Vector Mechanics for engineers Statics 7th - Cap 02 solution PDF

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PROBLEM 2.1 Two forces are app graphically the mag parallelogram law, (b

SOLUTION (a)

(b)

PROBLEM 2.2 The cable stays AB and AD he tension is 500 N in AB and 16 magnitude and direction of the re at A using (a) the parallelogram la

SOLUTION

We measure: (a)

D

51.3 q, E

59

PROBLEM 2.3 Two forces P and Q Knowing that P magnitude and direct (b) the triangle rule.

SOLUTION (a)

PROBLEM 2.4 Two forces P and Q are applied Knowing that P 45 lb and Q magnitude and direction of their r (b) the triangle rule.

SOLUTION (a)

PROBLEM 2.5 Two control rods are knowing that the fo (a) the required forc forces exerted by t corresponding magni

SOLUTION

Graphically, by the triangle law We measure:

F2 R

PROBLEM 2.6 Two control rods are attached at knowing that the force in the ri (a) the required force F1 in the forces exerted by the rods on corresponding magnitude of R.

SOLUTION

Using the Law of Sines F1

R

80

PROBLEM 2.7 The 50-lb force is to b-bc. (a) Using trigo component along athe component along

SOLUTION

Using the triangle rule and the Law of Sines sin E

si

PROBLEM 2.8 The 50-lb force is to be resolved b-bc. (a) Using trigonometry, de component along b-bc is 30 lb. the component along a-ac ?

SOLUTION

Using the triangle rule and the Law of Sines

PROBLEM 2.9 To steady a sign as it at A. Using trigonom required magnitude applied at A is to be v

SOLUTION

Using the triangle rule and the Law of Sines Have:

D

180

PROBLEM 2.10 To steady a sign as it is being low at A. Using trigonometry and kno determine (a) the required angle applied at A is to be vertical, (b) t

SOLUTION

Using the triangle rule and the Law of Sines (a) Have:

360 N sin D sin D

300 N sin 35 q 0.68829

PROBLEM 2.11 Two forces are appl and knowing that the angle D if the resulta horizontal, (b) the co

SOLUTION Using the triangle rule and the Law of Sines

(a) Have:

20 lb sinD

sin D

PROBLEM 2.12 For the hook support of Problem that the magnitude of P is 25 lb, the force Q if the resultant R o vertical, (b) the corresponding ma Problem 2.3: Two forces P and hook support. Knowing that P graphically the magnitude and d parallelogram law, (b) the triangle

SOLUTION Using the triangle rule and the Law of Sines

(a) Have:

Q

25 lb

PROBLEM 2.13 For the hook suppo (a) the magnitude a resultant R of the (b) the corresponding Problem 2.11: Two trigonometry and kn (a) the required angl support is to be horiz

SOLUTION (a) The smallest force P will be perpendicular to R, that

 20 lb

P

10 lb (b)

R

 20

PROBLEM 2.14 As shown in Figure P2.9, two ca the sign as it is being lowered. magnitude and direction of the sm of the two forces applied at A is v of R.

SOLUTION We observe that force P is minimum when D is 90q, that is, P is ho

Then:

(a) P

 360 N  sin 35q

And:

(b) R

 360 N  cos 35q

PROBLEM 2.15 For the hook support magnitude and direc support knowing tha

Problem 2.11: Two trigonometry and kn (a) the required angl support is to be horiz

SOLUTION Using the force triangle and the Law of Cosines

R2

10 lb 

2

 20 lb   2 2

ª¬100  400  400  0.3

636.8 lb 2

PROBLEM 2.16 Solve Problem 2.1 using trigonom

Problem 2.1: Two forces are ap Determine graphically the magn using (a) the parallelogram law, (b

SOLUTION Using the force triangle, the Law of Cosines and the Law of Sines

D

We have:

180q   50q  25q 105q

Then:

R2

 4.5 kN

2

  6 kN   2 4.5 kN 2

PROBLEM 2.17 Solve Problem 2.2 us

Problem 2.2: The ca that the tension is 50 the magnitude and d stays at A using (a) th

SOLUTION From the geometry of

T

Now:

And, using the Law o R2

500 N 331319

PROBLEM 2.18 Solve Problem 2.3 using trigonom

Problem 2.3: Two forces P and hook support. Knowing that P graphically the magnitude and d parallelogram law, (b) the triangle

SOLUTION Using the force triangle and the L We have:

J

180 135

Then:

R2

15 lb 

2

 25

1380.3 lb 2 or

R

PROBLEM 2.19 Two structural mem Knowing that both 30 kN in member trigonometry, the ma applied to the bracke

SOLUTION Using the force triang We have: Then:

R2

30 171

R

and

41.3

PROBLEM 2.20 Two structural members A and Knowing that both members are 20 kN in member A and 30 trigonometry, the magnitude and applied to the bracket by member

SOLUTION Using the force triangle and the L

J

We have: Then:

R2

 30 kN 

180q 2

  20

1710.4 kN2 R

41.357 kN

and 30 kN sinD i

§ 3

PROBLEM 2.21 Determine the x and y

SOLUTION 20 kN Force: Fx

  20 kN  co

Fy

  20 kN sin

Fx

  30 kN co

Fy

  30 kN sin

30 kN Force:

42 kN Force:

PROBLEM 2.22 Determine the x and y component

SOLUTION 40 lb Force: Fx

 40 lb sin 50 q,

Fy

  40 lb cos 50q,

Fx

  60 lb cos 60q,

Fy

 60 lb sin 60 q,

60 lb Force:

PROBLEM 2.23 Determine the x and y

SOLUTION We compute the follo

Then: 204 lb Force:

PROBLEM 2.24 Determine the x and y component

SOLUTION We compute the following distan

500 N Force:

OA

 70 

OB

 210 

OC

 120

2

2

PROBLEM 2.25 While emptying a w force P directed alo horizontal componen vertical component.

SOLUTION

(a)

P

PROBLEM 2.26 Member BD exerts on member A Knowing that P must have a 960magnitude of the force P, (b) its h

SOLUTION

(a)

P

Py sin 35 q 960 N

PROBLEM 2.27 Member CB of the v line CB. Knowing determine (a) the ma

SOLUTION

We note: CB exerts force P on B along CB, and the horizontal com Then: (a)

Px

P sin

PROBLEM 2.28 Activator rod AB exerts on crank Knowing that P must have a 25-lb the crank, determine (a) the mag along line BC.

SOLUTION

Using the x and y axes shown.

PROBLEM 2.29 The guy wire BD e along BD. Knowing determine (a) the m direction perpendicul

SOLUTION

PROBLEM 2.30 The guy wire BD exerts on the along BD. Knowing that P has determine (a) the magnitude of line AC.

SOLUTION

PROBLEM 2.31 Determine the resulta Problem 2.24: Dete shown.

SOLUTION

From Problem 2.24: F500

  140 N

PROBLEM 2.32 Determine the resultant of the thr Problem 2.21: Determine the x shown.

SOLUTION

From Problem 2.21: F20

15.32 kN i  12.86 kN 

F30

 10.26 kN  i   28.2 kN

PROBLEM 2.33 Determine the resulta Problem 2.22: Dete shown.

SOLUTION The components of the forces were determined in 2.23. Force

x comp. (lb

40 lb

30.6

60 lb

30

80 lb

72.5 Rx

71.9

PROBLEM 2.34 Determine the resultant of the thre

Problem 2.23: Determine the x a shown.

SOLUTION The components of the forces were determined in Problem 2.23.

F204

 4

F212

 112

F 400 Thus

 3

PROBLEM 2.35 Knowing that D shown.

SOLUTION 300-N Force:

400-N Force:

600-N Force:

PROBLEM 2.36 Knowing that D shown.

65q, determ

SOLUTION 300-N Force: Fx

 300 N

Fy

 300 N

Fx

 400 N

Fy

 400 N

Fx

 600 N

400-N Force:

600-N Force:

PROBLEM 2.37 Knowing that the ten the three forces exert

SOLUTION Cable BC Force:

100-lb Force:

PROBLEM 2.38 Knowing that D shown.

50 q, determ

SOLUTION The resultant force R has the x- an

140 lb cos 50

Rx

6Fx

Rx

7.6264 lb

Ry

6Fy

Ry

289.59 lb

and

140 lb sin 50q

PROBLEM 2.39 Determine (a) the re shown is to be vertic

SOLUTION For an arbitrary angle D , we have: 6Fx

Rx

 140 lb cosD   60 lb

(a) So, for R to be vertical: Rx

6Fx

 140 lb cosD   60 lb c

Expanding,  cos D  3  cos D cos 35 q Then:

PROBLEM 2.40 For the beam of Problem 2.37, de BC if the resultant of the three fo (b) the corresponding magnitude

Problem 2.37: Knowing that the the resultant of the three forces ex

SOLUTION We have: Rx

6Fx

or



84 12 TBC  156 lb   116 13 Rx

0.724 TBC  84 lb

and 80 5 4 TBC  156 lb   116 13 5

Ry

6Fy

Ry

0.6897TBC  140 lb

PROBLEM 2.41 Boom AB is held in t tensions in cables A determine (a) the te exerted at point A (b) the corresponding

SOLUTION

Choose x-axis along bar AB. Then (a) Require

PROBLEM 2.42 For the block of Problems 2.35 an of D of the resultant of the thre incline, (b) the corresponding mag Problem 2.35: Knowing that D three forces shown. Problem 2.36: Knowing that D three forces shown.

SOLUTION

Selecting the x axis along aac, we write Rx

6Fx

300 N  400 N cos D 









PROBLEM 2.43 Two cables are tied t tension (a) in cable A

SOLUTION Free-Body Diagram

From the geometry, we calculate the distances:

AC

 16 in.

BC

 20 i 2  

2



PROBLEM 2.44 Knowing that D rope BC.

25q , determi

SOLUTION Free-Body Diagram

Law of Sines:

Force Tria

PROBLEM 2.45 Knowing that D directed long line AC tension in cable BC.

SOLUTION Free-Body Diagram

PROBLEM 2.46 Two cables are tied together at C D 30q , determine the tension (a

SOLUTION Free-Body Diagram

Force Tria

Law of Sines: T

T

2943 N

PROBLEM 2.47 A chairlift has been chair weighs 300 N a that weight of the ski

SOLUTION Free-Body Diagram Point B

In the free-body diag

Thus, in the force tria Force Triangle

PROBLEM 2.48 A chairlift has been stopped in chair weighs 300 N and that the s the weight of the skier in chair E.

SOLUTION Free-Body Diagram Point F

In the free-body diagram of point

TEF

tan

T DF

tan

Thus, in the force triangle, by the Force Triangle

TEF sin100.3 TBC In the free-body diagram of point

PROBLEM 2.49 Four wooden membe equilibrium under th F A 510 lb and FB forces.

SOLUTION Free-Body Diagram

Resolving the forces into x and y components:

PROBLEM 2.50 Four wooden members are joined equilibrium under the action of F A 420 lb and F C 540 lb, de forces.

SOLUTION

Resolving the forces into x and y components: 6F

0

F

15 q

540 lb  420 lb 

15 q

0

PROBLEM 2.51 Two forces P and Q Knowing that the co 520 lb, determi Q A and B.

SOLUTION Free-Body Diagram

Resolving the forces

Substituting compon R

  400 l

 FB i 

In the y-direction (on 400

PROBLEM 2.52 Two forces P and Q are applie Knowing that the connection is in the forces exerted on rods A and determine the magnitudes of P an

SOLUTION Free-Body Diagram

Resolving the forces into x and y R

P

Substituting components: R

 320 lb i

 ª¬ 600 lb

 Pi  Q cos 55q

In the x-direction (one unknown f 320 lb

600 lb

PROBLEM 2.53 Two cables tied to W 840 N, determin

SOLUTION Free-Body Diagram

From geometry: The sides of the trian The sides of the trian Thus: 6Fx or

and

PROBLEM 2.54 Two cables tied together at C are of values of W for which the te cable.

SOLUTION Free-Body Diagram

From geometry: The sides of the triangle with hyp The sides of the triangle with hyp Thus: 3 0:  TC 5

6 Fx or

1  TCA  5 and 6F y

0:

4 TCA  5

PROBLEM 2.55 The cabin of an aeria roll freely on the su speed by cable DE combined weight of 24.8 kN, and assu determine the tensio cable DE.

SOLUTION Note: In Problems 2. considered as a rigid gravity should be loc vertical. Now 6Fx

0

or 0

PROBLEM 2.56 The cabin of an aerial tramway is roll freely on the support cable speed by cable DE. Knowing tha in cable DE is 20 kN, and ass negligible, determine (a) the com system, and its passengers, (b) the

SOLUTION Free-Body Diagram

PROBLEM 2.57 A block of weight W springs of which the constants of the spr determine (a) the ten

SOLUTION Free-Body Diagram At A

First note from geom The sides of the trian The sides of the trian The sides of the trian Then:

and LAB So:

PROBLEM 2.57 CONTINU (b) and 6F y

0:

3 150 N   5

PROBLEM 2.58 A load of weight 400 are attached to block constant of the sprin unstretched length of

SOLUTION Free-Body Diagram At A

First note from geom The sides of the trian The sides of the trian The sides of the trian 12:35:37. Then: 6F or

PROBLEM 2.58 CONTINU (b) Have spring force Fs

k

FAB

k

Where

and LAB

 0.360 m  2

So: 281.74 N

800

PROBLEM 2.59 For the cables and lo for which the tensi corresponding value

SOLUTION The smallest TBC is when TBC is perpendicular to the dir

Free-Body Diagram At C

PROBLEM 2.60 Knowing that portions AC and BC the shortest length of cable which if the tension in the cable is not to

SOLUTION Free-Body Diagram: C  For T 725 N 

6Fy

0: Ty T x2 

Tx2   500

Tx

By similar triangles:

PROBLEM 2.61 Two cables tied tog maximum allowable magnitude of the la corresponding value

SOLUTION Free-Body Diagram: C

Force triangle is isoceles with 2E

180 q

PROBLEM 2.62 Two cables tied together at C a maximum allowable tension is 30 determine (a) the magnitude of th at C, (b) the corresponding value

SOLUTION Free-Body Diagram: C

(a) Law of Cosines: 2



2 

2





PROBLEM 2.63 For the structure and D for which the ten corresponding value

SOLUTION T BC must be perpendicular to FAC to be as small as poss Free-Body Diagram: C

PROBLEM 2.64 Boom AB is supported by cable B boom exerts on pin B a force direc in rope BD is 70 lb, determine (a) cable BC is as small as possibl tension.

SOLUTION Free-Body Diagram: B

(a) Have:

TBD

where magnitude and direction of FAB is known.

Then, in a force triangle: By observation, TBC is minim (b) Have

T BC

 70 68.9

PROBLEM 2.65 Collar A shown in vertical rod and is a spring is 660 N/m, Knowing that the sy the weight of the col

SOLUTION Free-Body Diagram: Collar A Have: where: LcAB

PROBLEM 2.66 The 40-N collar A can slide on a as shown to a spring. The sprin Knowing that the constant of the of h for which the system is in equ

SOLUTION 6Fy

Free-Body Diagram: Collar A

or Now..

0: W

hFs

Fs

PROBLEM 2.67 A 280-kg crate is su shown. Determine fo The tension in the ro can be proved by the

SOLUTION Free-Body Diagram of pulley

6F

(a)

(b)

6F

(c)

6F

PROBLEM 2.68 Solve parts b and d of Problem rope is attached to the crate.

Problem 2.67: A 280-kg crate i arrangements as shown. Determi the rope. (Hint: The tension in simple pulley. This can be proved

SOLUTION Free-Body Diagram of pulley and crate (b) 6F y

0: 3T  T

(d)

PROBLEM 2.69 A 350-lb load is sup Knowing that E 2 force P which shoul equilibrium. (Hint: T simple pulley. This c

SOLUTION Free-Body Diagram: Pulley A and co For 6F y

PROBLEM 2.70 A 350-lb load is supported by th Knowing that D 35q, determin the force P which should be e maintain equilibrium. (Hint: The side of a simple pulley. This can b

SOLUTION Free-Body Diagram: Pulley A

6 Fx

0: 2

sin E

1 cos 2

Hence: (a)

PROBLEM 2.71 A load Q is applied t pulley is held in the p over the pulley A a determine (a) the ten

SOLUTION Free-Body Diagram: Pulley C

6Fx

(a) Hence

0: TACB  cos 30 q

TACB

PROBLEM 2.72 A 2000-N load Q is applied to th ACB. The pulley is held in the p which passes over the pulley A a tension in the cable ACB, (b) the m

SOLUTION Free-Body Diagram: Pulley C

6Fx or

0: TACB  cos 30 q  cos 50 q P

0.3473TACB

PROBLEM 2.73 Determine (a) the x angles T x, T y, and Tz t

SOLUTION Fx

(a)

Fy

Fz

200 lb cos30

 200 lb  sin 3

  200 lb cos30 q si

156

PROBLEM 2.74 Determine (a) the x, y, and z co angles T x, Ty, and Tz that the force

SOLUTION (a)

 420 lb  sin 20 qsin 70 q

Fx

Fy

F...