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Unformatted text preview: PROBLEM 2.7 Two forces are applied as shown to a hook support. Using trigonometry \and knowing that the magnitude of P is 600 N, determine (a) the required ' angle a if the resultant R of the two forces applied to the support is to be
vertical, (b) the corresponding magnitude of R. ' sOLUTION ' _ . I G 00 N . Lising the triangle ruleand the oi‘ Cosines,  I Have:p'=180°—45°’ ' I "
p =135° ' 11min: ' 112' = (900)2 + (600)2 — 2(900)(600)003135“. 1:0. 6:11: 1390.57_N 1 Using the Iaw of Sines,
' 600 _.1390.57 sin'y, 'sin135° _ ' or y =17.76l12r‘ and a =‘90°:'~ 17.764?
' a —.= 72.2391 (a) 43:7“th
(b) R=J.391kN4 PROBLEM 2.19 Two structural members A and B are b'olted to a bracket as shown.
Knowing that both members are in compression and that the force is '
30 kN in member A and 20 kN in member B, determine, using
trigonometry, the magnitude and direction of the resultant of the forces
applied to the bracket by members A and B. Using the force triangle and the Laws of Cosines and Sines
We have: 7 = 180° — (45° + 25°) = 110°
Then: R2 = (30 ka + (20 IN)2 ~2(30 kN)(20 kN)cosllO°
= 1710.42 sz .
R = 41.357 kN 20 kN = 41.357 kN
sina sin110° sine ='(—2%N—)siu110° 41.357kN J = 0.45443
(1 = 27.028“ Hence:  ¢ = a + 45° = 72.0w
" R = 41.4 kN V: 72.0°< PROBLEM 2.38 A 600 N Knowing that a = 50", determine the resultant of the three forces shown. SOLUTION
211*}:
‘ R, = >25 53 R, = (600 N)cos 50° + (300 N)cos 85° — (700 N)cos50° R, = —38.132N
2F}:
Ry = 2”} gay = (600 N) sin 50° + (300 N)sin 85° + (700 'N)sinso°' B x . _ Ry =_l294.72N R = (—38.132 N)2 + (1294.72 N)2 R=1295N a = 1294.72 N
38.132 N a = 88.3° R =1.295kN bx 883° 4 PROBLEM 2.70 '  t A load Q is applied to the pulley C, which can roll on the cable ACB. The
“' pulley is held in the position shown by a second cable CAD, which paSSes
over the pulley A and supports a load P. Knowing that P = 800 N,
detemtine (a) the tension in cable ACB, (b) the magnitude of load Q. SOLUTION .
FreeBody Diagram: Pulley C . (a) _+. = o: TACB(00530° — cos50°) — (800 N)cosSO° =' 0
Hence T AC3 = 2303.5 N
Tm9 = 2.30 kN 4
(b)  +1 IF, = o: TAw(sm30° + sin50°) + (800 N)sin50° — Q = 0 (2303.5 N) (sin30° + sin50°) + (300 N)sin50° — Q = 0 or Q = 3529.2N . Q = 3.53kN4 ...
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This note was uploaded on 09/26/2009 for the course ENGR 102 taught by Professor Eric during the Spring '09 term at Canterbury Christ Church University.
 Spring '09
 eric

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