Probability, Statistics & Random Processes Chapter 8 Solution Manual – 2025 Hypothesis Testing & Decision Theory Solutions

CHAPTER 8
g

, PROBLEMg8.1

Determinegwhethergthegblockgshowngisgingequilibriumgandgfindgthegmagnitud
egandgdirectiongofgthegfrictiongforcegwhenP=g150gN.




SOLUTION

Assumegequilibrium:




Fxg =g0:g g Fg +g(500g N)gsing 20g−g(150g N)gcosg20g
=g 0
Fg =g −30.056g Ng Fg =g30.056g N

F =g0:g g Ng −g(500gN)gcosg20g−g(150gN)gsing20g
Ng =g +521.15g Ng Ng =g521.15g
N

Maximumgfrictiongforce:
Fmg =gsgN
=g0.30(521.15gN)
=g156.345gN
SincegFgis blockgisgingequilibrium◀
andg Fg gF
Fg =g30.1gN 20.0g◀




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, PROBLEMg8.2

Determinegwhethergthegblockgshowngisgingequilibriumgandgfindgthegmagnitud
egandgdirectiongofgthegfrictiongforcegwhengP=g400gN.




SOLUTION

Assumegequilibrium:




Fxg =g0:g g Fg +g(500g N)gsing 20°g−g(400g N)gcosg 20°g=g0

Fg =g+204.87g N
Fg =g204.87g
N

Fyg =g0:g g Ng −g(500gN)gcosg20g−g(400gN)gsing20g=g0


Maximumgfrictiongforce: Fmg =gsgN
=g0.30(606.65gN)
=g181.995gN

Since Fg gFm blockgmovesgup◀
,
Actualgfrictiongforce: Fg =g Fkg =g kgNg =g0.25(606.65g Fg =g151.7g N 20.0g◀
N)




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, PROBLEMg8.3

Determineg whetherg theg blockg showng isg ing equilibriumg andg findg t
hegmagnitudegandgdirectiongofgthegfrictiongforcegwhengPg=g120glb.




SOLUTION

Assumegequilibrium:




Fxg =g0:g g Fg +g(50g lb)gsing30g−g(120g lb)gcosg40g=g0

Fg =g+66.925g l
b

Fyg =g0:g g Ng −g(50glb)gcosg30g−g(120glb)gsing40g=g0

N = +120.436 lb
Maximumgfrictiongforce:
Fmg =gsgN
=g0.40(120.436g lb)
=g48.174glb
Wegnotegthatg FggFmg.g Thus,gAct blockgmovesgup◀
ualgfrictiongforce:
Fg =g Fkg =g kgNg =g0.30(120.436g lb)g=g36.131glb, Fg =g36.1gl 30.0g ◀
b




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