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Unformatted text preview: PROBLEM 3.35 Given the vectors P = 7i — 2j + 5k, Q 2 3i — 4j + 6k, and S : 8i +
j — 9k, compute the scalar products P.Q, PS, and Q6. SOLUTION P~Q:(7i—2j+5k)(—3i—4j+6k)
: (7)(3) + (2)(41) + (5)(6)
=17 or PQ=17‘ PS :(7i—2j+5k)~(8i+j—9k)
= (7)(8)+(—2)(1)+(5)(9)
= 9 or PS=94
Q~S=(~3i—4j+6k)v(8i+j—9k) = (3)(8) + (40(1) + (6)(9) = —82 or QSz—824 PROBLEM 3.45 The 0732 x 1.2m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the
tension in the cord is 54 N, determine the moment about each of the
coordinate axes of the force exerted by the cord at D. SOLUTION («u—RN First note 2 2 (0.732)2 — (0.132)2 m
1 D 0.732m
\H 1 0.13% : 7 = 0.720m
‘ T
L—— 2 —>l Then dDE = (0.360)2 + (0.720)2 + (0.720)2 m =1.08m and r5.D = (0.360 m)i + (0.720 m)j — (0.720 m)k Have TDE = %(&m)
DE . : %(0.360i + 0.720j — 0.720k) = (18.0 N)i + (36.0 N)j — (36.0 N)k Now MA = [but X Tut where rpm = (0.132 m)j + (0.720 m)k
i j k Then MA = 0 0.132 0.720 Nm 18.0 36.0 736.0 PROBLEM 3.45 CONTINUED
MA ={[(0.132)(736.0)—(0.720)(36.0)]i+[(0.720)(18.0)—0]j +[0  (0.132)(18.0)]k} Nm or MA =w(30.7Nm)i+(12.96Nm)j~(2.38Nm)k M1 = —3o.7 Nm, My = 12.96 Nm, M, = 72.38Nm 4 PROBLEM 3.52 of an arbor press. Knowing that P lies in
plane and that My = —180ibin.and 30 ibvm.,detennine the moment M“, of P about the x axis when Based on M: = (Pcos¢)[(9 in.)sin€] — (Psin¢)[(9 in.)cos€] (I)
My = —(Pcos¢)(5m.) (2)
M2 = ~(Psin¢)(5 in.) (3)
Equation (3). ﬂ _ —(Psin¢)(5)
Then Equation (2)' My ‘ —(Pcos¢)(5)
or —;138% = tan¢
¢ : 9.4523°
f From Equation (3), —301b~in. : —(Psin9i4623°)(5 in.) P : 36.497 lb From Equation ('1), Mx = (36.497 1b)(9 in.)(cos9.4623°sin 60° — sin 9.4623”cos 60°) = 253.601bin. or 1Mr = 254lbin.‘ ...
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 Spring '08
 Jenkins
 Statics

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