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Unformatted text preview: PROBLEM 3.35 Given the vectors P = 7i — 2j + 5k, Q = —3i — 4j + 6k, and S = 8i +
j — 9k, compute the scalar products P‘Q, PS, and QS. SOLUTION P~Q=(7i—2j+5k)(—3i—4j+6k) = (7)(3) + (2)(—4) + (5X6) =17
or PQ=17<
PS=(7i—2j+5k)~(8i+j—9k)
= (7)(8) + (2)(1) + (5)(9)
=9
or P'S=94 QS=(—3i—4j+6k)(8i+j—9k) = (3)(8) + (4)(1) + (6)(9) = —82 PROBLEM 3.45 The 0.732 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 excited by the cord at D. SOLUTION ' =J . 22— .122
D/‘V\O‘732m Firstnote z (073) (0 3) m T A
L——« 2 *4 Then dDE = ,/(0.360)2 + (0.720)2 + (0.720)2 m = 1.08m
and rm = (0.360 m)i + (0.720 m) j — (0.720 m)k
Have TDE = TO—E(rE,D)
dDE _54N (0.360i + 0.720j — 0.720k)
1.08 = (18.0 N)i + (36.0 N)j — (36.0 N)k Now MA = rD/A X TDE where rm, = (0.132 m) j + (0.720 m)k
i j k Then MA = 0 0.132 0.720 N~m 18.0 36.0 —36.0 ‘_______.___________.____ PROBLEM 3.45 CONTINUED M A = {[(0.132)(—36.0) — (o.720)(36.0)]i + [(0.720)(18.0) — 0] j +[0 — (o.132)(18.0)]k} Nm
MA = —(30.7 Nm)i + (12.96 Nm)j — (2.38 Nm)k M = —30.7N.m, M =12.96Nm, M = —2.38Nm {
x y z PROBLEM 3.52 A force P is applied to the lever of an arbor press. Knowing that P lies in
a plane parallel to the yz plane and that M y = —180 lbin. and M z = —30 lb‘in., determine the moment Mx of P about the x axis when
0 = 60°. SOLUTION Mx = (Pcos¢)[(9 in.)sin0] — (Psin¢)[(9 in.)cos9] (1) My = —(Pcos¢)(5 in.) (2) Ml = —(Psin¢)(5 in.) (3) Equation (3), £ _ —(Psin¢)(5) Equation (2). My _ —(Pcos¢)(5) —3o
__ =t
—180 a” ¢ 2 9.4623“
From Equation (3),
—301b~in. = —(Psin9.4623°)(5 in.) P = 36.497 1b
From Equation (1), Mx = (36.497 1b)(9 in.)(cos9.4623°sin 60° — sin 9.4623°cos 60") = 253.601bin. or Mx = 2541bin.< ...
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 Spring '08
 Jenkins

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