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Pearson Education Asia Limited 2009

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NSS Physics in Life Full Solution of Textbooks(Force and Motion)

Chapter 8 Forces in a Plane and Moent

Checkpoint (p!"##)

1. (a)

(b)

(c)

(d)

2.

From the diagram, magnitude of the resultant force N61.332 22 +=

1

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Checkpoint (p!"#$)

1. B

2. x-component of1

F 2 ! cos 6"# 1 N

y-component of 1F 2 ! sin 6"# 1.$32 Nx-component of

2F " N

y-component of2

F %1 N

sum of thex-components 1 & " 1 N

sum of they-components 1.$32 & (%1) ".$32 N

'ence, the magnitude of the resultant ( ) ( ) N2.1$32."1 22 +=

uppose the resultant ma*es an angle +ith thex-ais. hen,

( )( )1$32."

tan = , 36.2#

%xercise (p!"#&)

1. /

2. /

3. (a) No. he resultant is non-0ero.

(b) No. he resultant is non-0ero.

(c) es. he resultant is 0ero.

. (a) he t+o forces are acting in the same direction.(b) he t+o forces are perpendicular to each other.

(c) he t+o forces are acting in the opposite direction.

. (a) (i)

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(ii) x-component of the resultant 2 & 2 ! cos # 3 N

y-component of the resultant 2 ! sin # 1 N

magnitude of the resultant N16.331 22 +=

et be the angle bet+een the resultant and the &x-direction.

=

.14

3

1tan

he resultant is 3.16 N (at 14.# to the &x-direction).

(b) (i)

(ii) x-component of the resultant + cos4cos4 N

y-component of the resultant "sin4sin4 =

magnitude of the resultant N" 22 =+=

et be the angle bet+een the resultant and the &x-direction.

=

=

"

"tan

he resultant is N (to the &xdirection).

(c) (i)

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(ii) x-component of the resultant 2 % cos2 1 N

y-component of the resultant sin2 1 N

magnitude of the resultant N1.111 22 +=

et be the angle bet+een the resultant and the &xdirection.

=

=

1

1tan

he resultant is 1.1 N at (# to the &xdirection).

6. B5 s5mmetr5, +e onl5 consider the nail in the top left-hand corner. +o forces

are eerted on the nail, one b5 the hori0ontal segment of the rubber band and one

b5 the ertical segment of the rubber band. ince each of them has a magnitude

of ". N, +e hae

magnitude of the resultant force N$"$."."." 22 +=

et be the angle bet+een the resultant force and the hori0ontal.

=

=

."

."tan

'ence the resultant force points to the centre of the board.

imilarl5, the resultant forces on the other three nails are ".$"$ N and the5 all

point to the centre of the board.

$. a*e the direction of the pull from tugboatBas the &x-direction.

x-component of the resultant force 1" """ & 1" """! cos $"# 13 2" N

y-component of the resultant force 1" """ ! sin $"# 737$ N

resultant force N""16737$2"13 22 +=

et be the angle bet+een the resultant force and the &x-direction.

=

3

2"13

737$tan

he resultant is 16 "" N (at 3# to the &xdirection).4. a*e the direction of the pulling forceFBas the &x-direction.

x-component of the resultant force 12"" & 1""! cos $# 144 N

y-component of the resultant force 1"" ! sin $# 17 N

magnitude of the resultant force N21"17144 22 +=

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$. (a)

F; pulling forceR; normal reactionf; frictionW; +eight

(b) ince the tram moes uphill at a constant elocit5, the component of the

resultant force parallel to the rails is 0ero. a*e the uphill direction as

positie. 9onsidering the force components parallel to the rails, +e hae

N6""26

"12sin)1"4"""("""1"

"12sin

=

=

F

F

WfF

he pulling force proided b5 the cable is 26 6"" N.

(c) ince the tram moes do+nhill at a constant elocit5, the component of the

resultant force parallel to the rails is 0ero. a*e the do+nhill direction as

positie.

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4. (a)

F; pulling forceR; normal reactionf; frictionW; +eight

(b) here is no acceleration along the direction perpendicular to the ramp.

'ence the normal reaction acting on the bloc* is

N4""276cos)1"3"""(cos == WR

(c) (i) he acceleration along the ramp is 0ero, hence +e hae

N7""1461

"6sin)1"3"""("""14

"6sin

=

=

=

f

f

fWF

he friction acting on the bloc* is 1 7"" N.

(ii)

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N$"1"$

N2$$

===

===

mgF

maF

y

x

(b) magnitude of the resultant N71$"2 22 +=

et be the angle bet+een the resultant and the hori0ontal.

=

".

2

$"tan

he force eerted on the pilot b5 the seat is 71 N (at # to the hori0ontal).

11. (a) he reactions are the forces eerted on the g5mnast b5 the bars, +hich point

to the directions opposite to the forcesF, i.e. to+ards the g5mnast at an

angle of $"# to the hori0ontal.

(b) he g5mnast does not accelerate along the ertical direction. 'ence his

+eight is balanced b5 the ertical components of the t+oF=s.

N3$2

1"$"$"sin2

$"sin2

=

=

F

F

mgF

12. 9onsider the ertical direction, +e hae

Fcos mg (1)

9onsider the hori0ontal direction, +e hae

Fsin ma (2)

:iiding (2) b5 (1),

=

=

=

4".21

1"

tan

cos

sin

g

a

ubstituting into (1),

N$""14".21cos

1"4""

cos

==

mgF

Checkpoint (p!"')

1. :

2. B

%xercise (p!"')

1. 9

2. 9

3. 9

4

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. 9

. :

6. (a)

T; tension in the stringW1; +eight of ".2 *g massW2; +eight of ". *g massa; common acceleration

(b) the ". *g mass

(c) 9onsider the ".2 *g alone. /ppl5ingF ma, +e hae

T% W1 m1a

T% 2 ".2a (1)

9onsider the ". *g alone. /ppl5ingF ma, +e hae

W2% T m2a

% T ".a (2)

/dding (1) and (2), +e hae

2sm33.3333.3

6."2

=

=

aa

ubstituting ainto (1),

N6$.2

333.32."2

=

T

T

'ence the common acceleration is 3.33 m s%2and the tension in the string is

2.6$ N.

$. (a) he free-bod5 diagrams of the three bloc*s are sho+n belo+. >nl5 the

forces of interest are dra+n.

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9onsiderAalone. /ppl5ingF ma, +e hae

WAB% TAB mAa

6 % TAB ".6a (1)

9onsiderBalone. /ppl5ingF ma, +e hae

TAB% TBC%fB mBa

TAB% TBC% 2 a (2)

9onsider Calone. /ppl5ingF ma, +e hae

TBC%WC mCa

TBC% 3 ".3a (3)

/dding (1), (2) and (3), +e hae

2sm26."263."

3."6."326

=

++=

a

aaa

he common acceleration of the bloc*s is ".26 m s%2. ince ais positie,B

moes to the right in this case.

Alternative method:

9onsider the three bloc*s as a single s5stem.

he net force acting on the s5stem is

F WA % WC%fB (".6 ! 1") % (".3 ! 1") % 2 1 N

he common acceleration is

2sm".26263."

)7."3."1(

1 =

++==

m

Fa

(b) From (3) in (a), +e hae

N16.3

263."3."3

=

=

BC

BC

CCBC

T

T

amWT

From (1) in (a), +e hae

N64.

263."6."6

=

=

AB

AB

AABA

T

T

amTW

he tensions in the string are 3.16 N and .64 N respectiel5.

4. (a) friction on the locomotief1 ".2 ! (1.2 ! 1"! 1") 2. ! 1"N

1"

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(i) 9onsidering diceA, the forceFBAeerted onAb5Bis gien b5

"

)1"("1.")1""1."(

=

=

=

BA

BA

AABA

F

F

amWF

(ii) 9onsidering diceB, the forceFp eerted onBb5the plate is gien b5

"

)1"("1.")1""1."("

p

p

p

=

=

=

F

F

amWFFBBBA

1". (a) inceAis at rest, the component of its +eight WAperpendicular to the

inclined plane must be balanced b5 the normal reactionR.

N"."2.12cos)1"."(12cos === AWR

he component of the +eight WAparallel to the inclined plane must be

balanced b5 the frictionfin the direction parallel to the slope.

N".736736."12sin)1"."(12sin === AWf

(b) magnitude of the resultant force eerted onAb5B

N.

736.""2. 22

22

=

+=

+= fR

et be the angle bet+een the resultant force and the direction

perpendicular to the inclined plane.

12

"2.

736."tan

=

=

'ence the resultant force is at 12# to the direction perpendicular to the

inclined plane, i.e. in the up+ard ertical direction.

?ithout +or*ing through (a), +e can still find this resultant force b5 using

the fact that it must cancel out the +eight WAof boo*A. 'ence it must hae

a magnitude e8ual to WA . N and a direction opposite to WA, i.e. in the

up+ard ertical direction.

Checkpoint (p!"$)

1. (a) moment of the force about O ! ".2 1 N m (cloc*+ise)

(b) moment of the force about O ( ! sin 3"#) ! ".2 ". N m (cloc*+ise)

(c) moment of the force about O ( ! sin 3"#) ! ".2 ". N m (cloc*+ise)

2. ?hen the leer is balanced, b5 the principle of moments, +e hae

sum of cloc*+ise moments sum of anti-cloc*+ise moments

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2. First, the net force acting on the beam is 0ero. 'ence, +e hae

63"1"4" =+=+ BA FF (1)

hen, considering the moments aboutA, +e hae

sum of cloc*+ise moments sum of anti-cloc*+ise moments

N23

3).11"()14"(

=

=+

B

B

F

F

ubstitutingFBinto (1), +e hae N3763" B/ == FF

%xercise (p!"$#)

1. 9

2. B

3. B. /

. /

6. (a)T1; tension in the left cableT2; tension in the right cableWp; +eight of the platform

F+; force eerted b5 the +or*er

(b) (1) Net force acting on it is 0ero.

(2) Net moment on the obect about an5 point is 0ero.

(c) /ppl5ing condition (2) in (b), the net moment about the left end of the

platform is 0ero, hence +e hae

N$333.$33

.11""1)1"$"(3

.113

2

2

p+2

=

+=

+=

T

T

WFT

1

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/ppl5ing condition (1) in (b), the net force acting on the platform is 0ero,

hence +e hae

N76$

)1"1""()1"$"(3.$33

1

1

p+21

+=+

+=+

T

T

WFTT

$. (a)

F5force eerted b5 the bo5 Wp+eight of the plan*

FAforce eerted b5 trestleA FBforce eerted b5 trestleB

(b) /s the plan* is in e8uilibrium, the net moment on the plan* about an5 point

is 0ero. 9onsidering the moments aboutA, +e hae

N3"

2)1"2"(.")1""(

2." p5

=

=+

=+

B

B

B

F

F

WFF

(c)

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1. (a) he cord ma*es an angle of 1"# +ith the ertical.

+eight of the buc*et 2""" ! 1" 2" """ N (1M)

N3""2"

"""2"1"cos

=

T T(1M)

uppose the helicopter accelerates hori0ontall5 +ith an acceleration a. ince

the ertical acceleration of the helicopter isg, i.e. the acceleration due to

grait5, +e hae

2sm1.$6$63.1

1"1"tan

=

=

a

a

(1M)

he tension in the cord is 2" 3"" N and the acceleration of the helicopter is

1.$6 m s%2. (1A+1A)

(b) 9onsider the helicopter and the buc*et as a single s5stem.

/ppl5ingF ma, +e hae (1M)

hori0ontal component of the lifting force

N$73

)$63.12""()$63.12"""(

+=(1M)

ertical component of the lifting force

N"""

)1"2""()1"2"""(

=

+=(1M)

magnitude of the lifting force N""$"""$73 22 +=

et be the angle bet+een the lifting force and the ertical.

=

1"

"""

$73tan

(1M)

he lifting force generated b5 the rotor is $"" N (at 1"# to the ertical).

(1A)

16. (a) (i) (ii)(1A+1A)

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(b) height of the slope h ! tan 2"# 1.42" m

he distancestraelled b5 impson on the slope is gien b5

m321.

2"cos

=

=

s

s (1M)

he component of +eight Wparallel to the slope is

Wp 3 !1" ! sin 2"# 117.$ N

he component of +eight Wperpendicular to the slope is

Wn 3 !1" ! cos 2"# 324.7 N (1M)

ince the normal reactionRbalances Wn, +e hae

R Wn 324.7 N

he frictionfacting on impson is

f ".1 !R ".1 ! 324.7 32.47 N

he net forceFacting on impson is

F 117.$ % 32.47 46.41 N (1M)

he acceleration aof impson is

2sm4".23

41.46 ==a

/ppl5ing asuv 222 += , +e hae (1M)

1

2

sm1.134.

321.4".22"

=

+=

v

v

he elocit5 of impson +hen it reaches the bottom of the slope is

.1 m s%1. (1A)

(c) ?hen impson is on the hori0ontal ground,

normal reactionR W 3 !1" 3" N

a*e the right as the positie direction.

frictionf %(".1 !R) %(".1 ! 3") %3 N

net forceFf %3 N

acceleration a2sm1

33 ===

mF (1M)

From (b), the elocit5 of impson +hen it reaches the bottom of the slope is

.134 m s%1.

/ppl5ing asuv 222 += , +e hae

( )

"sm3.2

1"12134.

1

22

>=

+=

v

v(1M)

'ence impson fails to stop before hitting Bart. (1A)

2"

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1$. he free-bod5 diagrams of the three bloc*s are sho+n belo+. >nl5 the forces of

interest are dra+n.

9onsiderAalone. /ppl5ingF ma, +e hae

TAB%fA mAa

TAB% a (1) (1M)

9onsiderBalone. /ppl5ingF ma, +e hae

TBC%TAB%fB mBa

TBC%TAB% 2 ".a (2) (1M)

9onsider Calone. /ppl5ingF ma, +e hae

WC% TBC mCa

2" % TBC 2a (3) (1M)

/dding (1), (2) and (3), +e hae

2sm

.31

2"2"2

=

=

++=+

a

a

aa.a

(1A)

ubstituting ainto (3), +e hae

N12

22"

=

=

BC

BC

T

T(1A)

ubstituting ainto (1), +e hae

N4

=

=

AB

AB

T

T(1A)

he common acceleration of the bloc* is m s%2. he tensions TBCand TABin the

strings are 12 N and 4 N respectiel5.

14. (a) et m1and m2be the mass of the fireman and the pac*et respectiel5.

et T1and T2be the tensions of the upper and lo+er ropes respectiel5.

T1 m1g& m2g 6 ! 1" & 1" ! 1" $" N (1M)

T2 m2g 1" ! 1" 1"" N

he tensions of the upper and lo+er ropes are $" N and 1"" N

respectiel5.

(1A+1A)

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(b) uppose the hori0ontal force acting on the deice is 7"" N.

et Tbe the maimum tension and abe the maimum acceleration.

N1"37

7""3"cos

=

=

T

T (1M)

?hen the tension is maimi0ed, +e hae

2

2121

sm46.3

1")1"6()1"6(1"37

)()(

+++=

+++=

a

a

gmmammT

(1M)

he maimum acceleration is 3.46 m s%2. (1A)

17. (a) etFAandFBbe the supporting forces proided b5 +or*ersAandB

respectiel5.

9onsidering the moments about the point +here +or*erBsupports the pipe,

( )

N2.$6

2.2

1"4"

=

=

A

A

F

F(1M)

ince the ertical net force acting on the pipe is 0ero, +e hae

( )

AB FF

F

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(b) he net moment on the board about an5 point is 0ero. (1M)

9onsidering the moments aboutA,

( )[ ]

N6"2

3sin1"123sin

=

=

R

R(1M)

he normal reaction from the ground is 6" N. (1A)

(c) he net force on the board is 0ero.

9onsidering the direction parallel to the board, +e hae

( )

N.23.2

sin1"12sin6"

sinsin

1

1

1

=

=+

=+

F

F

WRF

(1A)

9onsidering the direction perpendicular to the board, +e hae

( )

N.23.2

sin1"12sin6"

sinsin

2

2

2

=

=+

=+

F

F

WRF

(1A)

magnitude of the resultant N6".2.2 22222

1 =+=+= FFF

he angle bet+eenFandF1is gien b5

=

==

6"

3.2cos 1

F

F

(1M)

he resultantFis 6" N (at # toF1, i.e. in the up+ard direction). (1A)

22. (LE-GCE AS-level May 2004 a!er "#40$02 %1)

23. (&'EC AS-level 'an 200# a!er #41$01 %2)

2. (A%A AS-level 'an 200 a!er A02 %)

2. (*CEE 200 1 %4)

26. (a) (i) centre of grait5 (of the upper bod5) (1A)

(ii) he net moment on his spine about an5 point is 0ero. (1M)

9onsidering the moments about pointJ, +e haesum of cloc*+ise moments sum of anti-cloc*+ise moments

N2$7"

2.")6"sin""(.")6"sin1""(2."1"sin

+=

F

F

(1M+1A)

(b) he net moment on his spine about an5 point is 0ero. (1M)

9onsidering the moments about pointJ, +e hae

sum of cloc*+ise moments sum of anti-cloc*+ise moments (1M)

N161"2.")3"sin""(.")3"sin1""(2."1"sin

+=

FF (1M+1A)

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