1. No category

PROBLEM 6.126
An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°,
determine (a) the vertical force exerted on the block at D, (b) the force
exerted on member ABC at B.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC:
BD = (7) 2 + (24)2 = 25 in.
We have
( FBD ) x =
7
24
FBD , ( FBD ) y =
FBD
25
25
ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(16) = 0
7
24
FBD (24) +
FBD (7) = 84(16)
25
25
336
FBD = 1344
25
FBD = 100 lb
tan α =
24
α = 73.7°
7
FBD = 100.0 lb
(b)
Force exerted at B.
(a)
Vertical force exerted on block.
( FBD ) y =
73.7°
24
24
FBD =
(100 lb) = 96 lb
25
25
(FBD ) y = 96.0 lb
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
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928
PROBLEM 6.131
A couple M of magnitude 1.5 kN ⋅ m is applied to the crank of the engine system shown. For each of the two
positions shown, determine the force P required to hold the system in equilibrium.
SOLUTION
(a)
FBDs:
50 mm
175 mm
2
=
7
Dimensions in mm
Note:
tan θ =
FBD whole:
ΣM A = 0: (0.250 m)C y − 1.5 kN ⋅ m = 0 C y = 6.00 kN
FBD piston:
ΣFy = 0: C y − FBC sin θ = 0 FBC =
Cy
sin θ
=
6.00 kN
sinθ
ΣFx = 0: FBC cos θ − P = 0
P = FBC cos θ =
6.00 kN
= 7 kips
tan θ
P = 21.0 kN
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.
935
PROBLEM 6.131 (Continued)
(b)
FBDs:
Dimensions in mm
2
as above
7
Note:
tan θ =
FBD whole:
ΣM A = 0: (0.100 m)C y − 1.5 kN ⋅ m = 0 C y = 15 kN
ΣFy = 0: C y − FBC sin θ = 0 FBC =
ΣFx = 0:
Cy
sin θ
FBC cos θ − P = 0
P = FBC cos θ =
Cy
tan θ
=
15 kN
2/7
P = 52.5 kN
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.
936
PROBLEM 6.143
A small barrel weighing 60 lb is lifted by a pair of tongs as shown. Knowing
that a = 5 in., determine the forces exerted at B and D on tong ABD.
SOLUTION
We note that BC is a two-force member.
Free body: Tong ABD:
Bx By
=
15
5
Bx = 3By
ΣM D = 0: By (3 in.) + 3By (5 in.) − (60 lb)(9 in.) = 0
By = 30 lb
Bx = 3By : Bx = 90 lb
ΣFx = 0: −90 lb + Dx = 0
D x = 90 lb
ΣFy = 0: 60 lb − 30 lb − D y = 0
D y = 30 lb
B = 94.9 lb
18.43°
D = 94.9 lb
18.43°
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.
949