The law of conservation of energy is of
no use here.
Try again. Remember that none of the
blocks are in motion in our situation.
2. Let’s begin by analyzing the hanging block B.
Which of the following free body diagrams correctly
depicts all of the forces acting on block B?
T
A)
g=9.8m/s2
mBg
mAg+mC
g
B)
mBg
T
C)
T= tension
mAg+mC
g
mBg
Choice A
Correct
There are no forces acting on the
hanging block in the positive or negative
x-direction.
This diagram correctly shows the only two
forces acting on the block.
Choice B
Incorrect
The weight of blocks A and C do not act on the
hanging block directly. They pull on the rope
that is connected to block B.
The only forces that act on block B are the
tension and the block’s weight.
Choice C
Incorrect
There are no forces acting on the
hanging block in the positive or negative
x-direction.
3. Using the free body diagram for block B,
and applying Newton’s 2nd Law we get which
one of the following relations?
A)
T  m Bg
B)
T  m B g
C)
m A g  m C g  m Bg


Choice A
Correct
Using Newton’s 2nd Law and our
free body diagram we find:
F
y
 T  m Bg  0
T  m Bg
After writing down Newton’s 2nd Law
resolved into x and y components, we only
find a useful equation from the y-component,

because
none of the forces have xcomponents.
Choice B
Incorrect
Using Newton’s 2nd Law and our
free body diagram we find:
F
y
 T  m Bg  0
T  m Bg

The two forces are
subtracted because they
act in opposite directions.
Choice C
Incorrect
Check the free body diagram that we
found to be correct in the previous
question and try again.
4.
Let’s now analyze blocks A and C.
Which of the following free body diagrams correctly depicts all
of the forces acting on blocks A and C if we treat them as one
system?
Ff=frictional force
N
A)
N=normal force
T
Ff
(mA+mC)g
N
B)
Ff
T
(mA+mC)g
N
C)
Ff
T
(mA-mC)g
Choice A
Incorrect
The directions of the forces of friction and
tension should be switched. The force of
static friction opposes the sliding which would
be to the right if block C were not massive
enough.
The rope is attached to block A on the right
side, so the tension would act away from the
block in the positive x-direction.
Choice B
Correct
This diagram correctly depicts all of
the forces acting on the system of
blocks A and C.
Choice C
Incorrect
The combined weight W of blocks A and
C should be considered. We are
treating them together as one system.
W=(mA-mC)g
W=(mA+mC)g
5. Use the free body diagram from the previous question.
Which of the following relations do we get by applying
Newton’s 2nd Law to blocks A & C together as one system?
(Resolve the forces into x and y components.)
F
A)

B)

F
0
y
0
N  m A  m C g
T  Ff

T  Ff
N  m A  m C g

C)

x
T  m A  m C g

N  m A  m C g
Choice A
Correct
Using Newton’s 2nd Law and our
free body diagram we find:
F
x
F
 T  Ff  0
y
 N  (m A  m C )g  0
N  (m A  m C )g
T  Ff

Choice B
Incorrect
Using Newton’s 2nd Law and our
free body diagram we find:
F
x
F
 T  Ff  0
y
 N  (m A  m C )g  0
N  (m A  m C )g
T  Ff

Choice C
Incorrect
The weight of the system of blocks A
and C has no x-component.
6. Use the information from the last question, to find another
way to express T in terms of known quantities and the
quantity that we are interested in finding, which is the
minimum mass (mC) that block C can have to keep the
system in equilibrium.
Which one of the following expressions is correct?


A)
T  m A  m B  m C g
B)
T  m s m A  m B  m C g
C)
Note:
The force of static friction that will
keep the blocks from slipping must
satisfy the following condition:
F f  m sN
We want to find the minimum mass of
block C that allows this condition to be
satisfied. This occurs when:
T  m s m A  m C g

F f  m sN
Choice A
Incorrect
Combine several
equations that we have
found in previous
questions,
N  m A  m C g

F f  m sN  m s m A  m C g

T  F f
T  m s m A  m C g

Choice B
Incorrect
Combine several
equations that we have
found in previous
questions.
N  m A  m C g

F f  m sN  m s m A  m C g

T  F f
T  m s m A  m C g

Choice C
Correct
We combine several
equations that we have
found in previous
questions.
N  m A  m C g

F f  m sN  m s m A  m C g

T  F f
T  m s m A  m C g

7. We now have two equations for the tension in
the rope T. If we equate the two, we can solve
for the minimum mass of block C.
What is the minimum mass that block C must
have in order to keep block A from sliding on
the table top?
A) mC=10kg
B) mC=6kg
C) mC=4kg
Choice A
Incorrect
Begin by equating these
two expressions for
tension.
T  m B g

T  m s m A  m C g
Please try again. You should
find

the following expression for the
minimum mass of block C:
mC 
mB
ms
 mA
Choice B
Correct
T  m B g

T  m s m A  m C g
m B g  m s m A  m C g
m B  m s m A  m C 
mB
ms
mB
ms
 m A  m C 
 m A  mC


The force of friction on the combined mass
(mA+mC) is balanced by the tension force.
 have a mass of at least 6 kg in
Block C must
order to keep the other blocks from moving.
mC 
mB
mC 
2kg
ms
0 .2
 mA
 4kg  6kg
m C  6kg
Choice C
Incorrect
Begin by equating these
two expressions for
tension.
T  m B g

T  m s m A  m C g
Please try again. You should find

the following expression for the
minimum mass of block C:
mC 
mB
ms
 mA
Reflection Questions
• If the coefficient of static friction was
lower than 0.2, what would happen to
blocks A and C?
• If the coefficient of static friction was
higher than 0.2, what would happen to
More
blocks A and C?
Reflection Question:
• How do we know that the relation:
F f  m sN
gives the minimum mass required to
keep the blocks from slipping?

Help
Using the inequality we find:
T  m B g

T  m s m A  m C g
2kg
0.2
m B g  m s m A  m C g
m B  m s m A  m C 
ms
mB
ms
6kg  m C
 m A  m C 
This means that the mass of
block C must be greater than
or equal to 6 kg in order to
keep the blocks from
slipping.
 m A  mC
Obviously, 6kg is the
minimum mass to satisfy
this condition, so if we just
use the equal sign originally,
we find the minimum mass.

mB
 4kg  m C