Chapter 8 Solutions - Probability, Statistics & Random Processes | Solved Exercises & Explanations

CHAPTER8
j

, PROBLEMj8.1

Determinejwhetherjthejblockjshownjisjinjequilibriumjandjfindjthejmagnitudejan
djdirectionjofjthejfrictionjforcejwhenP=j150jN.




SOLUTION

Assumejequilibrium:




Fxj =j0:j j Fj +j(500j N)jsinj 20j−j(150j N)jcosj 20j=j0

Fj =j −30.056j Nj Fj =j30.056j N

Fyj =j0:j j Nj −j(500jN)jcosj20j−j(150jN)jsinj20j=j0

Nj =j +521.15j Nj Nj =j521.15j N


Maximumjfrictionjforce:
Fmj =jsjN
=j0.30(521.15jN)
=j156.345jN
SincejFjis andj Fj j Fmj , blockjisjinjequilibrium◀
Fj =j30.1jN 20.0j◀




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, PROBLEMj8.2

Determinejwhetherjthejblockjshownjisjinjequilibriumjandjfindjthejmagnitudejan
djdirectionjofjthejfrictionjforcejwhenjP=j400jN.




SOLUTION

Assumejequilibrium:




Fxj =j0:j j Fj +j(500j N)jsinj 20°j−j(400j N)jcosj 20°j=j0

Fj =j +204.87j N Fj =j 204.87j N

Fyj =j0:j j Nj −j(500j N)jcosj20j−j(400j N)jsinj20j=j0

Nj =j +606.65j Nj Nj =j606.65j N


Maximumjfrictionjforce: Fmj =jsjN
=j0.30(606.65jN)
=j181.995jN

Since Fj j Fmj, blockjmovesjup◀

Actualjfrictionjforce: Fj =j Fkj =j kjNj =j0.25(606.65j N) Fj =j151.7j N 20.0j◀




Copyrightj©jMcGraw-HilljEducation.jPermissionjrequiredjforjreproductionjorjdisplay.

, PROBLEMj8.3

Determinej whetherj thej blockj shownj isj inj equilibriumj andj findj thej
magnitudejandjdirectionjofjthejfrictionjforcejwhenjPj=j120jlb.




SOLUTION

Assumejequilibrium:




Fxj =j0:j j Fj +j(50j lb)jsinj 30j−j(120j lb)jcosj40j=j0

Fj =j +66.925j lb

Fyj =j0:j j Nj −j(50jlb)jcosj30j−j(120jlb)jsinj40j=j0

Nj =j +120.436j lb
Maximumjfrictionjforce:
Fmj =jsjN
=j0.40(120.436j lb)
=j48.174jlb
Wejnotejthatj FjjFmj.j Thus,jActual blockjmovesjup◀
jfrictionjforce:

Fj =j Fkj =j kjNj =j0.30(120.436j lb)j=j36.131jlb, Fj =j36.1jlb 30.0j ◀




Copyrightj©jMcGraw-HilljEducation.jPermissionjrequiredjforjreproductionjorjdisplay.