the force applied at the handle of the strict lever causes thelever to revolve clockwise about the pen B v an angle of 2degrees. Determine the mean normal strain emerged in eachwire. The wires space unstretched once the bar is in the horizontalposition. Please display work.
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the force applied at the manage of the rigid lever reasons the bar to rotate clockwise about the pin B v an edge of 2 degrees. Determine the typical normal strain arisen in every wire. The wires space unstretched when the lever is in the horizontal position. Please display work.
Angular displacement: that the as the angle through which an object moves top top a one path. It is the angle, in radians, between the initial and also final positions.
Strain: The change in measurement over the original measurement caused due to the used load is dubbed strain.
Strain is the ratio of adjust in length over the initial length.
ε=δL\\varepsilon = \\frac\\delta Lε=Lδ
The relation for angular displacement is,
δ=Lθ\\delta = L\\theta δ=Lθ
Conversion of level to radians,
θ=2∘(π180)rad=0.03491rad\\beginarrayl\\\\\\theta = 2^\\circ \\left( \\frac\\pi 180 \\right)\\;\\rmrad\\\\\\\\\\;\\;\\;\\rm = 0\\rm.03491\\;\\rmrad\\\\\\endarrayθ=2∘(180π)rad=0.03491rad
Calculate the displacement of point A.
δA=LABθ\\delta _A = L_AB\\theta δA=LABθ
Here size of lever abdominal is LABL_ABLAB and angle that tilt about the pen B is θ\\theta θ .
Substitute, 200mm200\\;\\rmmm200mm because that LABL_ABLAB , 0.03491rad\\rm0\\rm.03491\\;\\rmrad0.03491rad for θ\\theta θ in relation,
δA=(200×0.03491)=6.9813mm\\beginarrayl\\\\\\delta _A = \\left( 200 \\times 0.03491 \\right)\\\\\\\\\\;\\;\\;\\; = 6.9813\\;\\rmmm\\\\\\endarrayδA=(200×0.03491)=6.9813mm
Calculate the displacement of suggest C.
δC=LBCθ\\delta _C = L_BC\\theta δC=LBCθ
Here size of bar BC is LBCL_BCLBC and also angle of tilt around the pen B is θ\\theta θ .
Substitute, 300mm300\\;\\rmmm300mm for LBCL_BCLBC , 0.03491rad\\rm0\\rm.03491\\;\\rmrad0.03491rad for θ\\theta θ in relation,
δC=(300×0.03491)=10.4720mm\\beginarrayl\\\\\\delta _C = \\left( 300 \\times 0.03491 \\right)\\\\\\\\\\;\\;\\;\\; = 10.4720\\;\\rmmm\\\\\\endarrayδC=(300×0.03491)=10.4720mm
Calculate the displacement of suggest D.
δD=LBDθ\\delta _D = L_BD\\theta δD=LBDθ
Here length of lever BD is LBDL_BDLBD and also angle the tilt around the pin B is θ\\theta θ
Substitute 300mm300\\;\\rmmm300mm for LBDL_BDLBD , 0.03491rad\\rm0\\rm.03491\\;\\rmrad0.03491rad for θ\\theta θ in relation,
δD=(500×0.03491)=17.4533mm\\beginarrayl\\\\\\delta _D = \\left( 500 \\times 0.03491 \\right)\\\\\\\\\\;\\;\\;\\; = 17.4533\\;\\rmmm\\\\\\endarrayδD=(500×0.03491)=17.4533mm
Calculate the typical normal strain occurred in cable AH.
(∈avg)AH=δALAH\\left( \\in _avg \\right)_AH = \\frac\\delta _AL_AH(∈avg)AH=LAHδA
Here displacement of allude A of the lever is δA\\delta _AδA , unstretched length of the cable AH is LAHL_AHLAH .
Substitute 6.9813mm6.9813\\;\\rmmm6.9813mm because that δA\\delta _AδA , 200mm200\\rm mm200mm because that LAHL_AHLAH in relation,
(∈avg)AH=6.9813200=0.0349\\beginarrayl\\\\\\left( \\in _avg \\right)_AH\\; = \\frac6.9813200\\\\\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = 0.0349\\\\\\endarray(∈avg)AH=2006.9813=0.0349
Calculate the mean normal strain developed in wire CG.
(∈avg)CG=δCLCG\\left( \\in _avg \\right)_CG = \\frac\\delta _CL_CG(∈avg)CG=LCGδC
Here displacement of suggest C that the bar is δC\\delta _CδC , unstretched length of the wire CG is LCGL_CGLCG
Substitute 10.4720mm10.4720\\;\\rmmm10.4720mm for δC\\delta _CδC and 300mm300\\rm mm300mm because that LCGL_CGLCG in relation,
(∈avg)CG=10.4720300=0.0349\\beginarrayl\\\\\\left( \\in _avg \\right)_CG\\; = \\frac10.4720300\\\\\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = 0.0349\\\\\\endarray(∈avg)CG=30010.4720=0.0349
Calculate the median normal strain emerged in wire DF.
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(∈avg)DF=δDLDF\\left( \\in _avg \\right)_DF = \\frac\\delta _DL_DF(∈avg)DF=LDFδD
Here displacement of suggest D of the lever is δD\\delta _DδD , unstretched size of the wire DF is LDFL_DFLDF .
Substitute, 17.4533mm17.4533\\;\\rmmm17.4533mm because that δD\\delta _DδD and also 300mm300\\rm mm300mm for LDFL_DFLDF in relation,
(∈avg)DF=17.4533300=0.0582\\beginarrayl\\\\\\left( \\in _avg \\right)_DF = \\frac17.4533300\\\\\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = 0.0582\\\\\\endarray(∈avg)DF=30017.4533=0.0582Ans: