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Torque ....................................................................................................................................................... 196 Multiple Torques ................................................................................................................................... 197 Equilibrium and Torque ........................................................................................................................ 198 Teeter‐Totter Problems ........................................................................................................................ 200 Applying Torque ....................................................................................................................................... 203 Ladder Problems ................................................................................................................................... 205 Torque Wrap-up ........................................................................................................................210
AP Physics – Torque
Forces:
We’ve learned that forces change the velocity of an object. But what does it take to
change the angular velocity of a thing? Well, forces are involved, but the force has to be applied in
a special way. We call this special applied force a torque.
There are many ways to apply a force to a system that can rotate. In the drawing below we have a
turntable that can spin. If we just push sideways on the thing, as in the drawing to the left, we will
not make it spin. We basically would be trying to tip it over. But if we apply a force tangent to the
disc as in the drawing to the right, it will spin. This force is perpendicular to a radius of the circular
path. A force that is applied perpendicular to the circular path at some distance from the spin axis is
called a torque.
Torques change angular velocity. The symbol for torque is the Greek letter . Torque is given by
this equation:
  rF sin
r is the distance to the center of spin from where the force is applied. This
variable is often called the lever arm.
F sin  be the force component that is perpendicular to the lever arm.
F sin

F

r
If the angle  is 90, then the force is perpendicular to the lever arm, the sine is one, and the
equation for torque is simply:
  Fr
Note for some unknown reason, the force is written first and then the lever arm in this equation.
196
You can see that the unit for torque is going to be a newton meter (nm). We leave it like that.
This looks very similar to the unit for work, the joule, but it is quite different. So energy and work
are in joules and torque is left in newton meters.
Torque is a vector quantity.

125 N is applied to a nut by a wrench. The length of the wrench is 0.300 m. What is the torque?
  Fr  125 N  0.300 m  

A torque of 857 Nm is applied to flywheel that has a radius of 45.5 cm. What is the applied
force?
  Fr

37.5 Nm
F

r


1
 857 Nm 
 
0.455
m


1880 N
You push on the door as shown in the drawing. What is the torque?
  rF sin
330 N
  330 N 1.5 m  sin 55.0o

1.5 m
410 Nm
Multiple Torques:
55.0
What happens if two or more torques act on an object at the same time?
Two forces are applied to the object in the drawing to
the right. The object is free to rotate about the spin
axis. Both cause a torque.
F1 causes a CCW (counter clockwise) rotation around
the axis.
F1
spin axis
r1
r2
F2 causes a CW (clockwise) rotation around the axis.
If a torque causes a clockwise rotation, it is positive.
If a torque causes a counter clockwise rotation, it is
negative.
F2
197
The sum of the two torques would be:
  = 1   2
 F2r2  F1r1
Equilibrium and Torque: If an object is in angular equilibrium (sometimes called
rotational equilibrium), then it is either at rest or else it is rotating with a constant angular velocity:
If object is in rotational equilibrium, the net torque about any
axis is zero.
This means that the sum of the torques acting on the object must be zero.
=0
Static equilibrium exists when an object has no motion, either linear or angular. There are two
conditions which must exist in order to have your good old static equilibrium:
The net force must be zero and the net torque must be zero.
F=0
=0
This gives us some very powerful tools to solve static problems. We can analyze a system and look
at the forces acting on it, and we can also look at the torques that act on it. We’ll be able to do some
really cool stuff.
198

Two metal orbs are attached to a very lightweight rigid wire. They are suspended from a rigid
point on the overhead as shown. The system does not move. Calculate the distance from the
suspension line to the center of gravity on the
right sphere.
Since the system is at rest, the sum of the torques and
the sum of the forces must be zero.
Let’s look at a FBD and a drawing showing the two
torques:
4.0 kg
1.0 kg
F
45.0 cm
m1 g
r1
m2 g
Without using the torque equilibrium, we could not solve
the problem. The sum of forces would simply tell us that
the upward force would be equal to weight of the two
balls.
?
r2
m1 gr1
m2 gr2
Using torque, however, allows us to solve the problem.
All we have to do is add up d’ torques:
1   2  0
r2  
m1gr1  m2 gr2  0
1.0 kg  0.45 m 
4.0 kg

m2 gr2  m1gr1
r2  
m1 gr1
m2 g
0.11 m or 11 cm
Torque problems, as you have just seen, are fairly simple.
On average, we each have about 1500 dreams per year. The only
ones we remember are the ones interrupted by awakening and
then reviewed before they vanish: since dreams do not form
long-term memories (to our knowledge), any dream not reprocessed by the conscious mind is gone forever!!
Now we’ll do a classic teeter-totter beam problem.
199

A teeter-totter is in equilibrium as shown. The block on the left has a weight of 625 N. The
beam itself has a weight of 32.5 N. What is the mass of the second block?
625 N
?
This is a pretty simple problem, we can solve it
using the torques.
The sum of the torques must be zero:
=0
1.10 m
There are three torqes, 1 (from the 625 N rock) and
2 from the other rock. The weight of the beam
(Fbeam), even though it has a significant amount of
weight, does not cause a torque because the weight
acts at the CG of the beam which is also the center
of rotation. Thus the lever arm is zero.
1   2  0
F1
3.30 m
F2
Fbeam
 F1r1  F2r2  0
2 is positive (CW) and 1 is negative (CCW)
F1r1  F2 r2
F2 
F1r1
r2

685 N 1.10 m 
3.30 m
 228.3 N
To find the mass we use the second law:
F  ma m 
F
a

kg  m  1
 228.3

s 2  9.8 m

s2



 



23.3 kg
Another similar problem.

A 50.0 N seesaw supports two people who weigh 455 N and 525 N. The fulcrum is under the
CG of the board. The 525 N person is 1.50 m from the center. (a) Find the upward force n
exerted by fulcrum on the board. (b) Where does the smaller person sit so the seesaw is
balanced?
1.50 m
x
200
First, let’s draw a FBD
n
r2
r1
We know that the system is in static
equilibrium, so we can analyze the forces. In
the y direction, the sum of the forces must be
zero.
 Fy  0
(a) F1 and F2 are the weight of the two men,
FT is the weight of the teeter-totter, and n is
the normal force.
FT
F1
F2
We can write this out:
n  F1  F2  FT  0
Now we can solve for the normal force, this is the upward force exerted on the board by the support
stand.
n  F1  F2  FT
 0  525 N  455 N  50.0 N

1 020 N
(b) To find the distance the second man must be positioned from the center, we must analyze the
torques.
 y  0
r2 
F1r1
F2
1   2  0

525 N 1.50 m 
455 N

 F1r1  F2r2  0
F1r1  F2r2
1.73 N
201
Trivia Takes Over:
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Jimmy Carter was the first U.S. president to have been born in a hospital.
Eskimos use refrigerators to keep food FROM freezing.
The sentence "The quick brown fox jumps over the lazy dog." Uses every letter in the alphabet.
(developed by Western Union to test telex/twx communications)
In every episode of Seinfeld there is a depiction of Superman somewhere.
The average life span of a major league baseball: 7 pitches.
In the 1940s, the FCC assigned television's Channel 1 to mobile Services(two-way radios in
taxicabs, for instance) but did not Pre-number the other channel assignments. That is why TV
sets has channels 2 and up, but no channel 1.
The only 15 letter word that can be spelled without repeating a letter is uncopyrightable.
Hang On Sloopy is the official rock song of Ohio.
Did you know that there are coffee flavored PEZ?
The reason firehouses have circular stairways is from the days of yore when the engines were
pulled by horses. The horses were stabled on the ground floor and figured out how to walk up
straight staircases.
The airplane Buddy Holly died in was the "American Pie." (Thus the name of the Don McLean
song.)
When opossums are playing 'possum, they are not "playing." They actually pass out from sheer
terror.
The Main Library at Indiana University sinks over an inch every year because when it was built,
engineers failed to take into account the weight of all the books that would occupy the building.
The highest point in Pennsylvania is lower than the lowest point in Colorado.
Nutmeg is extremely poisonous if injected intravenously.
If you have three quarters, four dimes, and four pennies, you have $1.19.You also have the
largest amount of money in coins without being able to make change for a dollar.
No NFL team, which plays its home games in a domed stadium, has ever won a Superbowl.
The first toilet ever seen on television was on "Leave It to Beaver".
The only two days of the year in which there are no professional sports games (MLB, NBA,
NHL, or NFL) are the day before and the day after the Major League all-star Game.
Only one person in two billion will live to be 116 or older.
The name Wendy was made up for the book "Peter Pan."
In Cleveland, Ohio, it's illegal to catch mice without a hunting license.
It takes 3,000 cows to supply the NFL with enough leather for a year's supply of footballs.
There is an average of 178 sesame seeds on a McDonald's Big Mac bun.
The world's termites outweigh the world's humans 10 to 1.
Pound for pound, hamburgers cost more than new cars.
The 3 most valuable brand names on earth: Marlboro, Coca-Cola, and Budweiser, in that order.
When Heinz ketchup leaves the bottle, it travels at a rate of 25 miles per year.
202
AP Physics – Applying Torque
It is now time to go after some problems that are more complicated. You will find these to be a lot
of phun. Honest.

A uniform beam is supported by a stout piece of line as shown. The beam weighs 175 N. The
cable makes an angle of 75.0 as shown. Find (a) the tension in the cable and (b) the force
exerted on the end of the beam by the wall.
75.0
4.00 m
We can solve this problem by summing forces and adding up
torques.
T
R
75.0
First let’s draw a FBD:
We have three forces acting on the beam.
The weight of the beam which acts at the center of the
beam (its CG), FB.
FB
The tension in the cable, T.
And the force exerted by the wall on the beam, R. (The wall is pushing the beam up and
out.)
(a) Let us first sum the torques. The pivot point is the end of beam where it meets the wall.
Therfore R exerts no torque as its lever arm is zero. We only have two torques to deal with and, of
course, they add up to zero. Torque one is exerted by the tension in the cable and torque two is
caused by the weight of the beam. The force for this torque is applied at the CG, which is at the
center of the beam. Only the vertical component of the tension causes its torque so:
 cable  B  0
T
FB r
2 r sin 

r
Tr sin   FB    0
2
175 N
2 sin 75.0o

90.6 N
(b) Next we can sum up the forces:
x direction:
T cos  RX  0
y direction:
T sin   FB  RY  0
203
We can solve the x direction equation for RX:
  90.6 N  cos75.0o  23.4 N
RX  T cos
Next we solve for RY:
RY  FB  T sin 
 175 N   90.6 N  sin 75.0o
 87.5 N
We’ve found the x and y components for R, so now we can find the magnitude of the vector using
the Pythagorean theorem:
R  Ry 2  Rx 2

 87.5 N 2   23.4 N 2


90.6 N
A beam is supported as shown. The beam is uniform and weighs 300.0 N
and is 5.00 m long. A 635 N person stands 1.50 m from the building. (a)
What is the tension in the cable and (b) the force exerted on the beam by
the building?
We draw a FBD.
R
T
55.0
55.0
5.50 m
(a) Sum of torques:
 beam  man  cable  0
FB
Fm rm  FB rB  TrC sin   0
Fm
T
635 N 1.50 m    300.0 N  2.50 m 
sin 55.0  5.00 m
o

T
Fm rm  FB rB
 sin   rC
416 N
(b) We can resolve R and T into components and then sum the forces in the x and y direction. All
forces must add up to equal zero.
RX  T cos  0
Ry  T sin   FB  Fm  0
204
RX  T cos  0
RX  T cos
Ry  T sin   FB  Fm  0
  416 N  cos55.00
Ry  FB  Fm  T sin 
Ry  300.0 N  635 N   416 N  sin 
R  Ry 2  Rx 2

 238 N
 624 N
 624 N 2   238 N 2

668 N
Fabulous Ladder Problems: Ladder problems are very popular.
The basic idea is that
you have a ladder leaning against a wall (which is usually frictionless). The ladder is held in place
by the friction between its base and the deck it rests upon. We’re given the situation and then
required to figure out various things – the angle the ladder makes with the deck, the friction force,
the coefficient of friction, the force exerted on the top of the ladder by the wall, &tc.
Let’s go ahead and do a simple problem.

A uniform 250.0 N ladder that is 10.0 m long rests against a frictionless wall at
an angle of 58.0, the ladder just keeps from slipping. (a) What are the forces
acting on the bottom of the ladder? (b) What is the coefficient of friction of the
bottom of the ladder with the ground?
10.0 m
Draw a FBD.
The forces acting on the ladder are: the weight of the ladder FL, the frictional force f,
The force the deck pushes up on the ladder with F1, and the force exerted by the wall
on the top of the ladder F2.
58.0
Now we look at the forces acting on ladder - they have to add up to zero.
F2
 Fy  0
F1  FL  0
 Fx  0
f  F2  0
F1  FL

250 N
F1
f  F2
We have to find either F2 or else f. But we need more info, don’t we? You bet we
do. Blessed by good fortune as we are, we instantly recognize that we can make
use of the torque equilibrium deal.
First we make a drawing showing all the torques acting on the ladder. (Actually
we’re only looking at the forces that are perpendicular to the lever arm.)
FL

f
205
The pivot point is the base of the ladder.


Neither the friction or F1 cause a torque as their lever arm is zero.
F2
The weight of the ladder causes a CCW torque.
The lever arm from F2 causes a CW torque.
12.0 m

The torques add up to zero.
  0
6.00 m
 ladder   wall  0
FL
The angle  is, using geometry clearly going to be:
  90o  
 90o  58o
 32o
F2 cos  d 2  FL cos d1  0
The torques are:
F cos d1
F2  L
cos  d 2


250.0 N cos 58.0o  6.00 m 
Solve for F2:
 78.1 N
cos 32.0o 12.0 m 
The frictional force (the other force acting at the base of the ladder is therefore:
f 
78.1 N
(b) Find the coefficient of friction:
f n

f
n

78.1 N
250.0 N

0.312
Whew! Can we make it worse? You bet.
The longest recorded flight of a domestic chicken is
13 seconds.
206

A 15 m, 500.0 N uniform ladder rests against a frictionless wall. It
makes 60.0 angle with the horizontal. Find (a) the horizontal and
vertical forces on the base of the ladder if an 800.0 N fire fighter is
standing 4.0 m from the bottom. If the ladder is on the verge of
slipping when the fire fighter is 9.0 m from the bottom of the
ladder, (b) what is the coefficient of static friction on the bottom?
15 m
 FY  0
F1  FL  FF  0
4.0 m
F1  FL  FF
60.0
F1  500 N  800 N
F1  1300 N
 FX

F2
 F2
F2  f  0
0
f  F2
F1 FL 
Let’s look at torque to find F2:
  0

Pivot point is at the base of the ladder:
  90o  
 90o  60o
 30o
 2  L  F  0
FL

f
F2 cos  r2  FL cos rL  FF rF  0
F2 
F2 
FF
FL cos rL  FF rF
cos  r2

FF
500.0 N cos60.0o  7.50 m   800.0 N cos60.0o  4.00 m 
F2  270 N
cos30.0o 15.0 m 
so
f  270 N
207
So the force up is
F1 
1 300 N
The horizontal frictional force is:
f 
270 N
(b) When the fire man is at 9.0 m (we’ll figure this from the bottom of the ladder), then
We can use the same equation as we used to find F2 since the only thing that has changed is the
distance of the firefighter from the bottom of the ladder.
F2 
F2 
FL cos rL  FF rF
cos  r2
F2
500.0 N cos 60.0o  7.50 m   800.0 N cos 60.0o  9.00 m 
cos 30.0o 15.0 m 
F1
f  421 N
We can now find the coefficient of static friction for the bottom of the ladder
and the deck.
f  s n
s 
f
n

421 N
1300 N

0.32
FF
FL 
f
All there is to it.
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A superstition of yore involved a young unmarried woman taking a sprig of rosemary and a
sprig of thyme, sprinkling them three times with water, and placing each herb in a shoe. She
would then put the shoes at the foot of her bed. If she followed this ritual, she was sure to
dream of her next true love.
A U.S. federal law passed in 1994 requires that plastic six-pack ring holders disintegrate after
use. This is to prevent birds and marine life from getting tangled in them and strangling.
About 10 percent of the workforce in Egypt is under 12 years of age. Although laws
protecting children are on the books, they are not well enforced, partly because many povertystricken parents feel forced to send their children out to help support the family.
About a hundred years ago, it was the custom of sailors to put a tattoo of a pig on one foot
and a rooster on the other to prevent drowning.
According to “Emily Post’s Etiquette,” a tip at a family restaurant should be 15% of the bill
without tax. For a buffet a 10% tip is sufficient, but never leave less than a quarter even if you
only have a cup of coffee.
A in Providence, Rhode Island law makes it illegal to sell a toothbrush on the Sabbath Yet,
these same stores are allowed to sell toothpaste and mouthwash on Sundays.
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According to the Recruitment Code of the U.S. Navy, anyone "bearing an obscene and
indecent" tattoo will be rejected.
Almost every weekday morning, free Kleenexes are handed to the commuters in front of
Japan's rail and bus stations. The tissues are distributed by workers of the companies whose
messages and advertisement are printed on the packages. This is done because most public
bathrooms do not have paper towels or toilet paper.
An old Ethiopian tradition required the jewelry of a bride be removed after her wedding. Its
likeness would then be tattooed on her skin.
An old folk custom for selecting a husband from several suitors involved taking onions and
writing each suitor's name individually on each. Then all the onions were put in a cool dark
storeroom. The first onion to grow sprouts would determine which man the undecided
maiden should marry.
An old law in Delaware allowed public whipping for 24 crimes--and more than 1,600
people were publicly whipped.
Bad weather on the way to the wedding is thought to be an omen of an unhappy marriage;
some cultures, however, consider rain a good omen. Cloudy skies and wind are believed to
cause stormy marriages. Snow, on the other hand, is associated with fertility and wealth.
Baked goods made on Good Friday were thought to contain many virtues. A cross bun kept
from one Good Friday to the next was considered a lucky charm. It was not supposed to
grow moldy, and it was used as a charm against shipwreck. "Good Friday bread," when
hung over the chimney, was supposed to guarantee that all bread baked after that would be
perfect.
Because of heavy traffic congestion, Julius Caesar banned all wheeled vehicles from Rome
during daylight hours.
Before the enactment of the 1978 law that made it mandatory for dog owners in New York
City to clean up after their pets, approximately 40 million pounds of dog excrement were
deposited on the streets every year.
Body language differs from one country to another. For instance, grasping one's ears is a
sign of repentance or sincerity in India. A similar gesture in Brazil – holding the lobe of
one's ear between the thumb and forefinger – signifies appreciation.
By photographing the eyes of murder victims, early students of forensics hoped to see a
reflection of the murderer lingering in the victim’s eyes.
In Atlanta, Georgia, it is illegal to tie a giraffe to a telephone pole or street lamp.
In Breton, Alabama, there is a law on the town's books against riding down the street in a
motorboat.
In Britain, a horseshoe was not thought to be lucky traditionally. It was thought to be a
guardian against all evil forces, as inhabitants of the spirit world were supposed to flee from
the sight of cold iron.
In Britain, the law was changed in 1789 to make the method of execution hanging. Prior to
that, burning was the modus operandi. The last female to be executed by burning in
England was Christian Bowman. Her crime was making counterfeit coins.
In Clarendon, Texas, there is a law on the books that lawyers must accept eggs, chickens, or
other produce as payment of legal fees.
In colonial America, tobacco was acceptable legal tender in several Southern colonies, and
in Virginia, taxes were paid in tobacco.
209
AP Physics – Torque Wrapup
We got one really fine equation. Here it is:
  rF sin 
This is, of course, the good old torque equation. Recall that if the applied force is perpendicular to
the lever arm (r), then the torque equation becomes:   rF
Here’s the stuff that master you must (as Yoda would say).
You should understand the concept of torque so you can:
a. Calculate the magnitude and sense of the torque associated with a given force.
Use the equation, duh.
b. Calculate the torque on a rigid body due to gravity.
Use the equation. The force acting on the thing would be its weight, right?
You should be able to analyze problems in statics (meaning that nothing is moving) so you can:
a. State the conditions for translational and rotational equilibrium of a rigid body.
Translational equilibrium occurs when the object is either at rest or moving with a
constant velocity - the whole straight-line motion thing we studied to exhaustion.
Rotational equilibrium takes place when an object is either at rest or rotating at a constant
angular velocity.
b. Apply these conditions in analyzing the equilibrium of a rigid body under the
combined influence of a number of coplanar forces applied at different locations.
Torque is pretty straightforward stuff. The main idea is that the sum of the torques (i.e., the net
torque) acting on any object that is either at rest or rotating at a constant angular velocity must
be zero.
There are three types of problems that one can expect to have to deal with. You got your
basic teeter totter problem, the beam sticking out of the wall problem, and your ladder
problems.
AP Physics Test Problems:
Ah, so sad, the Physics Kahuna was unable to find any free response torque problems.
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