Probability, Statistics & Random Processes Chapter 10 Solution Manual – 2025 Nonparametric & Goodness-of-Fit Test Solutions

CHAPTER 10
g

, PROBLEMg10.1

Determineg theg verticalg forceg Pg thatg mustg beg appliedg atg Cg tog m
aintaingthegequilibriumgofgtheglinkage.




SOLUTION




Assumeg gyA

gyCg =g2gy
A


gyDg =ggyA
1g 1g
gyEg g =g gyDg g =g A
2 2
1g
gyGg g =ggyEg g =g A
2
gy
Also g =g ag=g0.3g
a m
VirtualgWork:gWegapplygbothgPgandgMgtogmembergABC.
Ug =g80gyAg −gPgyCg −gMg +g60gyDg +g20gyEg −g40gyG
=0 ggy  g1g g g1g g
80gyA −gP( gyAg )g−g  A  +g60gyAg g +g20g gAg
y  −g40
 g gy
2 M  a  =0 2  2 
Mg 
80g−g2Pg−g +g60g+g10g−g20g=g0
a
Mg
2Pg+g =g130g (1)
N
NowgfromgEq.g(1)gforg Mg =g a Pg=g65.0gNg g
0 ◀


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HillgEducation.gPermissiongrequiredgforgreproductiongorgdisplay.

, PROBLEMg10.2

DeterminegtheghorizontalgforcegPgwhichgmustgbegappliedgatgAgs
ogthatgtheglinkagegisgingequilibriumgforgthegpositiongshown.




SOLUTION

Assumeg 
gxAg =g10g in.
gyCg =g 4g in.
gyDg =gg yCg =g 4g i
n.
gy 2g
g =g Dg =g 
6 3

g2g g
gxGg g =g15g =g15g g =g10g i
n. 3 

VirtualgWork: AssumegthatgforcegPgisgappliedgatgA.
Ug =g0: Ug =g −gPgxAg +g 30g yCg +g 60g yDg +g 240g +g 80gxG
= 0
g2g
−gPg(10ggin.)g+g(30glb)(4g in.)g+g(60glb)(4gg)g+g(240glbin.g) 
 3 

+(80glb)(10ggin.)g=g0

−10Pg +g120g +g 240g +g160g +g800g =g 0

10Pg =g1320
P = 132.0 l




Copyrightg©gMcGraw-
HillgEducation.gPermissiongrequiredgforgreproductiongorgdisplay.

, PROBLEMg10.3

DeterminegthegcouplegMgthatgmustgbegappliedgtogmembergABCgtog
maintaingthegequilibriumgofgtheglinkage.




SOLUTION




Assumeg gyA

gyDg =ggy
A
1g 1g
gyEg g =g gyDg g =g A
2 2
1g
gyGg g =ggyEg g =g A
2
gy
Also g =g
0.3
VirtualgWork:gWegapplygbothgPgandgMgtogmembergABC.
Ug =g80gyAg −gPgyCg −gMg +g60gyDg +g20gyEg −g40gyG
=0 g gy  g1g g g1g g
80gyA −gP( gyAg )g−g g A +g60gyAg g +g20g gAgy  −g40
 g gy
2 M  0.3gmg =0 2  2 
M  
80g−g2Pg +g60g+g10g−g20g=g
− 0.3 gm
0
M
2Pg+ =g130g (1)
0.3gm N
M
NowgfromgEq.g(1)gforg Pg=g =g130g Mg=g39.0gNggm ◀
0 0.3gm N




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HillgEducation.gPermissiongrequiredgforgreproductiongorgdisplay.