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.

163

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 Į.

SOLUTION

Free-Body Diagram of Rod AB:

25α θ= − °

(20 lb)cosQ θ=

and ( )(9 in.)BM Q=

Therefore, 120 lb-in. (20 lb)(cos )(9 in.)

120 lb-in.cos

180 lb-in.

θ

θ

=

=

or 48.190θ = °

Therefore, 48.190 25α = ° − ° 23.2α = ° 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.

165

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. . .

2mg (80 kg)(9.81 m/s ) 784.8 NW = = =

(a) We have / (0.25 m)(784.8 N) 196.2 N mE H EM r W= = = ⋅ or 196.2 N mE = ⋅M W

(b) 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 / min( ) ( ) .E A E AM r F− =

Where 2 2/ (0.35 m) (0.5 m) 0.61033 mA Er = + =

then min196.2 N m (0.61033 m)( )AF⋅ =

or min( ) 321 NAF =

Also 0.35 m

tan0.5 m

φ = or 35.0φ = ° min( ) 321 NA =F 35.0° W

(c) 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

/ min

min

( ) ( )

196.2 N m (0.85 m)( )E D E vertical

vertical

M r F

F

− =⋅ = or min( ) 231 Nvertical =F at Point D 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 3.10

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):

1 1( ) ( )

7 24(2.24 in.) 500 lb (1.68 in.) 500 lb

25 25

492.8 lb in.

C AB x AB yM y F x F= − +

§ · § ·= − × + ר ¸ ¨ ¸© ¹ © ¹

= + ⋅

(a)

493 lb in.C = ⋅M W�

Using (b):

2 ( )

7(3.52 in.) 500 lb

25

492.8 lb in.

C AB xM y F=§ ·= ר ¸© ¹

= + ⋅

(b)

493 lb in.C = ⋅M W