Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium
Chapter 5. Torque and Rotational Equilibrium Unit Conversions
5-1. Dra and label the moment arm o! the !orce " about an a#is at point $ in "i%. 5-11a. &hat
Moment arms are drawn perpendicular to action line:
r $ ( )* !t+ sin *5
*5
r $
is the ma%nitude o! the moment arm'
$
0 !t
/
*5
r /
* !t F
r $ ( .5 !t
5-*. "ind the moment arm about a#is / in "i%. 11a. ) See figure above.+ r / ( )0 !t+ sin *5
r / ( 1.* !t
5-0. Determine the moment arm i! the a#is o! rotation is at point $ in "i%. 5-11b. &hat is the ma%nitude o! the moment arm' r / ( )* m+ sin 6
F
r / ( 1.0 m
5-. "ind the moment arm about a#is / in "i%. 5-11b. r / ( )5 m+ sin 0
6
*m $
5m
/
0
r $
r /
r / ( *.5 m
Torque
5-5. 2! the !orce " in "i%. 5-11a is equal to lb, hat is the resultant torque about a#is $ ne%lectin% the ei%ht o! the rod. &hat is the resultant torque about a#is /' Counterclockwise torques are positive, so that τ $ is - and τ / is 3.
)a+ τ
$
( ) lb+).5 !t+ ( -6.6 lb !t
)b+ τ / ( ) lb+)1.* !t+ ( 311 lb !t
5-6. The !orce " in "i%. 5-11b is 4 and the an%le iron is o! ne%li%ible ei%ht. &hat is the resultant torque about a#is $ and about a#is /' Counterclockwise torques are positive, so that τ $ is 3 and τ / is -. )a+ τ
$
( ) 4+)1.0* m+ ( 360 4 m
06
)b+ τ / ( ) 4+)*.5 m+ ( -1 4 m
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium
5-. $ leather belt is rapped around a pulley * cm in diameter. $ !orce o! 6 4 is applied to " the belt. &hat is the torque at the center o! the sha!t' r ( 7D ( 1 cm
τ
( )6 4+).1 m+ ( 36. 4 m
5-. The li%ht rod in "i%. 5-1* is 6 cm lon% and pi8oted about point $. "ind the ma%nitude and si%n o! the torque due to the * 4 !orce i! θ is )a+ , )b+ 6, )c+ 0, and )d+ . τ
( )* 4+ ).6 m+ sin θ
!or all an%les9
)a+ τ ( 1* 4 m
)b+ τ ( 1 4 m
)b+ τ ( 6 4 m
)d+ τ (
$ r
6 cm
* 4 θ
θ
5-. $ person ho ei%hs 65 4 rides a bicycle. The pedals mo8e in a circle o! radius cm. 2! the entire ei%ht acts on each donard mo8in% pedal, hat is the ma#imum torque' τ
( )*5 4+). m+
τ
( *6 4 m
5-1. $ sin%le belt is rapped around to pulleys. The dri8e pulley has a diameter o! 1 cm, and the output pulley has a diameter o! * cm. 2! the top belt tension is essentially 5 4 at the ed%e o! each pulley, hat are the input and output torques' 2nput torque ( )5 4+).1 m+ ( 5 4 m :utput torque ( )5 4+).* m+ ( 1 4 m
Resultant Torque
5-11. &hat is the resultant torque about point $ in "i%. 5-10. 4e%lect ei%ht o! bar. Στ
( 3)0 4+)6 m+ - )15 4+)* m+ - )* 4+)0 m+ τ
( . 4 m, Countercloc;ise.
0
15 4 m 0 4
*m
0m
$ * 4
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium
5-1*. "ind the resultant torque in "i%. 5-10, i! the a#is is mo8ed to the le!t end o! the bar. 15 4 Σ τ
( 3)0 4+)+ 3 )15 4+) m+ - )* 4+) m+ τ
$ m
( -1* 4 m, countercloc;ise.
0m
*m
* 4
0 4
5-10. &hat hori
about point / equal to
( P )* m+ = ) 4+)5 m+ )sin 0 + ( * P ( * 4
5m
/
0
*m
P ( 1 4
r /
P
5-1. To heels o! diameters 6 cm and * cm are !astened to%ether and turn on the same a#is as in "i%. 5-1. &hat is the resultant torque about a central a#is !or the shon ei%hts' r 1 ( 7)6 cm+ ( .0 m r * ( 7)0 cm+ ( .15 m τ
( )* 4+).0 m+ = )15 4+).15 m+ ( 0.5 4 m
( 0.5 4 m, cc
τ
5-15. >uppose you remo8e the 15-4 ei%ht !rom the small heel in "i%. 5-1. &hat ne ei%ht can you han% to produce
( )* 4+).0 m+ = & ).15 m+ (
& ( 4
5-16. Determine the resultant torque about the corner $ !or "i%. 5-15. Στ
( 3)16 4+).6 m+ sin - ) 4+).* m+ Στ
( 61. 4 m = 16. 4 m ( 5. 4 m τ
? @ A
? @AA
B @? ? ?
? @
? ?
( 5. 4 m
R
5-1. "ind the resultant torque about point C in "i%. 5-15. Στ
( - ) 4+).* m+ ( -16 4 m
* cm
4 6 cm
16 4
r
C
0
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium 5-1. "ind the resultant torque about a#is / in "i%. 5-15. F x = 16 cos Σ τ
F y (
/
4
* cm
16 sin
16 4
"y
6 cm
( = )1*0 4+).* m+ 3 )10 4+).6 m+ ( 0.* 4 m
"#
Equilibrium
5-1. $ uni!orm meter stic; is balanced at its midpoint ith a sin%le support. $ 6-4 ei%ht is suspended at the 0 cm mar;. $t hat point must a -4 ei%ht be hun% to balance the system' Στ
* cm
!he "#$ 4 weight is %# cm from the axis&
( )6 4+)* cm+ = ) 4+ x (
x
6 4
4
x ( 1* 4 cm or x ( 0 cm9 !he weight must be hung at the '#$cm mark( 5-*. &ei%hts o! 1 4, * 4, and 0 4 are placed on a meterstic; at the * cm, cm, and 6 cm mar;s, respecti8ely. The meterstic; is balanced by a sin%le support at its midpoint. $t hat point may a 5-4 ei%ht be attached to produce equilibrium. Στ
( )1 4+)0 cm+ 3 )* 4+)1 cm+
1 cm 0 cm x
= )0 4+)1 cm+ = )5 4+ x = # 1 4 * 4
5 x ( )0 3 * =0+ or x ( cm
0 4
54
!he )$ 4 weight must be placed at the *#$cm mark
5-*1. $n -m board o! ne%li%ible ei%ht is supported at a point * m !rom the ri%ht end here a 5-4 ei%ht is attached. &hat donard !orce at the must be e#erted at the le!t end to produce equilibrium' Στ = 0:
F
" )6 m+ = )5 4+)* m+ (
6 " ( 1 4 m
or
" ( 16. 4
6m
*m
5 4
0
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium
5-**. $ -m pole is supported at each end by hunters carryin% an -4 deer hich is hun% at a point 1.5 m !rom the le!t end. &hat are the upard !orces required by each hunter' Στ
/ ( 1* 4 or / ( 0 4 Σ
$ 1.5 m
( $ )+ = ) 4+)1.5 m+ 3 / ). m+ (
"y ( $ 3 / = lb (
/
*.5 m
$#is
4
$ ( 5 4
5-*0. $ssume that the bar in "i%. 5-16 is o! ne%li%ible ei%ht. "ind the !orces " and $ pro8ided the system is in equilibrium. Στ Σ
F
( ) 4+)1.* m+ = " ). m+ ( " ( 1 4
"y ( " = $ = 4 ( $ ( 1 4 = 4 ( *6. 4
0 cm
$#is
cm
4
$
" ( 1 4, $ ( *6. 4 5-*. "or equilibrium, hat are the !orces " 1 and "* in "i%. 5-1. )4e%lect ei%ht o! bar.+ Στ
F1
( ) lb+)5 !t+ = " * ) !t+ = )* lb+)5 !t+ (
!t
5 !t
"* ( .5 lb Σ "y ( "1 = "* = * lb = lb ( "1 ( "* 311 lb ( .5 lb 3 11 lb, " 1 ( 1 lb
$#is
1 !t
F2
lb
* lb
5-*5. Consider the li%ht bar supported as shon in "i%. 5-1. &hat are the !orces e#erted by the supports $ and /' Στ
/ ( .1 4
Α
A 0m
B *m
6m
( / )11 m+ = )6 4+)0 m+ = ) 4+) m+ ( $#is
Σ
"y ( $ 3 / = 4 = 6 4 (
$ ( 1 4 = / ( 1 4 = .1 4
6 4
4
/ ( 5. 4
5-*6. $ -belt is rapped around a pulley 16 in. in diameter. 2! a resultant torque o! lb !t is required, hat !orce must be applied alon% the belt' F
R ( 7)16 in.+ ( in. R ( )1* !t+ ( .66 !t
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium " ).66 !t+ ( lb !t
τ =
" ( 6. lb
5-*. $ brid%e hose total ei%ht is 5 4 is * m lon% and supported at each end. "ind the !orces e#erted at each end hen a 16-4 tractor is located m !rom the le!t end. Στ
Α
B
( / )* m+ = )16 4+) m+ = )5 4+) 1 A m+ ( m
/ ( * 4
Σ
"y ( $ 3 / = 16 4 = 5 4 (
$ ( 61 4 = / ( 61 4 = * 4
*m
1 m
$#is 16 4
/ ( 0*1 4
5 4
5-*. $ 1-!t plat!orm ei%hin% lb is supported at each end by stepladders. $ 1-lb painter is located !t !rom the ri%ht end. "ind the !orces e#erted by the supports. Στ
/ ( 1* lb
Α
( /)1 !t+ = ) lb+)5 !t+ = )1 lb+) 6 !t+ ( A
Σ
"y ( $ 3 / = lb = 1 lb (
$ ( ** lb = / ( ** lb = 1* lb
B
5 !t
1 !t
!t
$#is lb
$ ( *. lb
1 lb
5-*. $ hori
Ty
!
0m
( = 0 ( 50 Ty ( T sin 50
φ Στ
Α
$#is
( )T sin 50+).5 m+ = ) 4+)0 m+ = )1* 4+)6 m+ (
0.5 T ( 1* 4 3 * 4
B 1.5 m
Ty 1.5 m
4
1* 4
T ( *0 4
5-0. &hat are the hori
"# ( F = T# ( F = T cos 50 (
Σ
"y ( 3 T sin 50 = 4 = 1* 4 ( ( 16 4 = )*0 4+ sin 50 ( -*6 4
!hus, the components are9
F ( )*0 4+ cos 50 F ( 1 4
F ( 1 4 and ( -*6 4. !he resultant of these is:
1
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium + =
* + - * = 10
4 tanφ (
-*6 φ ( 1. > o! E 1
R ( 10 4, 0.*
Center o" #ravit$
5-01. $ uni!orm 6-m bar has a len%th o! 6 m and ei%hs 0 4. $ 5-4 ei%ht is hun% !rom the le!t end and a *-4 !orce is hun% at the ri%ht end. Fo !ar !rom the le!t end ill a sin%le F
upard !orce produce equilibrium' x Σ
"y ( " = 5 4 = 0 4 = * 4 ( " ( 1 4
Στ
$#is
( " x = )0 4+)0 m+ = )* 4+)6 m+ (
0m
0m
5 4
* 4
0 4
)1 4+ x ( *1 4 m x ( *.1 m 5-0*. $ -4 sphere and a 1*-4 sphere are connected by a li%ht rod * mm in len%th. Fo !ar !rom the middle o! the -4 sphere is the center o! %ra8ity'
F
x Σ
"y ( " = 4 = 1* 4 ( " ( 5* 4
Στ
* mm 4
( " x = ) 4+)+ = )1* 4+).* m+ (
1* 4
)5* 4+ x ( *. 4 m x ( .6* m or x = 6.* mm 5-00. &ei%hts o! *, 5, , and 1 4 are hun% !rom a 1-m li%ht rod at distances o! *, , 6, and m !rom the le!t end. Fo !ar !rom the le!t in is the center o! %ra8ity'
F
x Σ
"y ( " = 1 4 = 4 = 5 4 = * 4 ( " ( *5 4
*m
" x = )* 4+)* m+ = )5 4+) m+ = ) 4+)6 m+ = )1 4+) m+ ( *4
)*5 4+ x ( 15* 4 m x ( 6. m
*m
*m
*m *m
5 4 4 1 4
5-0. Compute the center o! %ra8ity o! sled%ehammer i! the metal head ei%hs 1* lb and the 0*in. supportin% handle ei%hs * lb. $ssume that the handle is o! uni!orm construction and ei%ht. Σ
"y ( " = * lb = 1* lb (
x F
16 in. 16 in.
" ( 1 lb 1* lb
*
* lb
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium " x = )1* lb+)+ = )* lb+)16 in.+ ( )1 lb+ x ( 0* lb in.
Fx ( 0* lb in.
x ( *.* in. !rom head.
Challen%e Problems
5-05. &hat is the resultant torque about the hin%e in "i%. -*' 4e%lect ei%ht o! the cur8ed bar. τ
6 cm
( ) 4+).6 m+ = )* 4+). m+ sin ( . 4 m = 51. 4 m
τ
4
r
( = 0.* 4 m
* 4
cm
5-06. &hat hori
F
r
4 5 6 cm
( - 0.* 4 m.
* 4
cm
( , then torque of 30.* 4 m must be added .
" ).6 m+ cos ( 30.5 4 m " ( .5 4 5-0. &ei%hts o! 1, *, and 5 lb are placed on a li%ht board restin% on to supports as shon in "i%. -*1. &hat are the !orces e#erted by the supports' Στ
$
( )1 lb+) !t+ 3 /)16 !t+ !t
= )* lb+)6 !t+ = )5 lb+)1* !t+ ( / ( *5 lb Σ
"y ( $ 3 / = 1 lb = * lb = 5 lb (
$ ( lb = / ( lb = *5 lb
/
6 !t
6 !t
!t
$#is 1 lb
* lb
5 lb
$ ( 05 lb
!he forces exerted by the supports are 9
$ ( 05 4 and / ( *5 4
5-0. $n -m steel metal beam ei%hs * 4 and is supported 0 m !rom the ri%ht end. 2! a -4 ei%ht is placed on the ri%ht end, hat !orce must be e#erted at the le!t end to balance the system'
F
m Στ
Α
1m
0m
( $ )5 m+ 3 )* 4+)1 m+ = ) 4+) 0 m+ ( A
0
* 4
4
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium $ ( * 4 Σ "y ( $ 3 / = * 4 = 4 ( / ( 11, 4 = $ ( 11, 4 = * 4
$ ( 6 4 /
5-0. "ind the resultant torque about point $ in "i%. 5-**. Στ Στ
( ) 4+).5 m+ sin 5 = )5 4+).16 m+ sin 55
5
5 cm
4
$
r
( *.6 4 m = 6.55 4 m ( =0. 4 m 16 cm Στ
( =0. 4 m
5 4
r
55
/ 5-. "ind the resultant torque about point / in "i%. 5-**. Στ
5 4
( ) 4+)+ = )5 4+) a 0 b+ First find a and b(
5 cm
a
a ( ).5 m+ cos 5 ( .*01 m b ( ).16 m+ sin 55 ( .101 m Στ
( = )5 4+).*01 m 3 .101 m+ ( =.16 4 m Στ
16 cm 55
5 4
b
( =.16 4 m
Criti&al Thin'in% (uestions
5-1. $ 0-lb bo# and a 5-lb bo# are on opposite ends o! a 16-!t board supported only at its midpoint. Fo !ar !rom the le!t end should a -lb bo# be placed to produce equilibrium' &ould the result be di!!erent i! the board ei%hed lb' &hy, or hy not' Στ
( )0 lb+) !t+ 3 ) lb+)#+ = )5 lb+) !t+ (
x ( . !t 1ote that the weight acting at the center of the board does 12! contribute to torque about
F
!t
!t
x
0 lb
lb
)
5 lb
the center, and therefore, the balance point is not affected, regardless of the weight(
5-*. :n a lab bench you ha8e a small roc;, a -4 meterstic; and a sin%le ;ni!e-ed%e support. E#plain ho you can use these three items to !ind the ei%ht o! the small roc;. a
b
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium Measure distances a and b3 determine F and then F
.5 m
&
4
calculate the weight 4 from equilibrium methods(
5-0. "ind the !orces "1, "*, and "0 such that the system dran in "i%. 5-*0 is in equilibrium. "1
1ote action$reaction forces + and +5(
R * !t
6 !t
* !t
5 lb
First, lets work with top board: Στ Στ
)about R+ ( Force R is upward .
( )0 lb+)6 !t+ = )5 lb+)* !t+ = " 1) !t+ ( 1ow, Σ F y = # gives9
0 !t
0 lb
R
"1 ( *10 lb
"0
"*
* !t
5 !t
RG G
* lb
*10 lb 3 R =0 lb = 5 lb ( R ( 10 lb ( RG
1ext we sum torques about F % with +5 = 6/' lb is directed in a downward direction: Στ
"
( )10 lb+)0 !t+ 3 " 0) !t+ = )* lb+)5 !t+ ( From which9 "0 ( 0. lb
Σ F y = # = "* 3 0. lb = 10 lb = * lb !he three unknown forces are:
" * ( =*5 lb
"1 ( *10 lb, "* ( =*5 lb, "0 ( 0. lb
5-. )a+ &hat ei%ht & ill produce a tension o! 4 in the rope attached to the boom in "i%. 5-*'. )b+ &hat ould be the tension in the rope i! & ( 4' 4e%lect the ei%ht 0
o! the boom in each case.
4
)a+ Σ τ = ( 4 00 Ν) (4 m+ sin 0+ = & )6 m+ cos 0 (
&
m $#is
& ( 15 4
*m
0
)b+ Σ τ ( T) m+ sin 0 = ) 4+)6 m+ cos 0 ( T ( 6 4 5-5. >uppose the boom in "i%. 5-* has a ei%ht o! 1 4 and the suspended ei%ht & is 0
equal to 4. &hat is the tension in the cord' Σ τ = Τ (4
T
m+ sin 0+ = ) 4+)6 m+ cos 0
&
m $#is
5
*m
0
1 4
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium = )1 4+)0 m+ cos 0 ( T ( 116 4
5-6. "or the conditions set in Problem 5-5, hat are the hori
"# ( F = 116 4 ( or
0 116 4
F ( 116 4
!
Σ
$#is
"y ( = 1 4 = 4 ( or ( 5 4 F ( 116 4
and
*m
m
4 1 4
0
( 5 4
5-. &hat is the tension in the cable !or "i%. 5-*5. The ei%ht o! the boom is 0 4 but its len%th is un;non. ) Select axis at wall, 7 cancels(& 5
Στ = !7 sin 5 − ) 0 1 +
7
*
T
sin 0 − 56 7 sin 0 =
5
H
T sin 5 ( 5. 4 3 *0 4
T ( 06 4
0
r
!
5-. &hat are the ma%nitude and direction o! the !orce e#erted by
56 4
0 4
the all on the boom in "i%. 5-*5' $%ain assume that the ei%ht o! the board is 0 4. +efer to the figure and data given in .roblem )$8 and recall that ! = 06 4. T ( 06 4
Σ
"# ( F - )06 4+ cos 5 (
Σ
"y ( 3 )06 4+ sin 5 = 0 4 = 56 4 ( ( 51 4 F ( *55 4
and
F ( *55 4
( 51 4
5 6 0
5-. $n car has a distance o! 0. m beteen !ront and rear a#les. 2! 6 percent o! the ei%ht rests on the !ront heels, ho !ar is the center o! %ra8ity located !rom the !ront a#le' " $#is x Σ τ ( .6&)+ 3 .&)0. m+ = F x (
6
.&
0. m
.6&
Physics, 6 th Edition
Chapter 5 Torque and Rotational Equilibrium /ut " ( &9
1.06 & = & x (
x ( 1.06 m from front axle
