基礎力學1作業（靜力學）1FNAFBFNBP450A第一章

P19：1-1（b）1-2（a）FNBA300PBFA21-2（b）FNBFAA450PBFBABCF300FAxFAyFBABCF300FA31-2（d）或41-5PABCEFDFNEFAFDxFDyFBFB/FDyFCxFNEFA/FCx/FCy/ABCFCyBD

FDxPFEACABCPFCFBCBBPFB/FA

AABPFB/FAxFAy51-6第二章

P35：P35：2-1

已知F1=150N，F2=200N，F3=250N，F4=100N，試分別用幾何法和解析法求這四個力的合力。600OxyF4F3F2600450F1F3F2F1F4FR

a

1厘米代表100N6量出FR的長度，折算出合力的大小，量出角度a

的值。幾何法解析法FROxyF4F3F2600a600450F1=

293.93NFRx=

Fx=

-F1cos

450

+F2cos600+F3+F4cos

600FRy

=

Fy

=

F1sin450

+F2sin60-F4sin600=

192.67NFR=

F

2+F

2

=

…

=351.45NRx

Rytana

=FRy/FRx

a

=33.167aaACBFFAFCFCFBBC二力構件FCFAFA8C22

F

=

0.707FF

=

F

=

F

sin

450

=FB

=

FC

=

0.707F450

450P35：2-4

圖示三鉸剛架由AB和BC兩部分組成，A、C為固定鉸支座，B為中間鉸。試求支座A、C和鉸鏈B的約束力。設剛架的自重及摩擦均可不計。解：【BC】【整體】【BC】P36：2-8

圖示梁AB，F=20KN.

試求支座A和B的約束力。梁重及摩擦均可不計。2m2mABCF450FBFAx

F =

0y

F =

025F

2

-

F

2

=

0AA4F

=

10

F

=

5 10

=

15.81kN25FB

+

FA1

-

F

2

=

0B9F

=

5 2

=

7.07kN解：【AB】P47：3-4

圖示折梁AB，試求支座A和B的約束力。第三章

P47：300FBl梁重及摩擦均可不計。ABFF/FAm

=

0A-F l

+

F

a

cos

300

=

0FA=

3Fa

=

0.866Fa2l

ll10AB=

0.866FaF

=

F解：【AB】第四章

P68：MeF1300F211F3O1F4O3O21m

2m2m2m1m

1m1m

5mP68：4-1

平板上力系如圖示，試求其合成結果。已知：F1

=80N，F2

=50N，F3

=60N，F4

=40N，M

e

=140N.m66.9F

￠q

=

arctan

Ry=

arctan

132.4

=

63.20FR￠xF

￠=

F

￠2

+

F

￠2R

Rx

Ry=

66.92

+132.42=148.3NO為左下角點MO

=

mO(

Fi

)

=-F1

cos300

×

1+F1

sin300

×9+F2

×

3

+F3

cos450

×

5

+

F2

sin

450

×1

+

Me

-

F4

×1=795.3N.m132

.4x

=

M

0

=

795

.3

=

6

.01mFR￠y

F

￠Rx

=

Fx=F1

cos300

–

F3

cos45o+

F4

=

66.9NF

￠Ry=

Fy1

2

3=

F

sin300

+

F

+F

sin45o=132.4NMeF1300F124F2O1O3O2F3

1m

2m2m2m1m1m

1m5mMO1

=

mO1(

Fi

)

=

F1

cos300

×

2

+

F1

sin300

×6+F3

cos

450

×

2

-

F3

sin

450

×2

+

Me

+

F4

×2=598.6N.m1132

.4Ryx=

M

O

1=

598

.6

=

4

.52

mF

￠MeF1300F134F2F3O1O3O21m

2m2m2m1m

1m1m

5m+

F4

×4MO2

=

mO2(

Fi

)=

F1

cos300

×

4

+

F1

sin300

×8+

F2

×

2

+

Me=997.1N.m2132

.4Ryx=

M

O

2=

997

.1

=

7

.53mF

￠MeF1300F144F2F3O1O3O21m

2m2m2m1m

1m1m

5mMO3

=

mO3(

Fi

)=

F1

sin300

×8+

F2

×

2

+F3

cos

450

×

4

+

Me=729.7N.m3132

.4Ryx=

M

O

3=

729

.7

=

5

.51mF

￠MeF1300F154F2F3O1O3O21m

2m2m2m1m

1m1m

5mP68：4-3

已知F=400N,桿重不計，試求C點處的約束力。FBFCx450F0.6m

0.5m0.5mABCDBCFCyDFCx

=

880

N-

FCx

0.5

+

F

1.1

=

0FCyFCy

=

-480

N0.5

+

F

0.6

=

0CF

=

8802

+

4802

=

1002.4N880a

=

arctan

480

=

28.60第四象限解：【DBC】

MA

=

0

MB

=

0F16P69：4-5(b)試求外伸梁支座處的約束力。梁重及摩擦均不計。aaFMeaaABCDqFAyFB22a

-

M

-

F

3a

+

qa

a

=

0=

0

F

MeBAFB=

Me

+

3F

-

qa2a

2

4FAy

+

FB

-

F

-

qa

=

0

Fy

=

0Ay17F=-

Me

-

F

+

5qa2a

2

4解：【AB】P69：4-6(a)試求構架A、B處的約束力。構件重及摩擦均不計。600ABFAxFAyFNB2.5m3m400kN2.5mA

M

=

02323

-

400

·

2.5

=

0NBNBF

1

5

+

FNBF

=

196.2kNAx18NBAxx2F

=

169.9kN

3

=

0

F

=

0

F

-

F2yAyNBAyF

-

400

+

F1

=

0

F

=

301.9kN

F

=

0解：【AB】P69：4-7

活塞機構如圖，F=800N，方向垂直于ED。設活塞D和缸壁間的接觸面光滑，各構件重均不計。試求活塞D作用于物塊C上的壓力。300BCDFE450A300BDFEFNDFCFB解：【EBD】C

C19x2

23FB

=

3312.9N1

2

FB

-

0.4

· -

0.4

·

+

F

1.6

=

02

2

2

F

=

0

-

F

+

3312.9

·

2

-800

·

1

=

0

F

=1943N

=1.94kN

MD

=

0活塞D作用于物塊C上的壓力與FC等值、反向、共線。P69：4-8

機構如圖，設A、C兩點在同一鉛垂線上，且AC=AB。試問桿保持平衡時角θ等于多少？AP2BC20180

-θP1θAB21800

-θP1θFAxFAy

M

A

=

02sinq

=

021800

-q

l

l

-

P1P2

sin2

212

sin

q

cos

q2

2q

PP2

cos

=1sin

=2

Pq

P2120Pq

=

2

arcsin

P2解：【AB】滑輪半徑R與桿長l相比很小。繩重及摩擦均不計。FTB=P2P70：4-10

圖示雙跨靜定梁，由梁AC和梁CD通過鉸鏈C連接而成。試求支座A、B、D處的約束力和中間鉸C處兩梁之間相互作用的力。梁的自重及摩擦均不計。ABCD2mMe=40kN.m2m2m1m1m解：有主次之分的物體系統【CD】q=10kN/mDMe=40kN.mCFAyFBFD21FDFCy

MC

=

0FD

4

-

40

-

20

·1

=

0FD

=15kNy

F

=

0CyF

+15

-

20

=

0CyF

=

5kN【整體】

MA

=

0FB

2

+15·8

-

40

-

40

·

4

=

0FB

=

40kN

Fy

=

0AyF

+

40

+15

-

40

=

0AyF

=

-15kNP71：4-17（a）試用截面法求圖示桁架中指定桿件的內力。FAyFBABC100kN

50kNE321FN3FN2FBBE1C

FN1D32【右半部分】50kN

Fy

=

0N

2F

=

53kN1mD4×1m=4m22-

F

cos

450

+

F

=

0N

2

B

MD

=

0BF

cos

450

1+

F

2

=

0N1F

=

-125kNN1

ME

=

0-FN

3

1+

FB

1

=

0FN

3

=

87.5kN解：【整體】

M

A

=

0FB

4

-100

·

2

-

50

·3

=

0FB

=

87.5kNqP84：5-1

已知θ=300，P=350N，FT=100N，fs=0.35，fd=0.25。試問物塊是否平衡？并求出摩擦力的大小和方向。yPxF

a1T

A23FfFNFT

cosq

-

P

sinq

+

Ff

=

0Ff

=

88.4NFN

=

353.1NFN

-

P

cosq

-

FT

sinq

=

0Ff

<

fs

FN

=

123.6N所以物塊平衡。由fs=0.35得

Fx

=

0

Fy

=

0φ=19.290

。P84：5-5物塊A置于水平面上，自重P=150N，物塊與水平面間的fs=0.25，試求使物塊滑動所需的力F1