Resoluçao mec vetorial p engenheiros cap 2-5

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PROBLEM 2.1
Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 75 N and Q = 125 N, 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:

R = 179 N, α = 75.1°

R = 179 N

75.1° W

PROPRIETARY MATERIAL. © 2010 The McGraw-HillCompanies, 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 2.2
Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 60 lb and Q = 25 lb, 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:

R = 77.1 lb, α = 85.4°

R = 77.1 lb

85.4° W

PROPRIETARY MATERIAL. © 2010 TheMcGraw-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 withoutpermission. 4

PROBLEM 2.3
The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

We measure: (a) Parallelogram law:

α = 51.3° β = 59.0°

(b)

Triangle rule:

We measure:

R = 139.1lb,

γ = 67.0°

R = 139.1 lb

67.0° W

PROPRIETARY MATERIAL. © 2010 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 coursepreparation. If you are a student using this Manual, you are using it without permission. 5

PROBLEM 2.4
Two forces are applied at Point B of beam AB. 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:

R = 3.30 kN, α = 66.6°

R = 3.30 kN

66.6° W

PROPRIETARYMATERIAL. © 2010 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 usingit without permission. 6

PROBLEM 2.5
The 300-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Determine the angle α by trigonometry knowing that the component along line a-a′ is to be 240 lb. (b) What is the corresponding value of the component along b-b′?

SOLUTION
(a) Using the triangle rule and law of sines:
sin β sin 60° = 240 lb 300 lb sin β = 0.69282 β = 43.854° α+ β + 60° = 180°

α = 180° − 60° − 43.854°
= 76.146°

α = 76.1° W
Fbb′ = 336 lb W

(b)

Law of sines:

Fbb′ 300 lb = sin 76.146° sin 60°

PROPRIETARY MATERIAL. © 2010 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...
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