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CHAPTER 3

PROBLEM 3.1 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into horizontal and vertical components.

SOLUTION Free-Body Diagram of Rod AB:

x = (9 in.) cos 65° = 3.8036 in. y = (9 in.)sin 65 ° = 8.1568 in.

F = F xi + F y j

rA /B

= (20 lb cos 25°)i + (−20 lb sin 25°) j = (18.1262 lb) i − (8.4524 lb) j )))& = BA = (− 3.8036 in.)i + (8.1568 in.)j

MB = rA /B × F = ( −3.8036i + 8.1568j) × (18.1262i − 8.4524j) = 32.150k − 147.852k = − 115.702 lb-in.

M B = 115.7 lb-in.

W

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PROBLEM 3.2 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into components along AB and in a direction perpendicular to AB.

SOLUTION Free-Body Diagram of Rod AB:

θ = 90° − (65° − 25° ) = 50°

Q = (20 lb) cos50° = 12.8558 lb M B = Q (9 in.) = (12.8558 lb)(9 in.) = 115.702 lb-in.

M B = 115.7 lb-in.

W

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PROBLEM 3.3 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that the moment of the force about B is 120 lb · in. clockwise, determine the value of α.

SOLUTION Free-Body Diagram of Rod AB:

α = θ − 25°

Q = (20 lb)cos θ

and Therefore,

M B = (Q )(9 in.)

120 lb-in. = (20 lb)(cosθ )(9 in.) cos θ =

120 lb-in. 180 lb-in.

or

θ = 48.190°

Therefore,

α = 48.190° − 25°

α = 23.2° W

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PROBLEM 3.4 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E.

SOLUTION

(a)

By definition, We have

W = mg = 80 kg(9.81 m/s2 ) = 784.8 N

ΣM E : M E = (784.8 N)(0.25 m) M E = 196.2 N ⋅ m

(b)

W

For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: We have

d = (0.85 m)2 + (0.5 m)2 = 0.98615 m

ΣM E : 196.2 N ⋅ m = FB (0.98615 m) FB = 198.954 N

and

§ 0.85 m · ¸ = 59.534° © 0.5 m ¹

θ = tan− 1 ¨

or

FB = 199.0 N

59.5° W

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PROBLEM 3.5 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E.

SOLUTION First note. . . W = mg = (80 kg)(9.81 m/s 2 ) = 784.8 N

(a)

We have

M E = rH /EW = (0.25 m)(784.8 N) = 196.2 N⋅ m

W

For FA to be minimum, it must be perpendicular to the line joining Points A and E. Then with FA directed as shown, we have (− M E ) = rA/ E ( FA ) min .

(b)

(c)

or M E = 196.2 N ⋅ m

Where

rA /E = (0.35 m)2 + (0.5 m)2 = 0.61033 m

then

196.2 N ⋅ m = (0.61033 m)(FA )min

or

( FA ) min = 321 N

Also

tan φ =

0.35 m 0.5 m

or

φ = 35.0°

( FA ) min = 321 N

35.0° W

For Fvertical to be minimum, the perpendicular distance from its line of action to Point E must be maximum. Thus, apply (Fvertical)min at Point D, and then (− M E ) = rD / E (Fvertical )min 196.2 N ⋅ m = (0.85 m)(Fvertical )min

or ( Fvertical ) min = 231 N

at Point D W

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PROBLEM 3.6 A 300-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part (a), determine the perpendicular distance from O to the line of action of P .

SOLUTION

x = (0.2 m)cos 40°

= 0.153209 m y = (0.2 m)sin 40° = 0.128558 m ∴ r A/O = (0.153209 m)i + (0.128558 m)j

(a)

F x = (300 N)sin 30°

= 150 N F y = (300 N) cos30 °

= 259.81 N F = (150 N) i + (259.81 N) j M O = rA/O × F = (0.153209i + 0.128558j ) m× (150i + 259.81j ) N = (39.805k − 19.2837k ) N⋅ m = (20.521 N⋅ m)k

(b)

M O = 20.5 N ⋅m

M O = Fd 20.521 N ⋅m = (300 N)( d ) d = 0.068403 m

W

d = 68.4 mm W

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PROBLEM 3.7 A 400-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into components along line OA and in a direction perpendicular to that line. (b) Determine the magnitude and direction of the smallest force Q applied at B that has the same moment as P about O.

SOLUTION

(a)

Portion OA of crank:

θ = 90° − 30° − 40° θ = 20° S = P sinθ = (400 N)sin 20° = 136.81 N M O = rO /A S = (0.2 m)(136.81 N) = 27.362 N ⋅ m

(b)

MO = 27.4 N⋅ m

W

Smallest force Q must be perpendicular to OB.

Portion OB of crank:

M O = rO /BQ M O = (0.120 m)Q 27.362 N ⋅m = (0.120 m)Q

Q = 228 N

42.0° W

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PROBLEM 3.8 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if α = 10°, (c) the smallest force P that creates the same moment about B.

SOLUTION (a)

We have

M B = rC /BF N = (4 in.)(200 lb) = 800 lb ⋅ in.

or MB = 800 lb ⋅ in. (b)

By definition,

Then

W

M B = r A/B P sinθ θ = 10° + (180° − 70° ) = 120°

800 lb ⋅ in. = (18 in.) × P sin120°

or P = 51.3 lb W (c)

For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus

MB = dPmin d = rA /B

or or

800 lb ⋅ in. = (18 in.)Pmin Pmin = 44.4 lb

Pmin = 44.4 lb

20° W

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PROBLEM 3.9 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

SOLUTION Using (a): MC = y1 ( FAB )x + x1 ( FAB )y § 7 · § 24 · = (2.24 in.) ¨ × 500 lb ¸ + (1.68 in.)¨ × 500 lb ¸ 25 25 © ¹ © ¹ = 1120 lb ⋅ in.

(a) MC = 1.120 kip ⋅ in.

W

Using (b): MC = y2 ( FAB ) x §7 · = (8 in.) ¨ × 500 lb ¸ © 25 ¹ = 1120 lb ⋅ in.

(b) M C = 1.120 kip ⋅ in. W

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 169

PROBLEM 3.10 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

SOLUTION Using (a): MC = − y1 ( FAB ) x + x1 ( FAB ) y § 7 · § 24 · = −(2.24 in.) ¨ × 500 lb ¸ + (1.68 in.) ¨ × 500 lb¸ © 25 ¹ © 25 ¹ = +492.8 lb ⋅ in.

(a) MC = 493 lb ⋅ in.

W

Using (b): MC = y2 (FAB )x § 7 · = (3.52 in.)¨ × 500 lb¸ © 25 ¹ = +492.8 lb ⋅ in.

(b)

MC = 493 lb ⋅ in.

W

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 170

PROBLEM 3.11 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E.

SOLUTION (a)

Slope of line:

Then

and

Then

EC =

0.875 m 5 = 1.90 m + 0.2 m 12

12 (T ) 13 AB 12 (1040 N) = 13 = 960 N 5 TABy = (1040 N) 13 = 400 N TABx =

(a)

MD = TABx (0.875 m) − TABy (0.2 m) = (960 N)(0.875 m) − (400 N)(0.2 m) = 760 N ⋅ m

(b)

We have

or M D = 760 N⋅ m

MD = TABx ( y ) + TABx (x ) = (960 N)(0) + (400 N)(1.90 m) = 760 N ⋅ m

W

(b) or M D = 760 N⋅ m

W

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PROBLEM 3.12 It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D.

SOLUTION

Slope of line:

EC =

0.875 m 7 = 2.80 m + 0.2 m 24

Then

TABx =

24 T AB 25

and

TABy =

7 T AB 25

We have

MD = TABx ( y ) + TABy ( x ) 24 7 TAB (0) + TAB (2.80 m) 25 25 TAB = 1224 N

960 N ⋅m =

or TAB = 1224 N W

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PROBLEM 3.13 It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D.

SOLUTION

The minimum value of d can be found based on the equation relating the moment of the forceT

ABabout

D:

MD = ( TAB max) y ( d)

where

MD = 960 N⋅ m (TAB max )y = T AB max sinθ = (2400 N)sinθ

Now

sinθ =

0.875 m 2 2 ( d + 0.20) + (0.875) m

ª º 0.875 » (d ) 960 N ⋅ m = 2400 N « «¬ ( d + 0.20) 2 + (0.875) 2 »¼

or

( d + 0.20) 2 + (0.875) 2 = 2.1875d

or

( d + 0.20)2 + (0.875)2 = 4.7852 d 2

or

3.7852 d 2 − 0.40 d − 0.8056 = 0

Using the quadratic equation, the minimum values of d are 0.51719 m and − 0.41151 m. Since only the positive value applies here, d = 0.51719 m or d = 517 mm W PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 173

PROBLEM 3.14 A mechanic uses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 485 N is exerted on the alternator at B. Determine the moment of that force about bolt C if its line of action passes through O.

SOLUTION We have

M C = r B/ C × F B

Noting the direction of the moment of each force component about C is clockwise, M C = xF By + yF Bx

where

and

x = 120 mm − 65 mm = 55 mm y = 72 mm + 90 mm = 162 mm F Bx = F By =

65 2

2 (65) + (72)

72 2

(65) + (72) 2

(485 N) = 325 N (485 N) = 360 N

MC = (55 mm)(360 N) + (162)(325 N) =72450 N ⋅m =72.450 N ⋅m

or MC = 72.5 N⋅ m

W

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 174

PROBLEM 3.15 Form the vector products B × C and B′′ × C, where B = B′, and use the results obtained to prove the identity 1 1 sin α cos β = sin ( α + β ) + sin ( α − β). 2 2

SOLUTION Note:

By definition,

Now

B = B (cos β i + sin β j ) B ′ = B (cos β i − sin β j) C = C (cos α i + sin α j)

| B × C| = BC sin (α − β )

(1)

| B′× C | = BC sin (α + β )

(2)

B × C = B (cos β i + sin β j) × C (cos αi + sin α j)

= BC (cos β sin α − sin β cos α ) k

and

(3)

B′× C = B (cos β i − sin β j) × C (cos αi + sin α j)

= BC (cos β sin α + sin β cos α ) k

(4)

Equating the magnitudes of B × C from Equations (1) and (3) yields: BC sin(α − β ) = BC (cos β sin α −sin β cos α)

(5)

Similarly, equating the magnitudes of B ′ ×C from Equations (2) and (4) yields: BC sin(α + β ) = BC (cos β sin α +sin β cos α)

(6)

Adding Equations (5) and (6) gives: sin(α − β ) + sin(α + β ) = 2cos β sin α

or sin α cos β =

1 1 sin(α + β ) + sin( α − β ) W 2 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 175

PROBLEM 3.16 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.

SOLUTION (a)

We have

A = |P × Q|

where

P = −7i + 3j − 3k Q = 2i + 2j + 5k

Then

i j k P × Q = −7 3 − 3 2

2

5

= [(15 + 6)i + ( −6 + 35) j + ( −14 − 6)k ] = (21)i + (29) j( −20)k 2

2

A = (20) + (29) + ( −20)

(b)

We have

A = |P × Q|

where

P = 6i − 5 j − 2 k<...