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# HW11 - PROBLEM 4.3 Two crates each weighing 250 lb are...

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Unformatted text preview: PROBLEM 4.3 Two crates. each weighing 250 lb, are placed as shown in the bed ofa 3000—lb pickup truck. Determine the reactions at each of the two (a) rear wheels A, (17) front wheels B. SOLUTION (a) From f.b.d. of truck 9211,? : 0: (250 1b)(1211 a) + {2501b)(65ﬁ) + (30001b)(3.9 11) — (ZFA)(9.8 a) : 0 217A 2 ﬂ = 16683711) 98 F1, 2 834116 34 ([1) From f.b.d. oftruck +3 EMA : 0: (ZFH){9.8 ft} —(30001b](5.9 ﬁ) — (2501b)(3.3 it) + (2501b)[2.3 ft} : 0 21“,? = 1150 = 1831.63 lb 9.8 FB = 9161b Id Check: +1 ZFr = 0: (7250 +1668.37 — 250 — 3000 + 1831.63) 1b = 0? (3500 — 3500}lb = 0 0k PROBLEM 4.15 l__1___q A follower ABCD is held against a Circular cam by a stretched spring, . which exerts a force of 21 N for the position shown. Knowing that the tension in rod BE is 14 N, determine (a) the force exerted on the roller at A, (b) the reaction at bearing C, l—SOLUTION Note: From f.b.d. ofABCD AI = AcoséO" : NIL J5 A) = Asin60O = — 2 (a) From f.b.d. ofABCD I) EMF = 0: (§](40mm) — 21 N(40 mm) +14N(20mm}= o ¢A=BN or A : 28.0N A: 60°4 (:5) From f.b.d. ofABCD t- 25v : 0: Cl, +14N + [28N)cosﬁ(]° : 0 -. q 2 —28 N or ex = 28.0 N -— +9211. = o: C_,. 7 21 N + (28 N)sin60” = o C), = —3.2487N or (3,, : 325 N l Then C = ta: + Cf, : (28)2 + (3.2487)2 = 28.188N C'_ 7 and 6' = tan {(3)}: tan '[ 32487] = 6.6182” —28 or C 2 28.2 N ‘7"6132‘J 4 PROBLEM 4.21 The required tension in cable AB is 800 N. Determine (a) the vertical force P which must be applied to the pedal, (b) the corresponding reaction at C. r SOLUTION (a) From f.b.d. ofpeda] P c 3) 2M. : o: P(0.4 m) — (800 N)[(0.18 m)sin 60°] : 0 3 D P=3ll.77N w Cx BOON 0r 9:312N14 A B ([2) From f.b.d.ofpedal ma. Lamaze: CK—SOON:0 Cx=800N or CX2800Nﬁ +121?) =0: q,-311.77N:0 C}.=311.77N or C) 2311477Nf Then C = «fo + Ci = (800)2 + (311.77)2 : 858.60 N C. and 6=tan" —-‘ :tem"[3]l'77]=21.291° 800 or C : 859N4213°4 ...
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