Solution Manual for Engineering Statics: Friction - Chapter 8 (Beer and Johnston Style)

CHAPTER 8
V

, PROBLEMV8.1

DetermineVwhetherVtheVblockVshownVisVinVequilibriumVandVfindVtheVma
gnitudeVandVdirectionVofVtheVfrictionVforceVwhenP=V150VN.




SOLUTION

AssumeVequilibrium
:




FxV =V0:V V FV +V(500V N)VsinV 20V−V(150V N)Vc
osV 20V=V0
FV =V −30.056V NV FV =V30.056V N

F =V0:V V NV −V(500VN)VcosV20V−V(150VN)Vsi
NV =V +521.15V NV NV =V521.
15V N

MaximumVfrictionVforc
e:
FmV =VsVN
=V0.30(521.15V
N)
= 156.345 N
SinceVFVis blockVisVinVequilibrium
andV FV VF ◀
FV =V30.1VN
20.0




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HillVEducation.VPermissionVrequiredVforVreproductionVorVdisplay.

, PROBLEMV8.2

DetermineVwhetherVtheVblockVshownVisVinVequilibriumVandVfindVtheVma
gnitudeVandVdirectionVofVtheVfrictionVforceVwhenVP=V400VN.




SOLUTION

AssumeVequilibrium
:




FxV =V0:V V FV +V(500V N)VsinV 20°V−V(400V N)VcosV 20°V=
V0

FV =V +204.87V N
FV =V 204.87
VN

F = 0: N − (500 N) cos 20 − (400 N) sin 20 =

MaximumVfrictionVforc FmV =VsVN
e: =V0.30(606.65V
N)
=V181.995VN
Since FV  V blockVmovesVup
F , ◀
ActualVfrictionVforc FV =V FkV =V kVNV =V0.25(606.6 FV =V151.7V N
e: 5V N) 20.0




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HillVEducation.VPermissionVrequiredVforVreproductionVorVdisplay.

, PROBLEMV8.3

DetermineV whetherV theV blockV shownV isV inV equilibriumV andV fi
ndV theVmagnitudeVandVdirectionVofVtheVfrictionVforceVwhenVPV=
V120Vlb.




SOLUTION

AssumeVequilibrium
:




FxV =V0:V V FV +V(50V lb)VsinV 30V−V(120V lb)VcosV40V=
V0

FV =V +66.925V
lb

FyV =V0:V V NV −V(50Vlb)VcosV30V−V(120Vlb)VsinV40V=V0
MaximumVfrictionVforce:
FmV =VsVN
=V0.40(120.436Vl
b)
=V48.174Vlb
blockVmovesVup
WeVnoteVthatV FVVFmV.V Thus, ◀
VActualVfrictionVforce:
FV =V36.1 30.0V
FV = V F =V VNV =V0.30(120.436V lb)V=V36.131Vlb, Vlb ◀




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HillVEducation.VPermissionVrequiredVforVreproductionVorVdisplay.