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Unformatted text preview: PROBLEM 4.102 A stack of several sheets of drywall of mass 170 kg rests on three wooden
blocks placed under the edges of the stack. Determine the force exerted on each block. SOLUTION
FreeBody Diagram: The weight Wis W = mg = (170 kg)(9.81m/s2) = 1667.7 N
ZMC20: rmXNA+rCBXNB+rccxw=O
or [(—0.3 m)i + (1.2 m)k] x NAj + [(1.8 m)i + (0.9 m)k]'x NBj +[(0.6 m)i + (0.6 m)k] x (W)j : 0 or —(0.3 m)NAk — (1.2 m) NAi + (1.8 m) NBk  (0.9 m)NBi — (0.6 m)Wk + (0.6 m)W1 = 0 Equating the coefﬁcients of the unit vectors to zero:
i: —1.2NA —0.9NB+0.6W= 0 4NA + 3N30 = 2W
0.3NA +1.8NB — 0.6W = 0
—NA + 6NBO = 2W —2 x + Eq. (2) gives
—9NA = 2W N = %W= 370.6ON continued I\
PROBLEM 4.102 CONTINUED Now (2) gives
N3 = l(2W+ 3W) = B
6 9 27
NB = 617.67 N,
and from (1) 370.6ON + 617.67 N + NC — 1667.7 N = 0
NC = 679.43 N Therefore the forces on the blocks are:
N A = 371 N l 4 NB=618N14 NC=679N14 PROBLEM 4.105 L4
For the pipe assembly of Prob. 4.104, determine (a) the largest [VJ permissible value of a if the assembly is not to tip, (b) the corresponding
D
\
C I B
///\ tension in each wire.
A Problem 4.104: Two steel pipes AB and BC each having a weight per
z 4 a unit length of 5 lb/ft are welded together at B and are supported by three 2 a>\x wires. Knowing that a = 1.25 ft, determine the tension in each wire. SOLUTION
FreeBody Diagram:
First note WAB = (5 lb/ﬁ)(2 ﬂ) = 10 lb
W3C = (51b/ft)(4 a) = 201b
From freebody diagram. of pipe assembly
ZFy =0: TA+TC+TD—101b—201b=0 Q+Q+a=wm m ZMx = 0: (101b)(1ﬁ) — TA(2 a) = 0
or TA = 5.001b (2)
From Equations (1) and (2) TC + T D m 25 lb (3) 2M2 = 0: TC(4 a) + TD(am) » 201b(2 a) = 0 (4 ft)TC + TDam = 401bﬁ continued PROBLEM 4.105 CONTINUED Using Equation (3) to eliminate TC 60
or By observation, a is maximum when T D is maximum. From Equation (3), (T D)m occurs when TC = 0. Therefore, (T D)max = 251b and Results: (a) @ am=1.600ft<
a=5wm4
Tc=0‘ 1b=ﬁﬂm4 PROBLEM 4.144 The frame ACD is supported by balland—socket joints at A and D and by
a cable that passes through a ring at B and is attached to hooks at G and
H. Knowing that the frame supports at point C a load of magnitude P = 75 lb, determine the tension in the cable. SOLUTION
F ree—Body Diagram: Express forces in terms of their rectangular components: 17‘; 40i+74j32k 20. 37. 16 TBG : TBG— = TBG—‘—"“— = TBG[_—l —] — —k)
BG (__40)2 + + (_32)2 EH 30i+60'—60k 1. 2. 2 TBA! = TBHE = T3H_‘2—"2”m“_2 : TBH(““1 + 3] — 3k) (30) + (60) + (60)
P = —(751b)j
MD = % _ —»———80‘_ 60’ = 0.8i—O.6j (go)2 + (—60)2
__#_________,______"_J PROBLEM 4.144 CONTINUED
rB/A = (40 in.)i
rm = (80 in.)i Now, ZMAD = 0: XAD(rBM x TBG) + XAD(rm x TBH)+ AAD(rC/A x P): 0 0.8 0 —0.6T 0.8 0 —O.6T 0.8 0 —O.6
4000%+4000—B§i+800 0:0
—20 37 —16 1 2 —2 o —75 0 888 48
—ETBG —?TBH + I 0 Noting that TBG = T 3}, = T and solving: T = 100.746 lb or T = 100.71b‘ ...
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This note was uploaded on 09/28/2008 for the course ENGRD 2020 taught by Professor Zehnder during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 ZEHNDER

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