This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PROBLEM 4.5 A hand truck. is used to move .two barrels, each weighing 80 1b.
Neglecting the weight of the hand truckgdetermine (a) the vertical force
P which should be applied to the handle to maintain equilibrium when
a = 35°, (1)) the corresponding reaction .at each of the twowheels; SOLUTION
FreeBody Diagram: ‘ a1 = (20 in.)sina — (8 in.)icosa
a2 = (32 in.)cosa — (20 in.)sina
(mesa:
4 A b = (64 in.)cosa
(95») Costa '
I P t From freebody diagram of handtruck
+‘) 2MB = 0: P(b)‘— W(a2)+ ”((01): 0‘ (1)
“ML Amaze: P—2w+23=0 (2)
a. 02 . For a = 35°
l:— b a, = 205111350 — 8cos35° = 4.9183‘in. 02 = 3200535" — 20$i1135° : 14.741311r.
b = 64cos350 = 752.426 in.
(a) From Equation (1)
P(52.426 in.).— 80 lb(14.7413 in.) + 301b(4.9183'in.) = 0":
,P =14.98961b or P“: 14.9911: [4'9 (b) From Equation (2) 14.9896 lb — 2(801b) + 23 = o:
B = 72.505115 or B = 72.5 1134‘ 4' t PROBLEM 4.9 A control rod is attached to a crank at A and cords are attached at B and
C. For the given force in the rod, determine the range of values of the tension in the cord at C knowing that the cords must remain taut and that
the maximum allowed tension in a cord is 36 lb. SOLUTION
FreeBody Diagram: 53cm: 1' For (Tam, TB = 0 +) 2M0 = 0: (gram (4.8 in) — (801b)(2.4 in.) 0
[(13)max = 401b]>[Tmax = 36113]
(TC)max = 36.01b 'For (TC) ,TB = Tm = 361b min +‘) 2M0 = 0: (Tam (4.8 in.) + (361b)(1.6 in.) — (801b)(2.4 in.) 0 (TC)min = 28.01b Therefore: 28.0 lb S TC 5 36.0 lb { .330 N I‘ZSO n1ni+1+250 mm" PROBLEM 4'17 riff; 11: 17 300 mm F; i 1—0 2' Determine the reactions at A and B when (a) a = 0, (b) a = 90°,
.1 ' '1 (c) a 2 30°. SOL UTI O N
FreeBody Diagram: Equations of equilibrium: J) EMA : 0: —(330 N)(0.25 m) + Bsina(0.3 m) + Bcosa(0.5 m) = 0 (1)
22.25%: AX—Bsina20 (2)
+i 2F) = 02 A,» — (330 N) + Bcosoc = 0 (3) (a) Substitution a = Ointo (l), (2), and (3) and solving forA and B:
B = 165.000 N, AX = 0, A}, = 165.0N or A=165.0N I! B:165.0N I 4
(1)) Substituting 0: = 90° into (1), (2), and (3) and solving for/1 and B: B = 275.00 N, Ax = 275.00 N, A), = 330.00 N A = .1213 + Af, = ./(275)2 + (330)2 = 429.56 N A.
9 = tan—117) = mfg—’37? = 50.1940 .'.A=430NA:50.2°, B=Z75N4 PROBLEM 4.17 CONTINUED
(c) Substituting 0: = 30° into (1), (2), and (3) and solving forA and B: B = 141.506 N, Ax = 70.753 N, A}, = 207.45 N, :> A = JAE + 43 = ./(70.753)2 + (207.45)2 = 219.18 N 9 = tan’1% : tan’l%77‘% = 71.168° A = 219 N A: 712°, B = 141.5N 3; 60° 4 7 PROBLEM 4.21 1 The required force to be exerted at A by lever ABC is 3 1b. Knowing that
// a = 30° and that the spring has been stretched 1.2 in, determine (a) the {cg/j/ f a constant k of the spring ([7) the reaction at B.
ifl‘ 1111+ C
0. S) m. 3 i<—2 4 111 SOLUTION
FreeBody Diagram: (a)
2 4 in.
2M = 0 e A — 0 9 in F = O
‘1") Br [ 00505 J ( ) 5p
. 8
or Esp w 1b:kx:k(121n)
c0530
Solving for k:
k = 7.69800 lb/in. k = 7.70 lb/in. {
(b)
i, 211; 2 o: (31b)sin30° + B‘. +( 81" J: 0
' ' 00530"
or Bx = —10.7376 lb
+1 >313, : 0: , (3 lb)cos30° + By = 0 or By 2 2.5981 lb (—10.7376)2 + (2.5981)2 = 11.047511), and ,1 2.5981
10.7376 Therefore: B = 11.051b b; 13.60” 4 0 = tan = 136020" ...
View
Full Document
 Fall '07
 Delph

Click to edit the document details