07torque.doc

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PHYS 2211L LAB 7

Torque & Equilibrium

Diagram 1

o be in translational equilibrium, the sum of the forces on the body must be ero, &ust as

for a single particle. o be in rotational equilibrium, the turning effectof all the forces

must also be ero. We call the turning effect of a force a torque(). 'or a rigid body tobe in static equilibrium we must have both!

() "=i

iF and =i

i "

Torque

he torquethat a force exerts on a rigid body depends on three things! the magnitude of

the force, the direction of the force, and where on the body the force is applied. $efer to

iagram , which depicts an arbitrary two%dimensional ob&ect with a fixed point (a metalplate nailed to the wall, say) being acted upon by a force applied at a particular point in

the ob&ect. We define torque as!

Fr=

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PHYS 2211L LAB 7


his is a vector that points in the direction of the axis of rotation. he vector rruns fromthe point about which the torque is to be evaluated to the point of application of the force

F. orques can be evaluated about any point, but we usually choose a convenient pointsuch as the center of mass or a point on the axis of rotation.

In the diagram below, using the axis of rotation as our reference point, the magnitude of

the torque can be expressed as

() F=

whereFis the magnitude of the force and is called the lever arm. he lever arm is the

perpendicular distancefrom the axis of rotation to the line of action of the force. 0ote

that in the diagram, sinr= , where ris the distance from the axis of rotation to the

point of application of the force. he angle depends on the direction of the force.

Diagram 2

he units of torque are newton-meters.

In this experiment we will use a meter stick (essentially a one%dimensional ob&ect)balanced on a support as the rigid body. #t equilibrium, the forces will always be

perpendicular to the stick, so that =0 degrees, andsin=!, depending on

whether the force is upward (1) or downward (%), so that except for the sign we canignore the angular dependence of the torques. (2ut do not ignore it in general.)

May 05 -3

l

r

F

"xis #out

of page$


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PHYS 2211L LAB 7


4ur set up is indicated in iagram 5. he triangle represents the support6 the axis ofrotation is at the tip of the triangle, perpendicular to the page. If we exert forces on the

meter stick (F!andF%in the diagram), then the lever arms associated with each force () and . ) are as indicated 7 the distances along the stick from the axis of rotation to

the point of application of the force.

We can set up a coordinate system along the meter stick, with position coordinatex

measured by the stick itself. We can express the lever arm for any force acting on the

meter stick as!

(5a) "xxii = (upward force)

(5b) ii xx = " (downward force)

wherex0is the position of the axis of rotation andxiis the point where forceFiacts.

Diagram 3

$otations have directions! they can be clockwise or counterclockwise. It is customary to

treat the counterclockwise direction as positive and the clockwise direction as negative.

orques are also positive or negative, depending on the direction of the rotation theyengender. hus, two or more torques can sum to ero, resulting in rotational equilibrium.

May 05 /"

l2

F2

F1

l1

*upport 8 axis ofrotation (out of

page)

FN


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PHYS 2211L LAB 7


he expressions for the lever arm (5a and 5b) take the direction of the torque intoaccount. We see that a downward force exerted at the low end of the stick (xi& x0)

results in a positive (counterclockwise) torque and a downward force at the high end (xi' x0) results in a negative (clockwise) torque. 9pward forces would reverse thesedirections.

#lso note that when the force is applied atthe axis of rotation (xi= x0), the lever arm isero and so is the torque. hus the support itself (which exerts an upward force on the

stick, indicated byF(in iagram 5) does not exert a torque.

A Meter Stick Mass Scale: te Metod o! Moments

We can use the principle of rotational equilibrium to weigh ob&ects with our meter stick

(a triple%beam balance works on &ust this principle). iagram : illustrates the set%up.

Diagram "

We first place the meter stick on the support stand at the stick;s balance point (this point

is called the center of gravity7 we will discuss this further below). When the stick is

balanced (horiontal and not rotating), it must be in rotational equilibrium! the sum of thetorques on it is ero.

We place a reference mass m!

at a known location on the stick. he weight of m!

exerts atorque on the stick and makes it rotate away from equilibrium!

))))) gm) ==

with )") xx = , the position of the support minus the position of m.

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l2

l1

#1

#2


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PHYS 2211L LAB 7


We then place our unknown mass (m%) on the other side of the stick and ad&ust itsposition until the stick balances again. he unknown mass exerts a torque in the opposite

sense on the stick!

..... gmg) ==

with .". xx = , the position of the support minus the position of the unknown mass.

When the meter stick returns to equilibrium, we must have!

"..))

=+= gmgmi

i

so that gmgm )).. = , or

(:).

))

.

mm =

his, of course, is nothing more than the


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PHYS 2211L LAB 7


weight of the stick acts as if it were concentrated at the balance point. In fact, we canprove that for anyrigid body there is a point at which, for the purposes of calculating

torques, all of its weight seems to act. his point is called the center of gravity.

#eiging te meter stick (it te meter stick

his will be brought home more clearly in the next procedure. $efer to iagram >,which depicts a meter stick in equilibrium, supported notat its center of gravity. he

center of gravity is indicated by the dark spot and has purposely been placed off%center to

emphasie the fact that the center of gravity is not necessarilythe center of the stick. (Infact, it is unlikely that the center of gravity for any of the meter sticks we will use will

turn out to be in the exact middle of the stick.)

he axis of rotation is above the triangular support. #t equilibrium, all torques must sumto ero, so!

"))

=+ gmgm ss

where msand s are the mass and lever arm of the stick itself, with ss xx = " , the

position of the axis of rotation minus the position of the center of gravity of the stick. We

can determine the position of the center of gravity of the stick by first balancing the stick

with no extra weight on it. hen we can solve the above equation for ms!

(>)s

s

m

m

))

=

which is exactly like equation (:) except for a change of subscripts. hus the meter stickacts as if it were a mass hanging off itself?

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PHYS 2211L LAB 7


Diagram )

*redicting an Equilibrium *osition

In our third procedure, we will hang two masses from the meter stick and use the

equilibrium principle topredictwhere the support must be placed so that the system will

balance. he set%up is illustrated in iagram @.

Diagram +

his time we will place two specified masses, m!and m%,atfixedpositions on the stick.

We want to find where the support must be placed (x0) so that the system is inequilibrium. It is left to you to show that

May 05 /:

c.g.

m1 #

s

#1

#1

#2

#s

c.g

.

,1

,-

,s

,2


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PHYS 2211L LAB 7


(@)s

ss

mmm

xmxmxmx

++

++=

.)

..))

"

Note! do not assume thatx0will be to the left of the center of gravity, as in the

illustration. It depends entirely on the masses and their location and could go either way.

he positionx0is the center of gravity for the meter stick and two masses. 0ote that it is

also the center of massfor the system. he center of mass for any system of masses isdefined as!

M

xm

x iii

cm

=

whereMis the total mass of the system. In our case then, the center of gravity is the

same as the center of mass. his will be the case whenever the gravitational field is

uniform in magnitude and direction, but it is not true in general. 'or instance, the centerof mass of the Aarth%Boon system about the *un is not the same as its center of gravity,

since the forces on the Aarth and Boon are not parallel.

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Torque & Equilibrium Lab 7

Experiments

We will use a meter stick balanced on a support to measure the torques exerted byhanging masses. #s described in the heory section, a torque can be calculated from

F=

(torque equals force times lever arm.) he forces will be the weights of the hanging

masses. he lever arms should be measured from the point of support to the hanging

mass. orques can be positive or negative6 see equations 5a 8 5b, above for a usefulconvention for the sign of the lever arms in this lab.

orques should be expressed in units of 0ewton%meters. 'or the method of moments,

however, the results depend only on the ratio of the masses involved, so one can work inunits of grams and centimeters and no unit conversion is necessary.

he experiments can be outlined as follows!

. 'ind the center of gravity of the meter stick.

. Weigh the unknown mass by the method of moments.5. Weigh the meter stick by the method of moments.

:. Credict the balance point when masses are hung from the meter stick and

calculate the individual torques.

Equipment

Beter stick

*upport stand

Dnife edge clamps with

hanging bail ()

Dnife edge clamp without

bail ()

*tring or small mass hangers

"" gram mass

>" gram mass

9nknown mass

riple%beam balance

Procedures

Note! It is important that you measure all positions to the nearest millimeter, as accurately as possible.


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Experiments

'ind the masses of the knife%edge clamps and any mass hangers used in the

experiment. $ecord your data.

Note! Deep track of which clamp is which. 0ot all clamps are alike and you must take the massof the clamps and hangers into account in your calculations.

2. Find te center o! gra&it' o! te meter stick.

Insert the meter stick into the center clamp (the knife%edge clamp that has no

hanging bail) and set the knife%edges of the clamp into the slots in the supportstand. he tightening screw on the clamp should be on the down side.

With the screw loose, shift the meter stick back and forth through the clamp until

the stick is balanced. ighten the screw.

$emove the stick from the stand and read the position of the knife%edges on theclamp. (he edge of the cut%away sections on either side of the clamp line up with

the knife edges and is easier to read.) his is the center of gravity of the meter

stick. $ecord this. Eeave the clamp tightened so that it stays at the center of

gravity.

2. #eig te unkno(n mass b' te metod o! moments.

Clace the meter stick back in the support stand. $ecord the position of the support

stand (the axis of rotation) asx0.

Clace a knife%edge clamp (with hanging bail down) at 5"." cm on the stick. +ang

a ""%g mass from the clamp using a piece of string or a small mass hanger.$ecord the totalmass hanging at this position as m!. $ecord the position asx!.

+ang the unknown mass from another knife%edge clamp on the other side of the

support stand.

Bove the clamp with the unknown mass back and forth until the system balances.

$ecord this position asx%.

Anal'sis:

raw a diagram of the meter stick and masses. Indicate the lever arms and the

forces for each mass on the diagram.

*et up the torque equations for the system at equilibrium ( "=i

i ) and solve it

algebraically for the unknown mass.

9se your equation to calculate the total mass hanging atx%. Falculate the

unknown mass by subtracting the mass of the clamp andGor hanger. $ecord your

result.


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Experiments

Falculate the range of error for the unknown mass, based on the uncertainties in

the lever arms, , and the uncertainty in the reference mass, )m . *ince we

measured the lever arms to the nearest millimeter, we can estimate = ! mm.*imilarly, estimate )m as one gram (or less if you measured it to greater

precision). Which uncertainty has the greater effectH

'ind the percent error between your calculated value for the unknown mass and

that determined using the triple%beam balance. oes the measured value fallwithin your range of errorH

3. Determine te meter stick mass b' te metod o! moments.

$emove the unknown mass and its hanger. Eeave m!where it is.

Eoosen the screw of the center clamp and slide the meter stick back and forththrough it until the system balances. ighten the screw and read the position of

the center clamp. $ecord this asx0, the position of the axis of rotation.

Anal'sis

raw a force diagram of the set%up with the forces and lever arms indicated.

$emember that the weight of the meter stick acts at its center of gravity, which

you determined in Crocedure .

*et up and solve the torque equation for the mass of the meter stick. Falculate the

mass of the stick and record your result.

Falculate the range of error in the mass of the stick based on the uncertainties inthe reference mass and the lever arms.

Falculate the percent error between your calculated value and that determined by

the triple%beam balance.

". *redict te balance /oint !or t(o masses on te stick.

9se equation (@) to predict the balance point when "" grams are hung at 5"." cm

and >" grams are hung at @>." cm. (hese masses do not include the masses of the

clamps andGor hangers. ou must include these extra masses in your calculations.)$ecord your prediction.

+ang these masses and determine the balance point experimentally. $ecord the

experimental result forx0.

Anal'sis

iagram the set%up, indicating all forces and lever arms. Include in the diagram any

forces that exert no torques.


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Experiments

erive equation (@) using the definitions of torque and lever arm and the fact that the

system is in equilibrium ( "=i

i ).

ake the percent difference between your prediction and the experimental result.

$eferring to your diagram, decide which forces create counterclockwise torques and

calculate the sum of these torques.

$eferring to your diagram, decide which forces create clockwise torques and

calculate the sum of these torques.

ake the percent difference between the absolute values of the clockwise sum and the

counterclockwise sum.

*ptional "nalysis

9se your data and your diagram to recalculate the net torque on the system using

the end of the meter stick as the point of reference for calculating torques. hat

is, take the end of the stick as the assumed axis of rotation (x0= 0) and calculatethe lever arms from this point. his time you must include the torque created by

the upward force of the support. (he support;s location has not changed 7 only

our reference point has changed. Why did we not consider this force beforeH)

*how algebraically that if the system is in equilibrium, the net torque will sum to

ero about any point. (#fter all, if the system is in equilibrium it is not rotating

and there is no


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ata

Note: +our data should ,e recorded in your la, note,oo. The following is a guideonly.

Preliminar! ata

0nkno(n mass $lam/anger masses:

1.

Mass o! meter stick: 2.

3.

$enter o! gra&it' ".

meter stick4

).

+.

"n#no$n %ass %et'od o( %oments)

m1

,-

,1 )

,2 .

m2 unkno(n mass

5m2 6 error


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ata

%eter Stic# %ass %et'od o( %oments)

m1

,-

,1 )

,s s

ms 5ms

6 error

Balance Point $it' T$o %asses

m1 m2

,1 ,2

*redicted ,- E,/erimental ,-

6 di!!erence

$lock(ise torques $ounterclock(ise torques

6 di!!erence (use absolute values)


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07torque.doc

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