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Unformatted text preview: PROBLEM 3.2 A 90N force is applied to the control rod AB as showu. Knowing that the
length of the rod is 225 mm, determine the moment of the force about
point B by resolving the force into horizontal and vettical components. SOLUTION
R I; = (90 N)cos25°
15° _
\F qu —81.568N
Y y CHEM Fy = (90 N)sin25° \ 95‘, = 38.036 N I x x = (0.225 m)cos65° = 0.095089 m
y = (0.225 m)sin65°
= 0.20392 m MBsz)_ijr : (0.095089 m)(38.036 N) — (0.20392 m)(81.568 N) = 43.0165 Nm MB = 13.02 Nm) 4
“W I: PROBLEM 3.9 A winch puller AB is used to straighten a fence post. Knowing that the
tension in cable BC is 260 1b, length a is 8 in., length b is 35 in., and
length d is 76 in., determine the moment about D of the force exerted by
the cable at C by resolving that force into horizontal and vertical
components applied (a) at point C, (b) at point E. SOLUTION
TA“ C (a) Slope of line EC = 2 i
T —260b l 761n.+81n. 12
AE“ / w TAB),
/ ,
ii 35 m.
E N If D Jr Then T ABX = %(TAB) em,
0’) = 2(2601b) = 2401b
13
5
and TM = E(2601b) = 100 lb
Then MD 2 Tm (35 in.) u TAB), (8 ins) = (240 1b)(35 in.) — (100 1b)(8 in.) = 76001bin.
or MD = 76001bin.)4 MD = TABx (y) + TABy (x)
= (240 1b)(0) + (100 1b)(76 in.) = 7600 lb  in. or MD = 76001brin.) 4 PROBLEM 3.12 It is known that the connecting rod AB exerts on the crank BC a 2.5kN force directed down and to the left along the centerline of AB. Determine
I the moment of that force about C. 56 mm with dAB = (42 mm)2 + (144 mm)2 =150 mm ‘Wmm 42 mm
sinH 2 150 mm 59mm COS6=l44mm 150 mm and FAB = —FAB Sinai “" FAB COSBj __ 2.5 kN
150 mm [(—42 mm)i n (144 mm)j] = —(700 N)i — (2400 N) j
Also rB/C = w(0.042 m)i + (0.056 m) j NOW 2 ram X FAB (—0.042i + 0.056 j) x (—700i — 2400 j)Nm (140.0 Nm)k or MC = 140.0 Nim ‘) 4
—_—__H—~_—_—____— PROBLEM 3.45 The rectangular platform is hinged at A and B and is supported by a cable
that passes over a frictionless hook at E. Knowing that the tension in the
cable is 1349 N, determine the moment about each of the coordinate axes
of the force exerted by the cable at C. Have rC : (2.25 m)k CE
= T .—
TCE Cb CE )[(0.90 m)i + (1.50 m) j — (2.25 m)k] (0.90) +(1.50) +(—2.25) m = (426 N)i + (710 N)j — (1065 N)k NOW M0 2 TC X TCE i j k
m 0 0 2.25 Nm
426 710 —1065 = — (1597.5 Nm)i + (958.5 N.m)j '. Mx : —1598 Nrn, M, = 959 Nm, M2 = 0 4 ...
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 Fall '06
 ZEHNDER

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