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Unformatted text preview: t PROBLEM 3.70 Two 60mmdiameter pegs are mounted on a steel plate at A and C. and '\; I two rods are attached to the plate at B and D. A cord is passed around the
' \~\ pegs and pulled as shown, while the rods exert on the plate ION forces as
l 5 indicated. (a) Determine the resulting couple acting on die plate when T7 . s 36 N‘ (b) If only the cord is used, in what direction should it be pulled to
. W ' create the same couple with the minimum tension in the cord? (1‘) Determine the value of that minimum tension.
SOLUTION
(u) Have M = 2(Fd)
: [36 N)((l.345 m) — (10 N){0.3so m)
= 8.62 Nm
M: 8.62 Nm 'j 4
. O. 0&0 Inn
(5) ' meow ——>l
llave M : Ta' : 8,62 Nm
For Tto be minimum, dmust be maximum.
‘. Twin must be perpendicular to line AC
tanH : 938E =1.33333
0.285 rn
and (3 = 53.130“
or 0 = 53104
[f] J” : Tﬂlllldlllllk
where M : 8.62 Nm {#01380}: + (0.285}1 + 201030)] n1 2 0.535 rn
8,62 Nm : Tmm(0.535 m)
T = 16.1121 N min or T =16711N‘ mm PROBLEM 3.81 A landscaper tries to plumb a tree by applying a 240—N force as shown.
Two helpers then attempt to plumb the same tree. with one pulling at B
and the other pushing with a parallel three at C. Determine these two
forces so that they are equivalent to the single 240N fore: shown in the
ﬁgure. SOLUTION Based on
217‘: [240 N)cos30° = 7175 com 7 ﬁcosa
or 7m + meow : —(240N]c0530° (1)
2F]: (24D N)sin30° = Fysina + Ffsina
or (FR + Eqsina : (240 N)sin30° (2)
From
%: tana : 1211130D
a 30° Based on EMF: [(240 N)cos(30° — 20°)][0_25 m}:(f‘5c0510°)(0.60 m)
F3 : 100 N
or F“ : 100.0 N A 3094
From Equation ( l), 7(100 N + F( )cos3U“ : 724Dc0530" 5.. :140 N or FL : 140.0 N is 3D°{ It] In L Dl— — lﬁin‘  PROBLEM 3.88 A rectangular plate is acted upon by the force and couple shown. This
system is to be replaced with a single equivalent force. (a) Fora : 40°i specify the magnitude and the line of action of the equivalent force.
(1;) Specify the value of a ifthe line of action of the equivalent force is
to intersect line CD l2 in. to the right of D. SOLUTION ((1) Have EFY: *(3 lb)sin40° + (3 lb)sin40D = F".
P“ : 0
Have 2F1 : —(3 lb}cos40° 710 lb + (3 lb)c0540° : E F. = #10113 or F : 10.001h < Note: The two 3lb forces form a couple and EM_,: r(v,‘1xP( +r,;l,xP,‘.:r\. IxF
i j kl li j kl i j k
316 elf) 0+100100:10d 00 sin40° COS40° 0 (l 71 0 ‘0—10 k: 3(lﬁ)eos40° 7 (710)35in40c' — 160 = 10d
36.770 + 19.2836 7 160 7100’
d : 10,3946 in.
or F : [0.00 lb l at 10.39 in. right nf/I or at 5.61 in. left ofB ‘ (1:) From pan (a), F = 10.00 lb 1 Have 2M1: rt.”l x Pr + r,“ x Pb, :(lZinJix F
‘ i j kl i j kl i j k‘
316 710 0'+1601 U 0 21201 O 0
sina cosa 0 0 7] 0 It) —1 0 k: 48mm: + 30§lna 7160 * —120 24mm: : 20 — lSsinD: PROBLEM 3.88 CONTINUED Squaring both sides of the equation, and using the identity (:052 a = 1 — sin2 0:, results in sinza' 7 03490651115: — 0.21973 = 0 Using quadratic formula sina = 0.97453 sina= — 0.22547 a = 77.0“ and a = —13.03°4 ...
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This note was uploaded on 05/18/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.
 Spring '08
 Jenkins
 Statics

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