ConcepTest 6.1a Tension I
You tie a rope to a tree and you
1) 0 N
pull on the rope with a force of
2) 50 N
100 N. What is the tension in
the rope?
3) 100 N
4) 150 N
5) 200 N
ConcepTest 6.1a Tension I
You tie a rope to a tree and you
1) 0 N
pull on the rope with a force of
2) 50 N
100 N. What is the tension in
the rope?
3) 100 N
4) 150 N
5) 200 N
The tension in the rope is the force that the rope
“feels” across any section of it (or that you would
feel if you replaced a piece of the rope). Since you
are pulling with a force of 100 N, that is the tension
in the rope.
ConcepTest 6.1b Tension II
Two tug-of-war opponents each
1) 0 N
pull with a force of 100 N on
2) 50 N
opposite ends of a rope. What
is the tension in the rope?
3) 100 N
4) 150 N
5) 200 N
ConcepTest 6.1b Tension II
Two tug-of-war opponents each
1) 0 N
pull with a force of 100 N on
2) 50 N
opposite ends of a rope. What
is the tension in the rope?
3) 100 N
4) 150 N
5) 200 N
This is literally the identical situation to the
previous question. The tension is not 200 N !!
Whether the other end of the rope is pulled by a
person, or pulled by a tree, the tension in the rope
is still 100 N !!
ConcepTest 6.1c Tension III
You and a friend can
each pull with a force of
20 N. If you want to rip
a rope in half, what is
the best way?
1) you and your friend each pull on
opposite ends of the rope
2) tie the rope to a tree, and you both
pull from the same end
3) it doesn’t matter -- both of the above
are equivalent
4) get a large dog to bite the rope
ConcepTest 6.1c Tension III
You and a friend can
each pull with a force of
20 N. If you want to rip
a rope in half, what is
the best way?
1) you and your friend each pull on
opposite ends of the rope
2) tie the rope to a tree, and you both
pull from the same end
3) it doesn’t matter -- both of the above
are equivalent
4) get a large dog to bite the rope
Take advantage of the fact that the tree can
pull with almost any force (until it falls down,
that is!). You and your friend should team up
on one end, and let the tree make the effort on
the other end.
ConcepTest 6.2 Three Blocks
Three blocks of mass 3m, 2m, and
1) T1 > T2 > T3
m are connected by strings and
2) T1 < T2 < T3
pulled with constant acceleration a.
3) T1 = T2 = T3
What is the relationship between
4) all tensions are zero
the tension in each of the strings?
5) tensions are random
a
3m
T3
2m
T2
m
T1
ConcepTest 6.2 Three Blocks
Three blocks of mass 3m, 2m, and
1) T1 > T2 > T3
m are connected by strings and
2) T1 < T2 < T3
pulled with constant acceleration a.
3) T1 = T2 = T3
What is the relationship between
4) all tensions are zero
the tension in each of the strings?
5) tensions are random
T1 pulls the whole set
of blocks along, so it
a
must be the largest.
largest
T2 pulls the last two
masses, but T3 only
pulls the last mass.
3m
T3
2m
T2
m
T1
Follow-up: What is T1 in terms of m and a?
ConcepTest 6.3 Over the Edge
In which case does block m experience 1) case 1
a larger acceleration? In (1) there is a
10 kg mass hanging from a rope and
2) acceleration is zero
3) both cases are the same
falling. In (2) a hand is providing a
constant downward force of 98 N.
4) depends on value of m
Assume massless ropes.
5) case 2
m
m
10kg
a
a
F = 98 N
Case (1)
Case (2)
ConcepTest 6.3 Over the Edge
In which case does block m experience 1) case 1
a larger acceleration? In (1) there is a
10 kg mass hanging from a rope and
2) acceleration is zero
3) both cases are the same
falling. In (2) a hand is providing a
constant downward force of 98 N.
4) depends on value of m
Assume massless ropes.
5) case 2
In (2) the tension is 98 N
due to the hand. In (1)
the tension is less than
98 N because the block
10kg
a
rest would the tension
be equal to 98 N.
a
F = 98 N
is accelerating down.
down
Only if the block were at
m
m
Case (1)
Case (2)
ConcepTest 6.4 Friction
on a frictionless truck bed.
1) the force from the rushing air
pushed it off
When the truck accelerates
2) the force of friction pushed it off
forward, the box slides off
3) no net force acted on the box
the back of the truck
4) truck went into reverse by accident
A box sits in a pickup truck
because:
5) none of the above
ConcepTest 6.4 Friction
on a frictionless truck bed.
1) the force from the rushing air
pushed it off
When the truck accelerates
2) the force of friction pushed it off
forward, the box slides off
3) no net force acted on the box
the back of the truck
4) truck went into reverse by accident
A box sits in a pickup truck
because:
5) none of the above
Generally, the reason that the box in the truck bed would move
with the truck is due to friction between the box and the bed.
If there is no friction, there is no force to push the box along,
and it remains at rest. The truck accelerated away, essentially
leaving the box behind!!
ConcepTest 6.5 Antilock Brakes
Antilock brakes keep the
car wheels from locking
and skidding during a
sudden stop. Why does
this help slow the car
down?
1) µk > µs so sliding friction is better
2) µk > µs so static friction is better
3) µs > µk so sliding friction is better
4) µs > µk so static friction is better
5) none of the above
ConcepTest 6.5 Antilock Brakes
Antilock brakes keep the
car wheels from locking
and skidding during a
sudden stop. Why does
this help slow the car
down?
1) µk > µs so sliding friction is better
2) µk > µs so static friction is better
3) µs > µk so sliding friction is better
4) µs > µk so static friction is better
5) none of the above
Static friction is greater than sliding friction,
friction so
by keeping the wheels from skidding, the static
friction force will help slow the car down more
efficiently than the sliding friction that occurs
during a skid.
ConcepTest 6.6 Going Sledding
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
1) pushing her from behind
2) pulling her from the front
3) both are equivalent
easiest way to
4) it is impossible to move the sled
accomplish this?
5) tell her to get out and walk
1
2
ConcepTest 6.6 Going Sledding
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
1) pushing her from behind
2) pulling her from the front
3) both are equivalent
easiest way to
4) it is impossible to move the sled
accomplish this?
5) tell her to get out and walk
In Case 1, the force F is pushing down
(in addition to mg), so the normal
force is larger.
larger In Case 2, the force F
1
is pulling up,
up against gravity, so the
normal force is lessened.
lessened Recall that
the frictional force is proportional to
the normal force.
2
ConcepTest 6.7 Will it Budge?
A box of weight 100 N is at
rest on a floor where µs = 0.5.
A rope is attached to the box
and pulled horizontally with
tension T = 30 N. Which way
does the box move?
1) moves to the left
2) moves to the right
3) moves up
4) moves down
5) the box does not move
Static friction
(µs = 0.4 )
m
T
ConcepTest 6.7 Will it Budge?
A box of weight 100 N is at
rest on a floor where µs = 0.5.
A rope is attached to the box
and pulled horizontally with
tension T = 30 N. Which way
does the box move?
1) moves to the left
2) moves to the right
3) moves up
4) moves down
5) the box does not move
The static friction force has a
maximum of µsN = 40 N.
N The
tension in the rope is only 30 N.
N
Static friction
(µs = 0.4 )
m
T
So the pulling force is not big
enough to overcome friction.
Follow-up: What happens if the tension is 35 N? What about 45 N?
ConcepTest 6.8a Sliding Down I
A box sits on a flat board.
You lift one end of the
board, making an angle
with the floor. As you
increase the angle, the
box will eventually begin
to slide down. Why?
1) component of the gravity force
parallel to the plane increased
2) coeff. of static friction decreased
3) normal force exerted by the board
decreased
4) both #1 and #3
5) all of #1, #2 and #3
Normal
Net Force
Weight
ConcepTest 6.8a Sliding Down I
A box sits on a flat board.
You lift one end of the
board, making an angle
with the floor. As you
increase the angle, the
box will eventually begin
to slide down. Why?
l
1) component of the gravity force
parallel to the plane increased
2) coeff. of static friction decreased
3) normal force exerted by the board
decreased
4) both #1 and #3
5) all of #1, #2 and #3
As the angle increases, the component
of weight parallel to the plane increases
and the component perpendicular to the
plane decreases (and so does the Normal
Normal
force). Since friction depends on Normal
force, we see that the friction force gets
smaller and the force pulling the box
down the plane gets bigger.
bigger
Net Force
Weight
ConcepTest 6.8b Sliding Down II
A mass m is placed on an
inclined plane (µ > 0) and
slides down the plane with
constant speed. If a similar
block (same µ) of mass 2m
were placed on the same
incline, it would:
m
1) not move at all
2) slide a bit, slow down, then stop
3) accelerate down the incline
4) slide down at constant speed
5) slide up at constant speed
ConcepTest 6.8b Sliding Down II
A mass m is placed on an
inclined plane (µ > 0) and
slides down the plane with
constant speed. If a similar
block (same µ) of mass 2m
were placed on the same
incline, it would:
1) not move at all
2) slide a bit, slow down, then stop
3) accelerate down the incline
4) slide down at constant speed
5) slide up at constant speed
The component of gravity acting down
N
f
the plane is double for 2m. However,
the normal force (and hence the friction
force) is also double (the same factor!).
This means the two forces still cancel
to give a net force of zero.
Wy
θ
Wx
W
θ
ConcepTest 6.9 Tetherball
In the game of tetherball,
1) Toward the top of the pole
2) Toward the ground
the struck ball whirls
3) Along the horizontal component of
the tension force
4) Along the vertical component of the
tension force
around a pole. In what
direction does the net
force on the ball point?
5) Tangential to the circle
T
W
ConcepTest 6.9 Tetherball
In the game of tetherball,
the struck ball whirls
around a pole. In what
direction does the net
force on the ball point?
1) Toward the top of the pole
2) Toward the ground
3) Along the horizontal component of
the tension force
4) Along the vertical component of the
tension force
5) Tangential to the circle
The vertical component of the
tension balances the weight.
weight The
horizontal component of tension
W
T
T
provides the centripetal force that
points toward the center of the
circle.
W
ConcepTest 6.10a Around the Curve I
You are a passenger in a
car, not wearing a seat
belt. The car makes a
sharp left turn. From
your perspective in the
car, what do you feel is
happening to you?
(1) You are thrown to the right
(2) You feel no particular change
(3) You are thrown to the left
(4) You are thrown to the ceiling
(5) You are thrown to the floor
ConcepTest 6.10a Around the Curve I
You are a passenger in a
car, not wearing a seat
belt. The car makes a
sharp left turn. From
your perspective in the
car, what do you feel is
happening to you?
The passenger has the tendency
to continue moving in a straight
line. From your perspective in the
car, it feels like you are being
thrown to the right, hitting the
passenger door.
(1) You are thrown to the right
(2) You feel no particular change
(3) You are thrown to the left
(4) You are thrown to the ceiling
(5) You are thrown to the floor
ConcepTest 6.10b Around the Curve II
During that sharp left turn,
you found yourself hitting
the passenger door. What
is the correct description of
what is actually happening?
(1) centrifugal force is pushing you
into the door
(2) the door is exerting a leftward
force on you
(3) both of the above
(4) neither of the above
ConcepTest 6.10b Around the Curve II
During that sharp left turn,
you found yourself hitting
the passenger door. What
is the correct description of
what is actually happening?
(1) centrifugal force is pushing you
into the door
(2) the door is exerting a leftward
force on you
(3) both of the above
(4) neither of the above
The passenger has the tendency
to continue moving in a straight
line. There is a centripetal force,
provided by the door, that forces
the passenger into a circular path.
ConcepTest 6.10c Around the Curve III
You drive your dad’s car
too fast around a curve
and the car starts to skid.
What is the correct
description of this
situation?
(1) car’s engine is not strong enough to
keep the car from being pushed out
(2) friction between tires and road is not
strong enough to keep car in a circle
(3) car is too heavy to make the turn
(4) a deer caused you to skid
(5) none of the above
ConcepTest 6.10c Around the Curve III
You drive your dad’s car
too fast around a curve
and the car starts to skid.
What is the correct
description of this
situation?
(1) car’s engine is not strong enough to
keep the car from being pushed out
(2) friction between tires and road is not
strong enough to keep car in a circle
(3) car is too heavy to make the turn
(4) a deer caused you to skid
(5) none of the above
The friction force between tires and
road provides the centripetal force
that keeps the car moving in a circle.
If this force is too small, the car
continues in a straight line!
Follow-up: What could be done to
the road or car to prevent skidding?
ConcepTest 6.11 Missing Link
A ping pong ball is shot into a
circular tube that is lying flat
(horizontal) on a tabletop. When
the ping pong ball leaves the
track, which path will it follow?
ConcepTest 6.11 Missing Link
A ping pong ball is shot into a
circular tube that is lying flat
(horizontal) on a tabletop. When
the ping pong ball leaves the
track, which path will it follow?
l
Once the ball leaves the tube, there is no longer
a force to keep it going in a circle. Therefore, it
simply continues in a straight line, as Newton’s
First Law requires!
Follow-up: What physical force provides the centripetal force?
ConcepTest 6.12 Ball and String
Two equal -mass rocks tied to strings are
whirled in horizontal circles. The radius of
1) T2 = 1/4 T1
2) T2 = 1/2 T1
circle 2 is twice that of circle 1. If the period
of motion is the same for both rocks, what
3) T2 = T1
is the tension in cord 2 compared to cord 1?
5) T2 = 4 T1
4) T2 = 2 T1
ConcepTest 6.12 Ball and String
Two equal -mass rocks tied to strings are
whirled in horizontal circles. The radius of
1) T2 = 1/4 T1
2) T2 = 1/2 T1
circle 2 is twice that of circle 1. If the period
of motion is the same for both rocks, what
3) T2 = T1
is the tension in cord 2 compared to cord 1?
5) T2 = 4 T1
4) T2 = 2 T1
The centripetal force in this case is given by the
tension, so T = mv2/r. For the same period, we find
that v2 = 2v1 (and this term is squared). However, for
the denominator, we see that r2 = 2r1 which gives us
the relation T2 = 2T1.
ConcepTest 6.13 Barrel of Fun
A rider in a “barrel of fun”
finds herself stuck with
her back to the wall.
Which diagram correctly
shows the forces acting
on her?
1
2
3
4
5
ConcepTest 6.13 Barrel of Fun
A rider in a “barrel of fun”
finds herself stuck with
her back to the wall.
Which diagram correctly
shows the forces acting
on her?
1
2
3
4
5
The normal force of the wall on the
rider provides the centripetal force
needed to keep her going around
in a circle. The downward force of
gravity is balanced by the upward
frictional force on her, so she does
not slip vertically.
Follow-up: What happens if the rotation of the ride slows down?
ConcepTest 6.14a Going in Circles I
You’re on a Ferris wheel moving in a
vertical circle. When the Ferris wheel is
at rest, the normal force N exerted by
your seat is equal to your weight mg.
How does N change at the top of the
Ferris wheel when you are in motion?
1) N remains equal to mg
2) N is smaller than mg
3) N is larger than mg
4) None of the above
ConcepTest 6.14a Going in Circles I
You’re on a Ferris wheel moving in a
vertical circle. When the Ferris wheel is
at rest, the normal force N exerted by
your seat is equal to your weight mg.
How does N change at the top of the
Ferris wheel when you are in motion?
You are in circular motion, so there
has to be a centripetal force pointing
inward.
inward At the top, the only two
forces are mg (down) and N (up),
(up) so
N must be smaller than mg.
mg
Follow-up: Where is N larger than mg?
1) N remains equal to mg
2) N is smaller than mg
3) N is larger than mg
4) None of the above
ConcepTest 6.14b Going in Circles II
A skier goes over a small round hill
1) Fc = N + mg
with radius R. Since she is in circular
2) Fc = mg – N
motion, there has to be a centripetal
3) Fc = T + N – mg
force. At the top of the hill, what is
4) Fc = N
Fc of the skier equal to?
5) Fc = mg
v
R
ConcepTest 6.14b Going in Circles II
A skier goes over a small round hill
1) Fc = N + mg
with radius R. Since she is in circular
2) Fc = mg – N
motion, there has to be a centripetal
3) Fc = T + N – mg
force. At the top of the hill, what is
4) Fc = N
Fc of the skier equal to?
Fc points toward the center of
the circle, i.e., downward in
this case. The weight vector
points down and the normal
force (exerted by the hill)
points up.
up The magnitude of
the net force, therefore, is:
Fc = mg – N
5) Fc = mg
v
mg
N
R
Follow-up: What happens when
the skier goes into a small dip?
ConcepTest 6.15c Going in Circles III
You swing a ball at the end of string
1) Fc = T – mg
in a vertical circle. Since the ball is
2) Fc = T + N – mg
in circular motion there has to be a
3) Fc = T + mg
centripetal force. At the top of the
4) Fc = T
ball’s path, what is Fc equal to?
5) Fc = mg
top
v
R
ConcepTest 6.15c Going in Circles III
You swing a ball at the end of string
1) Fc = T – mg
in a vertical circle. Since the ball is
2) Fc = T + N – mg
in circular motion there has to be a
3) Fc = T + mg
centripetal force. At the top of the
ball’s path, what is Fc equal to?
Fc points toward the center of the
circle, i.e. downward in this case. The
weight vector points down and the
4) Fc = T
5) Fc = mg
v
mg
T
tension (exerted by the string) also
points down.
down The magnitude of the
net force, therefore, is:
Fc = T + mg
Follow-up: What is Fc at the bottom of the ball’s path?
R