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Three blocks of masses M1=2 kg, M2=4 kg, and M3=6 kg are
connected by strings on a frictionless inclined plane of 60 o, as
shown in the figure below. A force of F=120N is applied upward
along the incline, causing the system of three masses to
accelerate up the incline. Consider the strings to be massless
and taut. What is the acceleration of M2? Find the tensions T1
and T2 in the two sections of string.
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Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
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Question 10
Question 11
Question 12
Question 13
Question 14
Question 15
Paired Problem 1
Notes
* For simplicity when analyzing this
system, we choose the the x-axis to be
parallel to the surface of the incline and
the y-axis to be perpendicular to the
surface. Let’s define the positive direction
as “upward” for both x and y.
*
Vectors quantities will be in bold type
1.Which one of these statements finishes
the following sentence correctly? (there is no
acceleration in the y-direction, so “a” refers to the acceleration in
the x-direction throughout this tutorial)
The acceleration of M2 is expected to be…
• A : the same as the acceleration of the system as a
whole.
• B : less than the acceleration of the system as a whole.
• C : greater than the acceleration of the system as a
whole.
Choice: A
Correct
Newton’s Second Law tells us: F=Ma
where: M = M1+M2+M3 (total mass of the system)
F is the net force on the system
a is the acceleration of the whole system
F = (M1+M2+M3 )a = M1a+M2a+M3a
Thus, each block accelerates with the same “a,” which
is also the acceleration of the system as a whole.
Choice: B
Incorrect
Remember that the strings are all
taut, which allows the masses to
move together.
Choice: C
Incorrect
Remember that the strings are all
taut, which allows the masses to
move together.
2. Which one of the following free body diagrams
is correct for the system of three blocks?
F
A:
Weight (W=Mg)
Force of Friction (Ff)
Normal Force (FN)
FN
F
B:
Mg
Mg
FN
F
C:
FN
F
D:
Ff
Mg
Mg
(the length of the arrows does not
represent the magnitudes of the forces here)
Choice: A
Incorrect
It is necessary to include the
normal force (FN).
Choice: B
Correct
This diagram shows all of the
forces acting on the system.
Choice: C
Incorrect
The gravitational force always acts
perpendicular to the ground, not
perpendicular to the surface of the plane.
Also, there is no friction between the
blocks and the inclined plane.
Choice: D
Incorrect
The gravitational force always
acts perpendicular to the
ground, not perpendicular to
the plane of contact.
3. A good choice of coordinate axes may be to consider
the x-axis parallel to the surface of the inclined plane and
the y-axis perpendicular to it. (The positive x and y
directions should be upwards.)
Which forces are aligned with these axes?
• A: FN and W
• B: F and W
• C: FN and F
Choice: A
Incorrect
FN is in the positive y-direction (always
perpendicular to the surface), but W is not along
either coordinate axis.
Remember that W=Mg, where g is gravitational
acceleration, which always points downward
(towards Earth).
Choice: B
Incorrect
F is in the positive x-direction, but W is not
along either coordinate axis.
Remember that W=Mg, where g is the
acceleration due to gravity, which always points
downward (towards Earth).
Choice: C
Correct
FN is directed in the positive ydirection and F is directed in the
positive x-direction.
4. When we resolve the gravitational force into its
x and y components, we get which one of the
following?
(the length of the arrows does not
represent the magnitudes of the forces here)
Mgcos()
A:
B:
Mgsin()
Mgsin()
Mgcos)
C:
D:
Mgsin()
Mgcos()
Mgcos()
Mgsin()
Choice: A
Incorrect
Mgcos() should point in the
negative y-direction.
Choice: B
Incorrect
Mgsin() should point in
the negative x-direction.
Choice: C
Correct
Here, the components are
correctly resolved.
Choice: D
Incorrect
Check the directions of both
components.
5. Which one of the following is the correct
equation of motion (x-component) for the
whole system (using Newton’s Second
Law)? Remember M=M1+M2+M3
• A:
• B:
• C:
F=Ma
F-Mgsin()=0
F-Mgsin()=Ma
Choice: A
Incorrect
Don’t forget the effect of gravity
Choice: B
Incorrect
The original problem statement tells us
that the applied force is causing upward
acceleration.
Therefore, there is a net force; it is not
equal to zero.
Choice: C
Correct
This equation is correct according to Newton’s
Second Law.
The applied force on the system of blocks is in
the opposite direction to the x-component of the
weight of the blocks. We know from the original
problem statement that the net force is not zero.
6. The acceleration of the system of
blocks can be expressed by which of
the following equations?
F
• A: a 
M
F
• B: a   gsin(60 )
M
 • C: a  F  Mgsin (60 )


Choice: A
Incorrect
This equation is incorrect
because it doesn’t take gravity
into consideration.

Choice: B
Correct
This is what we get when we
rearrange the equation that we
found for force in question 6.
F Mgsin( 60) Ma
F



 gsin( 60)  a
M
M
M
M
Choice: C
Incorrect
This is just the net force on the
system. We are looking for the
acceleration of the system.
7. Determine the value of the
acceleration a of the blocks up the
incline.
Reasoning:
F
a   gsin( 60 )
M
Where: F=120N
M= M1+M
2+M3= 2 kg +4 kg+6 kg= 12 kg
g= 9.8 m/s2
120N
a
 (9.8m/s2 )(0.866)  1.5 m/s2
12kg

• As discussed earlier, the acceleration
of M2 is the same as that of the
system as a whole because the
blocks move together.
• Now let’s focus on finding the
tensions T1 and T2.
8. Which one of these is the free body
diagram of M2? (the length of the arrows does not
represent the magnitudes of the forces here)
A:
FN
T1
C:
F+T2
T2
M2g
T2
B:
T1
M2 g
FN
M1 g
FN
D:
FN
T2
M2 g
T1
M2g
Choice: A
Incorrect
F only acts on M3.
Choice: B
Correct
This diagram shows all of the
forces acting upon M2.
Choice: C
Incorrect
The gravitational force on M1
does not act on M2. Therefore, it
does not belong in the free body
diagram for M2. Include T1.
Choice: D
Incorrect
The force on an object from
contact with a string in tension
always acts in the direction away
from the object. The directions of
T1 and T2 should be switched.
9. Which of the following is the correct
representation of the x-component of Newton’s
second law?
You must resolve the gravitational force into
components.
• A : T  T1  M2a
• B : T  T1  M2gsin( 60 )  M2a
• C : T  M2a
2
2
2



Choice: A
Incorrect
This expression does not include
the gravitational force.
Choice: B
Correct
This expression is consistent with
the free body diagram.
Choice: C
Incorrect
Gravitational force and tension T1
should also be considered because
they are acting on M2.
10. We have two unknowns, T1 and T2. We need two
equations. We have one equation from the free body
diagram of M2. Get the other one from the free body
diagram of M1.
The free body diagram for M1 is:
A:
FN
T1
B:
T1+T2
FN
M1 g
M1 g
C:
FN
F
D:
M1 g
FN
T1
M1g
(the length of the arrows does not
represent the magnitudes of the forces here)
Choice: A
Correct
All of the forces are shown
correctly.
Choice: B
Incorrect
T2 doesn’t act on M1.
Choice: C
Incorrect
F only acts on M3.
Choice: D
Incorrect
The normal force acts
perpendicular to the surface of
the plane of contact.
11. Resolving the gravitational force into its
components and applying Newton’s Second
Law gives us which of the following as the xcomponent of the equation of motion for M1?
• A : T1  M1a
• B : T1  M1gsin( 60 )  M1a
• C : T1 M1g  M1a



Choice: A
Incorrect
T1 is not the net force.
Remember the force in
Newton’s Second Law is the net
force. Gravity acts upon M1.
Choice: B
Correct
This equation is consistent with the
free body diagram for M1.
Choice: C
Incorrect
This would be correct if the
angle of inclination were 90
degrees.
12. Which equation should be solved first?
• A : T2  T1  M2gsin( 60 )  M2a
• B : T1  M1gsin( 60 )  M1a


Choice: A
Incorrect
We start with the other equation
because this one contains two
unknowns. It is not possible to
solve for two unknowns using a
single equation.
Choice: B
Correct
This equation should be solved first
since only T1 is unknown.
13. After solving for T1 you will get :
T1  M1gsin( 60 )  M1a
If you plug this back into the equation of motion
for M2( T2  T1  M2gsin( 60 )  M2a ) and simplify,
you get which of the following expressions for
T2?

• A : T2  M1gsin( 60 )  M2a
• B : T2  M1a  M2a
• C : T2  (M2  M1)(gsin( 60 )  a)



Choice: A
Incorrect
This expression omits several
terms. Check your algebra.
Choice: B
Incorrect
This expression does not include
the contribution of gravity.
Choice: C
Correct
We get this result using simple
algebra and factoring out
common variables.
Now all you have to do is insert the given
values and calculate the numerical values
of T1 and T2.
14) Find the value of T1
Answer
The correct answer is:
T1  M1(a  gsin( 60 ))
 2kg(1.51m/s  (9.8m/s )(0.866))
2
 20.0 N
2
15) Now find T2.
Answer
The correct answer is :
T2  T1  M2gsin( 60 )  M2a


T2  20N  4kg9.8m/s2 sin 60  4kg1.5m/s2 
T2  60.0 N

Paired Problem
1) Three blocks of masses M1=2 kg, M2=4 kg, and M3=6 kg are
connected by strings on a frictionless inclined plane of 60 o, as
shown in the figure below. A force of F=120N is applied upward
along the incline, causing the system of three masses to
accelerate up the incline. Consider the strings to be massless
and taut. What is the acceleration of M2? Find the tensions T1
and T2 in the two sections of string.