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Unformatted text preview: PROBLEM 2.5 Two control rods are attached at A to lever AB. Using trigonometry and
knowing that the force in the lefthand rod is F1 2 120 N, determine 172 (a) the required force F2 in the righthand rod if the resultant R of the
forces exerted by the rods on the lever is to be vertical, (1)) the
corresponding magnitude of R. §OLUTION Graphically, by the triangle law We measure: F2 5 108 N
'R s 77 N By trigonometry: Law of Sines F2 R  120 sina sin38° sinﬁ‘ a = 90°— 28° = 62°,ﬁ = 180° — 62° —38° = 80° F2 R _ 120N sin 62° sin38° sin80°
or (:1) F2 =107.6N 4
(b) R = 75.0 N 4 PROBLEM 2.10 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 300 N, determine (a) the required angle a 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 360N _ 300N (a) Have: . _
5111 a sm35° sina: = 0.68829 5 =180 — (35° + 435°) =101.5° R _300N sin101.5° sin35° or R=513Ni PROBLEM 2.28 Activator rod AB exerts on crank BCD a force P directed along line AB.
Knowing that P must have a 251b component perpendicular to arm BC of
the crank, determine (a) the magnitude of the force P, (17) its component
along line BC. SOLUTION Using the x and y axes shown. (a) or P = 25.91134 PROBLEM 2.41 Boom AB is held in the position shown by three cables. Knowing that the
tensions in cables AC and AD are 4 kN and 5.2 kN, respectively,
determine (a) the tension in cable AE if the resultant of the tensions exerted at point A of the boom must be directed along AB,
(5)) the corresponding magnitude of the resultant.  SOLUTION Choose xaxis along bar AB.
Then (a) Require R = 2F), = 0: (4 kN)cosZS° + (5.2 kN)sin35° — TAE sh165° = 0 01 TAE = 7.2909 kN TAE = 7.29 kN 4 (b) R = 2F:
= —(4 kN)sin25° — (5.2 kN)c0335° — (7.2909 kN)cos 65° = —9.03 kN PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the
tension (:1) in cable AC, (12) in cable BC. SOLUTION
FreeBody Diagram 60:: Ho From the geometry, we calculate the distances: AC = (16 in.)2 + (12 in.)2 = 20 in. BC = (20 in.)2 + (21 in.)2 = 29 in. Then, from the Free Body Diagramof point C: 16 21 29 4
T = __ x __T
BC 21 5 AC 12 20
and +1sz = o; EETAC + 39ch — 600 lb = o
12 20 29 4 '
or —T +—— —T —6001b=0
20 AC 29[21x 5 AC] TAG = 440.561b PROBLEM 2.57 : :4 ”3‘ A block of weight W is suspended ﬁ'om a SODmm long cord and two
5: "3°“ springs of which the unstretehed lengths are 450 mm. Knowing that the constants of the springs are kg = 1500 Min and kw = 500 N/m,
WW" determine (a) the tension in the cord, (17) the weight of the block. SOLUTION FreeBody Diagram At A Fixst note from‘geometry:
The sides of the tn'angle with hypotenuse AD are in the ratio 8:15: 17.
The sides of the triangle with hypotenuse AB are in the ratio 3:415.
The sides of the triangle with hypotenuse AC are in the ratio 7:24:25. Then: FAB = 19139143 — L0)
LAB = (0.44 m)2 + (0.33 m)2 = 0.55m FAB = 1500 N/m(0.55 m — 0.45 m) =150N Similarly, FAD = kAD(LAD — La) LAD = (0.66 m)2 + (0.32 m)2 = 0.68m FAD = 1500 N/m(0.63 m — 0.45 m) =115N' _+ 2s; = 0: ——:—(150 N) +§7§TAC — £015 N) = 0 PROBLEM 2.57 CONTINUED (b) and
24 8 +125, = 0: 2(150N) + —(66.18N) + —(115 N) —W = 0 5 25 17 or W=208N4 ...
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 Fall '08
 VALLE

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