PRO BLEM 2 - University of Massachusetts...

SOLUT

(a) P

(b) T

W

TION

arallelogram l

Triangle rule:

We measure:

PRO

Twomagn(b) th

law:

OBLEM 2.1

o forces are anitude and dirhe triangle rul

139R =

1

applied as shorection of theile.

1 kN, 47α =

own to a hooir resultant us

7.8°

ok. Determineing (a) the pa

1391 N=R

e graphically arallelogram la

N 47.8°

the aw,

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PROBLEM 2.3

Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

(a) Parallelogram law:

(b) Triangle rule:

We measure: 20.1 kN,R = 21.2α = ° 20.1 kN=R 21.2°


PROBLEM 2.7

A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines:

(a) 75 40 180

180 75 40

65

ββ

° + ° + = °= ° − ° − °= °

11000 lbsin 75° sin 65

T=

° 1 938 lbT =

(b) 1000 lbsin 75° sin 40

R=

° 665 lbR =


SOLUT

Using th

(a)

(b)

TION

he triangle rulee and the law

50β + °

4s

4s

P

Athmfom

of sines:

60 180

180

70

β° + ° = °

= °= °

425 lbsin 70 sin 6

P=

°

425 lbsin 70 sin 5

R=

°

PROBLEM

A steel tank is hat α = 20°, d

magnitude of forces applied magnitude of R

50 60− ° − °

0°

0°

2.11

to be positiondetermine by the force P iat A is to be v

R.

ned in an excavtrigonometry if the resultanvertical, (b) th

vation. Knowi(a) the requir

nt R of the the correspondi

392 lbP =

346 lbR =

ing red two ing


SOLUT

Comput

800-N F

424-N F

408-N F

TION

te the followin

Force:

Force:

Force:

ng distances:

PROBL

Determin

2

(600)

1000 m

(560)

1060 m

(480)

1020 m

OA

OB

OC

=

=

=

=

=

=

(800 NxF = +

(800 NyF = +

(424 NxF = −

(424 NyF = −

(408 NxF = +

(408 NyF = −

LEM 2.22

ne the x and y

2 2

2 2

2 2

(800)

mm

(900)

mm

(900)

mm

+

+

+

800N)

1000

600N)

1000

560N)

1060

900N)

1060

480N)

1020

900N)

1020

components oof each of the

F

forces shown.

640 NxF = +

480 NyF = +

224 NxF = −

360 NyF = −

192.0 NxF = +

360 NyF = −

.

N

N

N


SOLUT

Compon

TION

nents of the fo

F

2

5

5

orces were dete

Force x

29 lb

50 lb

51 lb

R

tan

PROBLE

Determine t

PROBLEMforces show

ermined in Pro

x Comp. (lb)

+21.0

–14.00

+24.0

31.0xR = +

(31.0 lb)

n

23.031.036.573

23.0 lbsin (36.57

x y

y

x

R R

RR

R

α

α

= +

=

=

=

= °

=

i j

i

R

38.601 lb=

EM 2.31

the resultant o

M 2.21 Determwn.

oblem 2.21:

y Comp

+20

+4

–4

2yR = +

(23.0 lb)

b73 )

+

°

i j

b

of the three for

mine the x and

p. (lb)

0.0

8.0

5.0

3.0

rces of Problem

d y componen

38.6=R

m 2.21.

nts of each of

6 lb 36.6°

the


SOLUT

Determi

Cable fo

500-N F

200-N F

and

Further:

TION

ine force comp

orce AC:

Force:

Force:

ponents:

(365 N

(365 N

= −

= −

x

y

F

F

(500 N)

(500 N)

x

y

F

F

=

=

(200 N)

(200 N

x

y

F

F

=

= −

2

(400 N

474.37

= Σ = −

= Σ = −

= +

=

=

x x

y y

x

R FR F

R R R

255tan

40032.5

α

α

=

= °

PRKnoresu

960N) 24

14601100

N) 271460

= −

= −

24) 480 N

257

) 140 N25

=

=

4) 160 N

53

N) 120 N5

=

= −

2

2

240 N 480

275 N 140 N

N) ( 255 N

N

− +

− +

+ −

yR

°

ROBLEM 2.owing that theultant of the th

40 N

75 N

N

2

N 160 N 4

N 120 N

N)

+ =

− = −

36 tension in rop

hree forces exe

400 N

255 N−

pe AC is 365 Nerted at point C

474=R

N, determine C of post BC.

4 N 32.5°

the


SOLUT

60-lb Fo

80-lb Fo

120-lb F

and

Further:

TION

orce:

orce:

Force:

(60 l

(60 lx

y

FF

=

=

(80 l

(80 lx

y

FF

=

=

(120

(12x

y

FF

=

= −

(20

202.

x x

y y

R FR F

R

= Σ

= Σ

=

=

29.tan

200α =

tan

8.46

α −=

= °

PROKnowiforces

lb)cos 20 5

lb)sin 20 20

° =

° =

lb)cos60 4

lb)sin 60 69

° =

° =

0 lb)cos30 1

20 lb)sin 30

° =

° =

2

200.305 lb

29.803 lb

00.305 lb) (

.510 lb

=

=

+

803.305

1 29.803200.305

°

BLEM 2.37ing that α =shown.

6.382 lb

0.521 lb

0.000 lb

9.282 lb

103.923 lb

60.000 lb= −

2(29.803 lb)

7 40°, determiine the result

203=R

ant of the th

3 lb 8.46°

hree


PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION

Free-Body Diagram Force Triangle

Law of sines:

400 lbsin 60 sin 40 sin80

AC BCT T= =

° ° °

(a) 400 lb

(sin 60 )sin80ACT = °

° 352 lbACT =

(b) 400 lb

(sin 40 )sin80BCT = °

° 261 lbBCT =


SOLUT

Law of s

(a)

(b)

TION

Free

sines:

e-Body Diagra

sin

PKAC

am

n 110 sin 5AC BCT T

=°

1200sin 6ACT =

1200sin 6BCT =

PROBLEM 2Knowing that α

C, (b) in rope

1200 lb5 sin 65

=° °

0 lbsin 110

65°

°

lbsin 5

65°

°

2.48 20 ,α = ° dete

BC.

Force Triang

rmine the ten

gle

T

T

sion (a) in ca

1244 lbACT =

115.4 lbBCT =

ble


SOLUT

xFΣ

For P =

yFΣ

Solvin

TION

0 CAT= −

200N= we ha

0=

0.8

ng equations (

PRO

Two KnowBC.

sin30 CBT+D

ve,

0.5 CAT−

coCAT

86603 0.8CAT +

1) and (2) sim

BLEM 2.49

cables are wing that P =

Free-B

sin30 cosP−D

0.5 21A CBT+ +

os30 coCBT° −

86603 21CBT −

multaneously g

9

tied togeth= 300 N, de

Body Diagram

s 45 200N° − =

12.13 200− =

s30 sin 4P−D

12.13 0= (2)

ives,

her at C anetermine the

m

0=

0 (1)

5 0=D

)

nd are loadtension in c

T

T

ded as showcables AC a

134.6 NCAT =

110.4 NCBT =

wn. and


SOLUT

Free-Bo (a)

(b)

(c)

(d )

(e)

TION

ody Diagram of Pulley

FΣ

FΣ

FΣ

FΣ

FΣ

0: 2yF T= −

0: 2yF T= −

0: 3yF T= −

0: 3yF T= −

0: 4yF T= −

PR

A 6andfor (Se

(600 lb) 0

12

T

− =

=

(600 lb) 0

12

T

− =

=

(600 lb) 0

1(

3T

− =

=

(600 lb) 0

1(

3T

− =

=

(600 lb) 0

14

T

− =

=

ROBLEM 2

600-lb crate id-pulley arrang

each arrangeee the hint for

(600 lb)

(600 lb)

(600 lb)

(600 lb)

(600 lb)

.67

is supported bgements as sh

ement the tensProblem 2.66

by several rophown. Determsion in the ro.)

300 lbT =

300 lbT =

200 lbT =

200 lbT =

150.0 lbT =

pe-ine pe.


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