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AP QUIZ #9 ADVANCED FORCES B3-RT54: PERSON IN A MOVING ELEVATOR—SCALE
READING
A person who weighs 600 N is standing on a scale in an elevator. The elevator is identical in all cases. The velocity and
acceleration of the elevators at the instant shown are given.
A
B
C
D
v = 3 m/s
v = 2 m/s
v = 3 m/s
v = 1 m/s
a = 2 m/s2
a = 2 m/s2
a=0
a = 2 m/s2
Rank the scale reading.
OR
1
Greatest
2
3
4
Least
All
the same
All
zero
Cannot
determine
Explain your reasoning.
Answer: A = D > C > B. The scale is providing the normal force on the person, and the scale reading indicates the magnitude of
this force. For cases A and D that normal force is larger than the person’s weight because the scale also has to accelerate the
person upward. For case C the normal force is equal in magnitude to the person’s weight, and for case B it is less than the
person’s weight since the person is accelerating downward.
B3-RT55: Person in an Elevator Moving Downward—Scale Reading
A person who weighs 600 N is standing on a scale in an elevator. The elevator is identical in all cases. The velocity and
acceleration of the elevators at the instant shown are given.
A
B
C
D
v = 3 m/s
v = 3 m/s
v = 3 m/s
v=0
a = 2 m/s2
a = 2 m/s2
a=0
a = 9.8 m/s2
Rank the scale reading.
OR
1
Greatest
2
3
4
Least
All
the same
All
zero
Cannot
determine
Explain your reasoning.
Answer: A > C > B > D. The scale is providing the normal force on the person, and the scale reading indicates the magnitude of
this force. For case A that normal force is larger than the person’s weight because the scale also has to accelerate the person
upward. For case C the normal force is equal, in magnitude, to the person’s weight, for case B it is less than the person’s weight
since the person is accelerating downward, and in case D the scale reading will be zero since the person is in free fall.
B3-CT58: PERSON IN AN ELEVATOR—SCALE READING
A person who weighs 500 N is standing on a scale in an elevator. The elevator is identical in both cases. In both cases the
elevator is moving at a constant speed, upward in case A and downward in case B.
A
B
v = 3 m/s
v = 3 m/s
Will the scale reading be (a) greater in case A, (b) greater in case B, or (c) the same in both cases?
Explain.
Answer: The scale reading will be the same in both cases.
By Newton’s Second Law the net force on the person must be zero in both cases, since the elevator is not accelerating. So the
normal force on the person by the scale (which is the reading on the scale) must be equal to the weight of the person in both cases.
B3-RT68: Hanging Blocks—Tension
Two blocks are connected by strings, and are pulled upward by a second string attached to the upper block. The lower block is the
same in all cases, but the mass of the upper block varies. The acceleration and velocity for each system at the instant shown is given.
A
B
C
D
E
F
15 g
15 g
20 g
20 g
10 g
10 g

v = 1 m/s

v = 1 m/s

v = 1 m/s

v = 1 m/s

v=0


a = 0

a = 0
a = 0

a = 2 m/s2
a = 2 m/s2
v = 1 m/s
a = 2 m/s2
Rank the tension in the string between the blocks.
OR
1
Greatest
2
3
4
5
6
Least
All
the same
All
zero
Cannot
determine
Explain your reasoning.
Answer: A > B = D = E > C = F. These tensions can be compared by considering a free-body diagram of the lower block. The
tension minus the weight of the lower block must equal the mass of the lower block times its acceleration in the upward direction.
Since the lower blocks are all identical, the tension must be greatest for blocks that are accelerating upward, less for blocks with no
acceleration, and least for blocks that are accelerating downward. The mass of the upper block does not matter, and the velocity of
the blocks at the instant shown does not matter.
B3-SCT69: HANGING STONE CONNECTED TO BOX—FREE-BODY DIAGRAMS
A massless rope connects a box on a horizontal surface and a hanging stone as shown below. The rope passes over a massless,
frictionless pulley. The box is given a quick tap so that it slides to the right along the horizontal surface. The figure below shows the
block after it has been pushed while it is still moving to the right. The mass of the hanging stone is larger than the mass of the box.
There is friction between the box and the horizontal surface. Free-body diagrams that a student has drawn to scale for the box and for
the hanging stone are shown.
v = 2 m/s
Fon stone by rope
240 g
Fon box by rope
Fon stone by earth
Fon box by surface
Fon box by surface
Fon box by earth
Four students discussing these free-body diagrams make the following contentions:
Ali:
Brianna:
Carlos:
Dante:
“I think there is a problem with the free-body diagram for the hanging stone. The two forces should have the
same magnitude.”
“But the stone is moving upward – there should be a larger force in that direction.”
“No, the diagram for the hanging stone is okay, but there is a problem with the diagram for the box. The
frictional force is in the wrong direction.”
“No, all three of you are wrong. Both free-body diagrams are correct because both show the way the objects
would be accelerating.”
With which, if any, of these students do you agree?
Ali _______ Brianna _______ Carlos ______ Dante ______ None of them _______
Explain your reasoning.
For the instant shown, Carlos is correct. Since the box is sliding to the right, the friction on the box by the surface will be to the
left. So all horizontal forces on the box would be to the left and the block would slow down (as we expect). If, as Ali claimed, the
forces on the stone had the same magnitude, the stone would have no net force and would not accelerate, and if (as Brianna
claimed) the force on the stone by the rope was greater than the weight of the stone then the stone would speed up.
B3-RT70: Hanging Stone Connected to Box on Rough Surface—Acceleration
In each case shown below, a box is sliding along a horizontal surface. There is friction between the box and the horizontal surface.
The box is tied to a hanging stone by a massless rope running over a massless, frictionless pulley. All these cases are identical except
for the different initial velocities of the boxes.
vo = 8 m/s
A
B
100 g
180 g
vo = 3 m/s
C
100 g
180 g
vo = 5 m/s
D
100 g
180 g
vo = 5 m/s
100 g
180 g
Rank the magnitudes of the accelerations of these boxes at the instant shown.
Box moving to the right
Box moving to the left
Explain your reasoning.
Answer: A = D > B = C.
Fon box by surface
Fon box by surface
Since the surfaces are identical and the blocks all
Fon box by rope
Fon box by rope
have the same mass, the magnitude of the friction
Fon box by surface
force is the same in all cases. However, when the
box is moving to the left the frictional force is to the Fon box by surface
Fon box by earth
Fon box by earth
right (opposite the direction of the tension force on
the box) and when the box is moving to the right
the frictional force is to the left, in the same
direction as the tension. The friction force doesn’t depend on the velocity. The acceleration is the same for all boxes that are
moving in the same direction, since the free-body diagrams for these boxes will be the same.
B3-RT71: MOVING STRING PASSING OVER A PULLEY—TENSION AT POINTS
A student pulls on a massless string that passes over a frictionless pulley
to a suspended mass. He is pulling the string horizontally so that, at the
the mass is moving upward at a constant speed.
A
and is attached
instant shown,
B
C
D
Rank the tension at the labeled points A, B, C, and D.
OR
1
2
3
4
All
All
Least
the same
zero de
Greatest
Explain your reasoning.
Answer: All the same. In the approximation that the string is massless, any small section of the string must have no net force
acting on it (since it is not accelerating). So the tension on one end of that small section must have the same magnitude as the
tension on the other end. By extension, the tension must be the same everywhere along the horizontal section of string and
everywhere along the vertical section of string. The tension in the horizontal section of string exerts a counterclockwise torque on
the pulley, and the tension in the vertical section of string exerts a clockwise torque on the pulley. Since the pulley has no angular
acceleration, the net torque on it is zero. The tension in the horizontal portion of the string between the pulley and the hand must
be equal to the tension in the vertical portion of the string. (The perpendicular distance between the line of action of the tension
force and the pulley pivot is the radius of the pulley, and is the same for both sections of string.)
B3-CT73: Ball Suspended from Ceiling by Two Strings—Tension
A 0.5-kg ball is suspended from a ceiling by two strings. The ball is at rest.
a) Is the tension in string 1 greater than, less than, or the same as the tension in
string 2?
Explain.
String 1
TB1
60°
30°
Answer: Greater since the forces on the ball must add to zero. Keeping the directions
of the vectors as shown constrains the relative sizes of TB1 and TB2 as shown in the
vector summation here. Note that TB1 and TB2 are perpendicular. So, TB1 must be greater than TB2.

String 2

TB2

WBE

Ball

TB1

WBE
Suppose that the ceiling in the picture is the ceiling of an elevator, and
that the
TB2
elevator is moving down at a constant speed of 2 m/s.
b) Is the tension in string 1 greater than, less than, or the same as the tension in string 1 in the previous question (a) where the
ball was at rest?
Explain.
Answer: The forces are the same as for the same system at rest. The net force is still zero since the ball is moving at a constant
speed. The weight does not change (the ball is still on the surface of the earth) and so the tensions will not change.
B3-SCT74: HANGING MASS—TENSION IN THREE STRINGS
A hanging mass is suspended midway between two walls.
attached to the left wall is horizontal while the string attached
wall makes an angle with the horizontal as shown. This angle
is larger than the angle () in Case B. Four students make the
claims about the tensions in the strings:
Abbie:
Bobby:
Che:
Damian:
Case A
The string
to the right
() in Case A
following
Case B


“I think the tensions in any string in case
A is going to
be the same as the equivalent string in
case B. The
weight is the same, and the weight is still
going to be
divided up among the three ropes.”
“I think the tensions in the horizontal and vertical strings are the same, because they are exactly the same in
both cases. But in Case B the diagonal rope is shorter, so the tension is more concentrated there.”
“The diagonal string still has to hold the weight up by itself, because the horizontal string can’t lift anything.
So the diagonal string still has the same tension. But in Case B it’s pulling harder against the horizontal
string because of the angle, so the tension in the horizontal string has to go up.”
“But the diagonal string is fighting harder against the weight in Case A – it is pointing more nearly opposite
the weight. So it has to have a greater tension in Case A. And since the tension in the diagonal string is
greater, and the tension in the vertical string is the same, the tension in the horizontal string must be less in
Case A. The tensions still have to balance out so that they are the same in both cases.”
With which, if any, of these students do you agree?
Abbie _____ Bobby _____ Che _____ Damian _____ None of them______
Explain your reasoning.
Answer: None of these responses is correct. The weight is the same for
both cases, and because the weight is at rest we can conclude from a freebody diagram of the weight that the tension in the vertical string must be
the same in both cases. But since the knot connecting the strings is at rest
in both cases vector sum of forces acting on the knot is zero in both cases.
Comparing vector sums, (with subscripts K, H, V, and D for knot,
horizontal, vertical, and diagonal) we can see that the horizontal tension
must be greater in case B. The tension in the diagonal string will also be
larger in case B


TKH
TKD



TKV


TKD
TKH
TKD
TKV

TKD
TKV


TKH

TKV

TKH
B3-CT80: Blocks Moving at Constant Speed—Tension in Connecting
String
Two identical blocks, 1 and 2, are connected by a
massless
A
10 cm/s
to block 2
string. In Case A, a student pulls on a string attached
so that the blocks travel to the right across a desk at a
constant
string
speed of 10 cm/s. In Case B, the student pulls on a
1
2
attached to block 1 so that the same blocks travel
across the
same desk to the left at a constant speed of 20 cm/s.
Will the tension in the diagonal string connecting
the two
B
blocks be greater in Case A, greater in Case B, or
the same
20 cm/s
in both cases?
Explain your reasoning.
1
2
Greater in case A. In both cases the net force acting
on the
trailing block is zero, since that block is moving at a
constant
speed. In case A, the tension in the connecting string
has a
component acting downward, on the trailing block, and since the net vertical force is zero, the normal force must be greater than
the weight. In case B, the tension in the connecting string has a component acting upward on the trailing block, and the normal
force on the block must therefore be less than the weight. Since the friction on the trailing block by the desk is proportional to the
normal force on that block, the friction force on the block by the desk is greater in Case A than it is in Case B. And since the
frictional force on the trailing block is equal to the horizontal component of the tension in the string connecting the blocks in both
cases, the tension in this string must be greater in Case A than in Case B.