CCHHAAPPTTEERR 22

PROPRIETARY MATERIAL. © 2007 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.

3

PROBLEM 2.1

Two forces P and Q are applied as shown at point A of a hook support. Knowing that P = 15 N and Q = 25 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

(a)

(b)

We measure: 37 N,R = 76α = °

37 N=R 76°t

N

N

N

N

PROPRIETARY MATERIAL. © 2007 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.

4

PROBLEM 2.2

Two forces P and Q are applied as shown at point A of a hook support. Knowing that P = 45 N and Q = 15 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

(a)

(b)

We measure: 57 N,R = 86α = °

57 N=R 86°t

N

N

N

N

PROPRIETARY MATERIAL. © 2007 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.

5

PROBLEM 2.3

Two forces are applied to an eye bolt fastened to a beam. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

(a) Parallelogram law:

(b) Triangle rule:

We measure:

10.5 kNR =

22.5α = °

10.5 kN=R 22.5° t

PROPRIETARY MATERIAL. © 2007 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.

6

PROBLEM 2.4

A disabled automobile is pulled by means of ropes subjected to the two forces as shown. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

(a) Parallelogram law:

We measure:

5.4 kN = 12R α= °

5.4 kN=R 12°t

(b) Triangle rule:

We measure:

5.4 kN = 12R α= °

5.4 kN=R 12° t

PROPRIETARY MATERIAL. © 2007 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.

7

PROBLEM 2.5

The 200-N force is to be resolved into components along lines -a a′ and - .b b′ (a) Determine the angle α using trigonometry knowing that the

component along -a a′ is to be 150 N. (b) What is the corresponding value of the component along - ?b b′

SOLUTION

Using the triangle rule and the Law of Sines

(a) sin sin45

150 N 200 Nβ °=

sin 0.53033β =

32.028β = °

45 180α β+ + ° = °

103.0α = °t

(b) Using the Law of Sines

200 N

sin sin45bbFα′ =

°

276 NbbF ′ = t

PROPRIETARY MATERIAL. © 2007 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.

8

PROBLEM 2.6

The 200-N force is to be resolved into components along lines -a a′ and - .b b′ (a) Determine the angle α using trigonometry knowing that the

component along -b b′ is to be 120 N. (b) What is the corresponding value of the component along - ?a a′

SOLUTION

Using the triangle rule and the Law of Sines

(a) sin sin45

120 N 200 Nα °=

sin 0.42426α =

25.104α = °

or 25.1α = °t

(b) 45 25.104 180β + ° + ° = °

109.896β = °

Using the Law of Sines

200 N

sin sin45aaF

β′ =

°

200 N

sin109.896 sin45aaF ′ =

° °

or 266 NaaF ′ = t

PROPRIETARY MATERIAL. © 2007 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.

9

PROBLEM 2.7

Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 600 N, determine (a) the required angle α if the resultant R of the two forces applied to the support is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the Law of Cosines,

Have: 180 45β = ° − °

135β = °

Then:

( ) ( ) ( )( )2 22 900 600 2 900 600 cos 135R = + − °

or 1390.57 NR =

Using the Law of Sines,

600 1390.57sin sin135γ

=°

or 17.7642γ = °

and 90 17.7642α = ° − °

72.236α = °

(a) 72.2α = °t

(b) 1.391 kNR = t

PROPRIETARY MATERIAL. © 2007 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.

10

PROBLEM 2.8

Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the left-hand rod is 1 30 N,F = determine (a) the required force F2 in the right-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

By trigonometry: Law of Sines

2 30sin sin38 sin

F Rα β

= =°

90 28 62 , 180 62 38 80α β= ° − ° = ° = ° − ° − ° = °

Then:

2 30 Nsin62 sin38 sin80

F R= =° ° °

or (a) 2 26.9 NF = t

(b) 18.75 NR = t

N

PROPRIETARY MATERIAL. © 2007 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.

11

PROBLEM 2.9

Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the right-hand rod is F2 = 20 N, determine (a) the required force F1 in the left-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the Law of Sines

1 20 Nsin sin38 sin

F Rα β

= =°

90 10 80 , 180 80 38 62α β= ° − ° = ° = ° − ° − ° = °

Then:

1 20 Nsin80 sin38 sin62

F R= =° ° °

or (a) 1 22.3 NF = t

(b) 13.95 NR = t

N

PROPRIETARY MATERIAL. © 2007 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.

12

PROBLEM 2.10

An elastic exercise band is grasped and then is stretched as shown. Knowing that the tensions in portions BC and DE of the band are 80 N and 60 N, respectively, determine, using trigonometry, (a) the required angle α if the resultant R of the two forces exerted on the hand at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the Law of Sines: 60 N 80 Nsin sin10α

=°

or = 7.4832α °

( )180 10 7.4832β = ° − ° + °

162.517= °

Then:

80 N

sin162.517 sin10R =

° °

or 138.405 NR = (a) 7.48α = °t (b) 138.4 NR = t

PROPRIETARY MATERIAL. © 2007 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.

13

PROBLEM 2.11

To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that α = 25°, determine (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the Law of Sines

Have: ( )180 35 25β = ° − ° + °

120= °

Then: 80 N

sin35 sin120 sin25P R= =

° ° °

or (a) 108.6 NP = t

(b) 163.9 NR = t

N

PROPRIETARY MATERIAL. © 2007 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.

14

PROBLEM 2.12

To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that the magnitude of P is 70 N , determine (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the Law of Sines

(a) Have: 80 N 70 Nsin sin35α

=°

sin 0.65552α =

40.959α = °

or 41.0α = °t

(b) ( )180 35 40.959β = − ° + °

104.041= °

Then: 70 N

sin104.041 sin35R =

° °

or 118.4 NR = t

N

N

PROPRIETARY MATERIAL. © 2007 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.

15

PROBLEM 2.13

As shown in Fig. P2.11, two cables are attached to a sign at A to steady the sign as it is being lowered. Using trigonometry, determine (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R.

SOLUTION

We observe that force P is minimum when 90 .α = °

Then:

(a) ( )80 sin35P = °

or 45.9 N=P t

And:

(b) ( )80 cos35R = °

or 65.5 N=R t

N ( N

N

PROPRIETARY MATERIAL. © 2007 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.

16

PROBLEM 2.14

As shown in Fig. P2.10, an elastic exercise band is grasped and then is stretched. Knowing that the tension in portion DE of the band is 70 N, determine, using trigonometry, (a) the magnitude and direction of the smallest force in portion BC of the band for which the resultant R of the two forces exerted on the hand at A is directed along a line joining points A and H, (b) the corresponding magnitude of R.

SOLUTION

For BCT to be a minimum,

R and BCT must be perpendicular.

Thus ( )70 N sin4BCT = °

4.8829 N=

And ( )70 N cos4R = °

69.829 N=

(a) 4.88 NBCT = 6.00°t

(b) 69.8 NR = t

PROPRIETARY MATERIAL. © 2007 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.

17

PROBLEM 2.15

Solve Prob. 2.1 using trigonometry.

Problem 2.1: Two forces P and Q are applied as shown at point A of a hook support. Knowing that P = 15 N and Q = 25 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

Using the force triangle and the Laws of Cosines and Sines

We have:

( )180 15 30γ = ° − ° + °

135= °

Then: ( ) ( ) ( )( )2 22 15 N 25 N 2 15 N 25 N cos135R = + − °

21380.33 N=

or 37.153 NR =

and

25 N 37.153 Nsin sin135β

=°

25 Nsin sin135

37.153 Nβ = °

0.47581=

28.412β = °

Then: 75 180α β+ + ° = °

76.588α = °

37.2 N=R 76.6°t

N