64

reVieW anD summarY

In this chapter we have studied the effect of forces on particles, i.e., on bodies of such shape and size that all forces acting on them may be assumed applied at the same point.

Forces are vector quantities; they are characterized by a point of application, a magnitude , and a direction , and they add according to the parallelogram law ( Fig. 2.35 ). The magnitude and direction of the resultant R of two forces P and Q can be determined either graphically or by trigonometry, using successively the law of cosines and the law of sines [Sample Prob. 2.1].

Any given force acting on a particle can be resolved into two or more components , i.e., it can be replaced by two or more forces which have the same effect on the particle. A force F can be resolved into two components P and Q by drawing a parallelogram which has F for its diagonal; the components P and Q are then represented by the two adjacent sides of the parallelogram ( Fig. 2.36 ) and can be determined either graphically or by trigonometry [Sec. 2.6].

A force F is said to have been resolved into two rectangular components if its components F x and F y are perpendicular to each other and are directed along the coordinate axes ( Fig. 2.37 ). Intro-ducing the unit vectors i and j along the x and y axes, respectively, we write [Sec. 2.7]

Fx 5 Fxi Fy 5 Fyj (2.6)

and

F 5 Fxi 1 Fyj (2.7)

where F x and F y are the scalar components of F. These components, which can be positive or negative, are defined by the relations

Fx 5 F cos u Fy 5 F sin u (2.8)

When the rectangular components F x and F y of a force F are given, the angle u defining the direction of the force can be obtained by writing

tan u 5

Fy

Fx (2.9)

The magnitude F of the force can then be obtained by solving one of the equations (2.8) for F or by applying the Pythagorean theorem and writing

F 5 2F2x 1 F2

y (2.10)

    resultant of  two  forces        resultant of  two  forces    

    components of a  force        components of a  force    

    rectangular  components unit  vectors    

    rectangular  components unit  vectors    

Q

R

P

A Fig. 2.35

QF

P

A

Fig. 2.36

F

x

y

Fy = Fy j

Fx = Fx i

j

i

�

Fig. 2.37

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65Review and SummaryWhen three or more coplanar forces act on a particle, the rectangular components of their resultant R can be obtained by adding algebra-ically the corresponding components of the given forces [Sec. 2.8]. We have

Rx 5 oFx Ry 5 oFy (2.13)

The magnitude and direction of R can then be determined from relations similar to Eqs. (2.9) and (2.10) [Sample Prob. 2.3].

A force F in three-dimensional space can be resolved into rectangular components Fx, Fy, and Fz [Sec. 2.12]. Denoting by ux, uy, and uz, respectively, the angles that F forms with the x, y, and z axes (Fig. 2.38), we have

Fx 5 F cos ux Fy 5 F cos uy Fz 5 F cos uz             (2.19)

resultant of  several  coplanar  forces

Forces  in  space

x

y

z

A

B

C

D

E

F

Fx

Fy

Fz

�x

�y

�z

(a)

x

y

z

A

B

C

D

E

F

Fx

Fy

Fz

x

y

z

A

B

C

D

E

F

Fx

Fy

Fz

(b) (c)

OOO

Fig. 2.38

Fig. 2.39

x

y

z

λ (Magnitude = 1)

F = F λ

Fy j

Fxi

Fzk

cos �y j

cos �zk

cos �xi

The cosines of ux, uy, uz are known as the direction cosines of the force F. Introducing the unit vectors i, j, k along the coordinate axes, we write

F 5 Fxi 1 Fyj 1 Fzk (2.20)

or

F 5 F(cos uxi 1 cos uyj 1 cos uzk) (2.21)

which shows (Fig. 2.39) that F is the product of its magnitude F and the unit vector

l 5 cos uxi 1 cos uyj 1 cos uzk

Since the magnitude of l is equal to unity, we must have

cos2 ux 1 cos2 uy 1 cos2 uz 5 1 (2.24)

When the rectangular components Fx, Fy, Fz of a force F are given, the magnitude F of the force is found by writing

F 5 2F2x 1 F2

y 1 F2z (2.18)

and the direction cosines of F are obtained from Eqs. (2.19). We have

cos ux 5

Fx

F  cos uy 5

Fy

F  cos uz 5

Fz

F (2.25)

direction  cosines

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66  Statics of Particles When a force F is defined in three-dimensional space by its magnitude F and two points M and N on its line of action [Sec. 2.13], its rectangular components can be obtained as follows. We first express the vector MN

¡ joining points M and N in terms of its components

dx, dy, and dz (Fig. 2.40); we write

MN¡

5 dxi 1 dyj 1 dzk (2.26)

We next determine the unit vector l along the line of action of F by dividing MN

¡ by its magnitude MN 5 d:

L 5

MN¡

MN5

1d

(dxi 1 dyj 1 dzk)

(2.27)

Recalling that F is equal to the product of F and l, we have

F 5 FL 5

Fd

(dxi 1 dy j 1 dzk)

(2.28)

from which it follows [Sample Probs. 2.7 and 2.8] that the scalar components of F are, respectively,

Fx 5

Fdxd

    Fy 5Fdyd

    Fz 5Fdzd

(2.29)

When two or more forces act on a particle in three-dimensional space, the rectangular components of their resultant R can be obtained by adding algebraically the corresponding components of the given forces [Sec. 2.14]. We have

Rx 5 oFx Ry 5 oFy Rz 5 oFz (2.31)

The magnitude and direction of R can then be determined from relations similar to Eqs. (2.18) and (2.25) [Sample Prob. 2.8].

A particle is said to be in equilibrium when the resultant of all the forces acting on it is zero [Sec. 2.9]. The particle will then remain at rest (if originally at rest) or move with constant speed in a straight line (if originally in motion) [Sec. 2.10].

To solve a problem involving a particle in equilibrium, one first should draw a free-body diagram of the particle showing all the forces acting on it [Sec. 2.11]. If only three coplanar forces act on the particle, a force triangle may be drawn to express that the particle is in equilib-rium. Using graphical methods of trigonometry, this triangle can be solved for no more than two unknowns [Sample Prob. 2.4]. If more than three coplanar forces are involved, the equations of equilibrium

oFx 5 0 oFy 5 0 (2.15)

should be used. These equations can be solved for no more than two unknowns [Sample Prob. 2.6].

When a particle is in equilibrium in three-dimensional space [Sec. 2.15], the three equations of equilibrium

oFx 5 0 oFy 5 0 oFz 5 0 (2.34)

should be used. These equations can be solved for no more than three unknowns [Sample Prob. 2.9].

resultant of  forces  in  space

equilibrium of a particle

Free-body diagram

equilibrium  in  space

Fig. 2.40

x

y

z

F

O

M(x1, y1, z1)

N(x2, y2, z2)

dy = y2 – y1

dx = x2 – x1

dz = z2 – z1 < 0 λ

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67

reVieW Problems

2.127 The direction of the 75-N forces may vary, but the angle between the forces is always 50°. Determine the value of a for which the resultant of the forces acting at A is directed horizontally to the left.

2.128 A stake is being pulled out of the ground by means of two ropes as shown. Knowing the magnitude and direction of the force exerted on one rope, determine the magnitude and direction of the force P that should be exerted on the other rope if the resul-tant of these two forces is to be a 400-N vertical force.

2.129 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 240-N vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

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

Fig. P2.127

240 N

75 N

75 N

50°

30°A

Fig. P2.128

300 N P

25°

40°Q

D

A B C

Fig. P2.129

1.7 m1 m

2.4 m 1.5 m

1.98 kN

A B

C 1.8 m

Fig. P2.130

2.131 Two cables are tied together at C and loaded as shown. Knowing that P 5 360 N, determine the tension (a) in cable AC, (b) in cable BC.

2.132 Two cables are tied together at C and loaded as shown. Determine the range of values of P for which both cables remain taut. Fig. P2.131 and P2.132

A B

P

Q = 480 N

C

34

600 mm

250 mm

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68  Statics of Particles 2.133 A force acts at the origin of a coordinate system in a direction defined by the angles ux 5 69.3° and uz 5  57.9°. Knowing that the y component of the force is 2800 N, determine (a) the angle uy, (b) the other components and the magnitude of the force.

2.134 Cable AB is 28 m long, and the tension in that cable is 15 kN. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles ux, uy, and uz defining the direction of that force.

2.135 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the ten-sion is 10 kN in cable AB and 7.5 kN in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

Fig. P2.13412 m

9.60 m

6 m

15 m

18 m

30°A

B

C

Fig. P2.135

x

y

zC

BO

D

A

α

50°

28 m

20°

Fig. P2.136

x

y

z

A

B

D

C

O

600 mm

320 mm

360 mm

500 mm

450 mm

2.136 A container of weight W 5 1165 N is supported by three cables as shown. Determine the tension in each cable.

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69Review Problems

Fig. P2.137

200 mm

x

y

y

z zB

Q

P

A

O

2.137 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P 5 (341 N)j is applied to collar A, determine (a) the tension in the wire when y 5 155 mm, (b) the magnitude of the force Q required to maintain the equi-librium of the system.

2.138 Solve Prob. 2.137 assuming that y 5 275 mm.

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70

comPuter Problems

2.c1 Write a computer program that can be used to determine the magnitude and direction of the resultant of n coplanar forces applied at a point A. Use this program to Solve Probs. 2.32, 2.33, 2.35, and 2.38.

Fi

Fn

F1

A

qi

q1qn

x

Fig. P2.c1

2.c2 A load P is supported by two cables as shown. Write a computer pro-gram that can be used to determine the tension in each cable for any given value of P and for values of u ranging from u1 5 b 2 90° to u2 5 90° 2 a, using given increments Du. Use this program to determine for the following three sets of numerical values (a) the tension in each cable for values of u ranging from u1 to u2, (b) the value of u for which the tension in the two cables is as small as possible, (c) the corresponding value of the tension:

(1) a 5 35°, b 5 75°, P 5 2000 N, Du 5 5° (2) a 5 50°, b 5 30°, P 5 3000 N, Du 5 10° (3) a 5 40°, b 5 60°, P 5 1250 N, Du 5 5°

2.c3 An acrobat is walking on a tightrope of length L 5 20.1 m attached to supports A and B at a distance of 20.0 m from each other. The combined weight of the acrobat and his balancing pole is 800 N, and the friction between his shoes and the rope is large enough to prevent him from slip-ping. Neglecting the weight of the rope and any elastic deformation, write a computer program to calculate the deflection y and the tension in portions AC and BC of the rope for values of x from 0.5 m to 10.0 m using 0.5-m increments. From the data obtained, determine (a) the maximum deflection of the rope, (b) the maximum tension in the rope, (c) the smallest values of the tension in portions AC and BC of the rope.

A

C

B

x

y

20.0 m

Fig. P2.c3

A B

C

P

ab

qFig. P2.c2

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71Computer Problems 2.c4 Write a computer program that can be used to determine the magni-tude and direction of the resultant of n forces Fi, where i 5 1, 2, . . . , n, that are applied at point A0 of coordinates x0, y0, and z0, knowing that the line of action of Fi passes through point Ai of coordinates xi, yi, and zi. Use this program to Solve Probs. 2.93, 2.94, 2.95, and 2.135.

x

y

z

O

A2(x2, y2, z2)A1(x1, y1, z1)

A0(x0, y0, z0)

Ai(xi, yi, zi)

An(xn, yn, zn)

F2

Fi

Fn

F1

Fig. P2.c4

x

y

z

OP

A3(x3, y3, z3)

A2(x2, y2, z2)

A1(x1, y1, z1)

A0(x0, y0, z0)

AP(xP, yP, zP)

Fig. P2.c5

2.c5 Three cables are attached at points A1, A2, and A3, respectively, and are connected at point A0, to which a given load P is applied as shown. Write a computer program that can be used to determine the tension in each of the cables. Use this program to Solve Probs. 2.102, 2.106, 2.107, and 2.115.

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